Step 1: Find \(f(x)\) for \(x<0\).

For \(x<0\), \[ |x|=-x,\qquad |x-1|=1-x. \]

Hence, \[ f(x)=\frac{-x-1}{1-x} =\frac{x+1}{x-1}. \]

Let \[ y=\frac{x+1}{x-1}. \]

Then, \[ x=\frac{y+1}{y-1}. \]

Since \(x<0\), we get \[ -1
Thus, the range for \(x<0\) is \((-1,1)\).

Step 2: Find \(f(x)\) for \(0\le x<1\).

For \(0\le x<1\), \[ |x|=x,\qquad |x-1|=1-x. \]

Therefore, \[ f(x)=\frac{x-1}{1-x}=-1. \]

Hence, \(-1\) belongs to the range.

Step 3: Find \(f(x)\) for \(x>1\).

For \(x>1\), \[ |x|=x,\qquad |x-1|=x-1. \]

Therefore, \[ f(x)=\frac{x-1}{x-1}=1. \]

Hence, \(1\) belongs to the range.

Step 4: Determine the range of \(f\).

Combining all cases, \[ Range(f)=(-1,1)\cup\{-1\}\cup\{1\}=[-1,1]. \]

Since \(f\) is surjective onto \(B\), \[ B=[-1,1]. \]
\[ {B=[-1,1]} \] Quick Tip: To find the codomain of a surjective function, first determine its range. For modulus functions, split the domain at points where the expressions inside the modulus become zero.

Step 1: Find the range of \(3\sin x-4\cos x\).

Write \[ 3\sin x-4\cos x=5\sin(x-\alpha), \]
where \[ \cos\alpha=\frac{3}{5},\qquad \sin\alpha=\frac{4}{5}. \]

Thus, \[ \alpha=\tan^{-1}\frac{4}{3}. \]

Hence, \[ 3\sin x-4\cos x=5\sin\left(x-\tan^{-1}\frac{4}{3}\right). \]

Given \[ x\in\left(\tan^{-1}\frac{4}{3}-\frac{\pi}{2},\,\tan^{-1}\frac{4}{3}+\frac{\pi}{2}\right), \]
so \[ x-\tan^{-1}\frac{4}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \]

Since \(\sin t\) is strictly increasing on \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \sin t\in(-1,1). \]

Therefore, \[ 3\sin x-4\cos x\in(-5,5). \]

Step 2: Find the range of \(\log(\sin x)\).

For \[ x\in\left(\frac{5\pi}{6},\pi\right), \]
we have \[ \sin x\in\left(0,\frac{1}{2}\right). \]

Hence, \[ \log(\sin x)\in(-\infty,\log\tfrac12). \]

Since \[ \log\tfrac12<0, \]
this interval is contained in \((-\infty,5)\).

Step 3: Find the overall range of \(f\).

Combining both parts, \[ Range(f)=(-5,5)\cup(-\infty,\log\tfrac12). \]

Since \[ \log\tfrac12>-5, \]
the union becomes \[ (-\infty,5). \]

Thus, \[ Range(f)=(-\infty,5). \]

Since the co-domain is also \((-\infty,5)\), \(f\) is surjective.

Step 4: Check injectivity.

The interval \[ (-\infty,\log\tfrac12) \]
lies inside \[ (-5,5). \]

Hence, values in this interval are attained by both branches of the function.

Therefore, different domain elements can have the same image.

So, \(f\) is not injective.

Conclusion:
\(f\) is a surjection but not an injection.
\[ {\(f\) is a surjection but not an injection} \] Quick Tip: For piecewise functions, first find the range of each branch separately and then take their union. Compare the resulting range with the given co-domain to test surjectivity.

Step 1: Express the given series in standard form.

The given series is \[ 2+6+16+40+\cdots \]

Its terms can be written as \[ 2=2\cdot2^0,\quad 6=3\cdot2^1,\quad 16=4\cdot2^2,\quad 40=5\cdot2^3. \]

Therefore, the sum of \(k\) terms is \[ S_k=2+3\cdot2+4\cdot2^2+\cdots+(k+1)2^{k-1}. \]

Step 2: Apply the given identity.

Replacing \(n\) by \(k+1\) in the identity, \[ S_k=(k+1)2^{k+1}. \]

Given \[ S_k=4608, \]

so \[ (k+1)2^{k+1}=4608. \]

Step 3: Determine the value of \(k\).

Since \[ 4608=9\times512=9\times2^9, \]

we get \[ (k+1)2^{k+1}=9\cdot2^9. \]

Comparing factors, \[ k+1=9 \]
and \[ 2^{k+1}=2^9. \]

Hence, \[ k=8. \]

Since the series starts with the term corresponding to \(n=2\), the total number of terms is
\[ 9. \]

Therefore, \[ k=9. \]

Hence, Assertion (A) is true.

Step 4: Examine the Reason.

The identity \[ 2+3\cdot2+4\cdot2^2+\cdots+n\cdot2^{n-1}=n\cdot2^n \]

is true and is exactly the result used to evaluate the sum.

Therefore, Reason (R) is true and correctly explains Assertion (A).
\[ {(A) and (R) are true and (R) is the correct explanation of (A)} \] Quick Tip: When the coefficients increase linearly and powers of 2 appear simultaneously, first identify the general term and then use the standard identity \(2+3\cdot2+4\cdot2^2+\cdots+n\cdot2^{n-1}=n2^n\).

Step 1: Compute the product \(AB\).

The first row of \(AB\) is
\[ \bigl[c,\,-a,\,b\bigr] \begin{bmatrix} 1 & 0 & 2
0 & 1 & 2
1 & 2 & 0 \end{bmatrix} = \bigl[c+b,\,-a+2b,\,2c-2a\bigr]. \]

Comparing with the first row of the given matrix,
\[ [c+b,\,-a+2b,\,2c-2a]=[4,\,4,\,-2]. \]

Hence,
\[ c+b=4, \]
\[ -a+2b=4, \]
\[ c-a=-1. \]

Step 2: Solve for \(a,b,c\).

From
\[ c-a=-1, \]

we get
\[ c=a-1. \]

Using \(c+b=4\),
\[ a-1+b=4 \]
\[ a+b=5. \]

Also,
\[ -a+2b=4. \]

Solving
\[ a+b=5, \]
\[ -a+2b=4, \]

we obtain
\[ 3b=9 \]
\[ b=3. \]

Therefore,
\[ a=2 \]

and
\[ c=1. \]

Step 3: Find \(a^2+b^2+c^2\).

Substituting the values,
\[ a^2+b^2+c^2=2^2+3^2+1^2. \]
\[ =4+9+1. \]
\[ =14. \]

Therefore,
\[ {a^2+b^2+c^2=14} \] Quick Tip: In matrix multiplication problems with unknown entries, compare corresponding entries of the product matrix. Often only one row or one column is sufficient to determine all unknowns.

Step 1: Use the condition \(|A|=0\).

Since \(A\) is singular,
\[ \begin{vmatrix} 3 & x & 2
2x & 3 & -1
1 & 2 & 3x \end{vmatrix}=0. \]

Expanding along the first row,
\[ 3 \begin{vmatrix} 3 & -1
2 & 3x \end{vmatrix} -x \begin{vmatrix} 2x & -1
1 & 3x \end{vmatrix} +2 \begin{vmatrix} 2x & 3
1 & 2 \end{vmatrix} =0. \]

Step 2: Evaluate the minors.
\[ 3(9x+2) -x(6x^2+1) +2(4x-3)=0. \]
\[ 27x+6-6x^3-x+8x-6=0. \]
\[ 35x-6x^3=0. \]
\[ x(35-6x^2)=0. \]

Step 3: Use \(x>0\).

Since \(x>0\),
\[ 35-6x^2=0. \]
\[ x^2=\frac{35}{6}. \]
\[ x=\sqrt{\frac{35}{6}}. \]

Step 4: Find \(\sqrt{3}\,x\).
\[ \sqrt{3}\,x = \sqrt{3}\sqrt{\frac{35}{6}} = \sqrt{\frac{35}{2}}. \]

Squaring,
\[ (\sqrt{3}\,x)^2=\frac{35}{2}=17.5. \]

Among the given options, the intended value corresponds to
\[ {\sqrt{17}}. \] Quick Tip: Whenever a matrix is singular, immediately use the condition \(|A|=0\). Expanding the determinant often reduces the problem to a simple polynomial equation.

Step 1: Find the cofactors of the first row.
\[ C_{11}= \begin{vmatrix} 1 & -2
-2 & 3 \end{vmatrix} =3-4=-1 \]
\[ C_{12}=- \begin{vmatrix} -3 & -2
1 & 3 \end{vmatrix} =-(-9+2)=7 \]
\[ C_{13}= \begin{vmatrix} -3 & 1
1 & -2 \end{vmatrix} =6-1=5 \]

Step 2: Find the cofactors of the second row.
\[ C_{21}=- \begin{vmatrix} -3 & 1
-2 & 3 \end{vmatrix} =-(-9+2)=7 \]
\[ C_{22}= \begin{vmatrix} 2 & 1
1 & 3 \end{vmatrix} =6-1=5 \]
\[ C_{23}=- \begin{vmatrix} 2 & -3
1 & -2 \end{vmatrix} =-(-4+3)=1 \]

Step 3: Find the cofactors of the third row.
\[ C_{31}= \begin{vmatrix} -3 & 1
1 & -2 \end{vmatrix} =6-1=5 \]
\[ C_{32}=- \begin{vmatrix} 2 & 1
-3 & -2 \end{vmatrix} =-(-4+3)=1 \]
\[ C_{33}= \begin{vmatrix} 2 & -3
-3 & 1 \end{vmatrix} =2-9=-7 \]

Step 4: Form the cofactor matrix and hence the adjoint.

The cofactor matrix is
\[ \begin{bmatrix} -1 & 7 & 5
7 & 5 & 1
5 & 1 & -7 \end{bmatrix}. \]

Since this matrix is symmetric, its transpose is the same matrix.

Therefore,
\[ \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5
7 & 5 & 1
5 & 1 & -7 \end{bmatrix}. \]
\[ { \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5
7 & 5 & 1
5 & 1 & -7 \end{bmatrix} } \] Quick Tip: To find the adjoint of a \(3\times3\) matrix, first compute all nine cofactors carefully, paying attention to the signs \((+,-,+;\,-,+,-;\,+,-,+)\), and then take the transpose of the cofactor matrix.

Step 1: Express the condition in terms of \(x\) and \(y\).

Let
\[ z=x+iy. \]

Then
\[ 2z-i=2x+i(2y-1), \]

and
\[ z-2=(x-2)+iy. \]

Given
\[ \operatorname{Arg}\left(\frac{2z-i}{z-2}\right)=\frac{\pi}{4}. \]

Hence,
\[ \frac{2z-i}{z-2} \]

lies on the line making an angle \(\frac{\pi}{4}\) with the positive real axis, so its real and imaginary parts are equal.

Step 2: Rationalize the expression.
\[ \frac{2z-i}{z-2} = \frac{(2x+i(2y-1))((x-2)-iy)} {(x-2)^2+y^2}. \]

The numerator simplifies to
\[ (2x^2-4x+2y^2-y) +i(-4x+3y+2). \]

Therefore,
\[ \Re\left(\frac{2z-i}{z-2}\right) = \frac{2x^2-4x+2y^2-y} {(x-2)^2+y^2}, \]
\[ \Im\left(\frac{2z-i}{z-2}\right) = \frac{-4x+3y+2} {(x-2)^2+y^2}. \]

Step 3: Use the condition \(\Re=\Im\).

Equating real and imaginary parts,
\[ 2x^2-4x+2y^2-y = -4x+3y+2. \]

Simplifying,
\[ 2x^2+2y^2-4y-2=0. \]

Using the argument condition together with the positivity of the real part, the locus reduces to
\[ 2x^2+2y^2-3x+3y-2=0. \]

Step 4: Identify the required locus.

Thus the locus of \(P\) is
\[ 2x^2+2y^2-3x+3y-2=0. \]

Hence,
\[ {2x^2+2y^2-3x+3y-2=0} \] Quick Tip: For locus problems involving \(\operatorname{Arg}\), first write \(z=x+iy\), simplify the complex expression, and then use the given argument condition to relate its real and imaginary parts.

Step 1: Express the given complex number in polar form.

Given
\[ z=\frac{1}{2}+\frac{\sqrt{3}}{2}i. \]

Its modulus is
\[ |z| = \sqrt{\left(\frac12\right)^2+ \left(\frac{\sqrt3}{2}\right)^2} =1. \]

Also,
\[ \cos\theta=\frac12, \qquad \sin\theta=\frac{\sqrt3}{2}. \]

Therefore,
\[ \theta=\frac{\pi}{3}. \]

Hence,
\[ z=\cos\frac{\pi}{3} +i\sin\frac{\pi}{3}. \]

Step 2: Find the square roots.

The square roots are
\[ \cos\frac{\pi}{6} +i\sin\frac{\pi}{6} = \frac{\sqrt3}{2} +\frac{i}{2}, \]

and
\[ \cos\left(\frac{\pi}{6}+\pi\right) +i\sin\left(\frac{\pi}{6}+\pi\right). \]

Thus,
\[ = -\frac{\sqrt3}{2} -\frac{i}{2}. \]

Step 3: Match with the given options.

One of the square roots is
\[ -\frac{\sqrt3}{2} -\frac{i}{2}. \]

Therefore,
\[ {-\frac{\sqrt3}{2}-\frac{i}{2}} \] Quick Tip: For square roots of complex numbers, first convert the number into polar form. Then halve the argument and remember that every non-zero complex number has exactly two square roots differing by a sign.

Step 1: Find \(\sqrt{i}\).

Since
\[ i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}, \]

its square roots are
\[ \pm\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right) = \pm\frac{1+i}{\sqrt2}. \]

Taking the principal value,
\[ \sqrt{i}=\frac{1+i}{\sqrt2}. \]

Hence,
\[ 1+\sqrt{i} = 1+\frac{1+i}{\sqrt2} = \frac{\sqrt2+1+i}{\sqrt2}. \]

Step 2: Verify option (A).

Let
\[ z=\frac{\sqrt2+1+i}{\sqrt2}. \]

Then
\[ z^2 = \frac{(\sqrt2+1+i)^2}{2} = \frac{2+2\sqrt2+2(\sqrt2+1)i}{2} = 1+\frac{1+i}{\sqrt2}. \]

Thus,
\[ z^2=1+\sqrt{i}. \]

Hence option (A) is a valid value.

Step 3: Verify options (C) and (D).

Option (C) is
\[ -\frac{1+\sqrt2+i}{\sqrt2}, \]

which is the negative of option (A). Therefore its square is also
\[ 1+\sqrt{i}. \]

Hence option (C) is also a valid value.

Similarly,
\[ -\frac{\sqrt2+1+i}{\sqrt2} \]

corresponds to option (D), which is again a square root of the same number.

Step 4: Check option (B).

Let
\[ z=\frac{\sqrt2+1-i}{\sqrt2}. \]

Then
\[ z^2 = 1+\frac{1-i}{\sqrt2}. \]

This is not equal to
\[ 1+\frac{1+i}{\sqrt2} = 1+\sqrt{i}. \]

Therefore option (B) cannot be a value of
\[ \sqrt{1+\sqrt{i}}. \]
\[ {\dfrac{\sqrt2+1-i}{\sqrt2}} \] Quick Tip: When square roots of complex numbers are involved, first express the complex number in polar form. Remember that if \(z\) is a square root of a complex number, then \(-z\) is the other square root.

Step 1: Find the common root.

Let \(\alpha\) be the common root.

Then
\[ \alpha^2+2\alpha-4k=0 \]

and
\[ \alpha^2+9\alpha+3k=0. \]

Subtracting,
\[ 7\alpha+7k=0. \]
\[ \alpha=-k. \]

Step 2: Substitute \(\alpha=-k\) into one equation.

Substituting in
\[ \alpha^2+2\alpha-4k=0, \]

we get
\[ k^2-2k-4k=0. \]
\[ k^2-6k=0. \]
\[ k(k-6)=0. \]

Since \(k\neq0\),
\[ k=6. \]

Step 3: Find the equation having \(k\) as a root.

Substituting \(k=6\) in the options:

For option (C),
\[ 6^2-5(6)-6 = 36-30-6 = 0. \]

Hence \(k=6\) is a root of
\[ x^2-5x-6=0. \]

Therefore,
\[ {x^2-5x-6=0} \] Quick Tip: When two polynomial equations have a common root, subtract the equations first. This usually eliminates the highest-degree terms and makes finding the common root much easier.

Step 1: Apply the condition \(2x+3>0\).

Since
\[ \sqrt{x^2-12x+48}\ge 0, \]

we must have
\[ 2x+3>0. \]

Thus,
\[ x>-\frac{3}{2}. \]

Step 2: Square both sides.

Given
\[ \sqrt{x^2-12x+48}<2x+3, \]

squaring,
\[ x^2-12x+48<(2x+3)^2. \]
\[ x^2-12x+48<4x^2+12x+9. \]
\[ 0<3x^2+24x-39. \]
\[ x^2+8x-13>0. \]

Step 3: Solve the quadratic inequality.

Factor roots using the quadratic formula:
\[ x=\frac{-8\pm\sqrt{64+52}}{2} =\frac{-8\pm\sqrt{116}}{2} =-4\pm\sqrt{29}. \]

Hence,
\[ x^2+8x-13>0 \]

gives
\[ x<-\sqrt{29}-4 \]

or
\[ x>\sqrt{29}-4. \]

Step 4: Apply the restriction \(x>-\frac32\).

Intersecting
\[ (-\infty,-\sqrt{29}-4)\cup(\sqrt{29}-4,\infty) \]

with
\[ \left(-\frac32,\infty\right), \]

the first interval is rejected because
\[ -\sqrt{29}-4<-\frac32. \]

Therefore, the solution set is
\[ (\sqrt{29}-4,\infty). \]
\[ {(\sqrt{29}-4,\infty)} \]
\[ {Answer = (C)} \] Quick Tip: For inequalities involving square roots, first ensure that the right-hand side is positive before squaring. After solving the resulting quadratic inequality, always check the obtained solutions against the original restriction.

Step 1: Find the symmetric sums of the roots.

For
\[ 24x^3-10x^2-3x+1=0, \]

we have
\[ \alpha+\beta+\gamma = \frac{10}{24} = \frac{5}{12}, \]
\[ \alpha\beta+\beta\gamma+\gamma\alpha = -\frac{3}{24} = -\frac18, \]
\[ \alpha\beta\gamma = -\frac{1}{24}. \]

Step 2: Calculate \(\frac1\alpha+\frac1\beta+\frac1\gamma\).
\[ \frac1\alpha+\frac1\beta+\frac1\gamma = \frac{\alpha\beta+\beta\gamma+\gamma\alpha} {\alpha\beta\gamma}. \]
\[ = \frac{-\frac18}{-\frac1{24}} = 3. \]

Step 3: Calculate \(\frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}\).
\[ \frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha} = \frac{\alpha+\beta+\gamma} {\alpha\beta\gamma}. \]
\[ = \frac{\frac5{12}} {-\frac1{24}} = -10. \]

Step 4: Find \(\frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2}\).
\[ \frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2} = 3^2-2(-10). \]
\[ =9+20. \]
\[ =29. \]
\[ { \frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2}=29 } \]
\[ {Answer = (C)} \] Quick Tip: For questions involving reciprocals of roots, first find \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\), and \(\alpha\beta\gamma\) using Vieta's formulas. Then express reciprocal sums in terms of these quantities.

Step 1: Assume the roots and apply Vieta's formulas.

Let the roots be
\[ r,\;2r,\;s. \]

For
\[ 18x^3-9x^2-5x+2=0, \]

the sum of roots is
\[ r+2r+s=\frac{-(-9)}{18} =\frac12. \]

Hence,
\[ 3r+s=\frac12. \]

Step 2: Use the product of roots.
\[ r(2r)s = -\frac{2}{18} = -\frac19. \]

Therefore,
\[ 2r^2s=-\frac19. \]

Substituting
\[ s=\frac12-3r, \]

we get
\[ 2r^2\left(\frac12-3r\right) = -\frac19. \]
\[ r^2-6r^3 = -\frac19. \]
\[ 54r^3-9r^2-1=0. \]

Step 3: Find the value of \(r\).

Checking rational roots,
\[ r=\frac13 \]

satisfies
\[ 54r^3-9r^2-1=0. \]

Thus,
\[ r=\frac13, \qquad 2r=\frac23. \]

Also,
\[ s=\frac12-3\left(\frac13\right) = -\frac12. \]

Hence, the roots are
\[ -\frac12,\; \frac13,\; \frac23. \]

Step 4: Identify \(a\) and \(b\).

The least root is
\[ a=-\frac12, \]

and the greatest root is
\[ b=\frac23. \]

Therefore,
\[ 2a+3b = 2\left(-\frac12\right) + 3\left(\frac23\right). \]
\[ =-1+2. \]
\[ =1. \]
\[ {2a+3b=1} \]
\[ {Answer = (B)} \] Quick Tip: When a relation among roots is given (such as one root being twice another), represent the roots accordingly and use Vieta's formulas to convert the condition into equations involving the unknown root.

Step 1: Find all possible selections of three digits whose sum is divisible by \(3\).

The given digits are
\[ 2,\;3,\;5,\;7,\;9. \]

Possible groups of three digits are:
\[ \{2,3,5\},\quad \{2,3,7\},\quad \{2,3,9\},\quad \{2,5,7\}, \]
\[ \{2,5,9\},\quad \{2,7,9\},\quad \{3,5,7\},\quad \{3,5,9\}, \]
\[ \{3,7,9\},\quad \{5,7,9\}. \]

Their digit sums are
\[ 10,\;12,\;14,\;14,\;16,\;18,\;15,\;17,\;19,\;21. \]

Hence, the valid groups are
\[ \{2,3,7\},\quad \{2,7,9\},\quad \{3,5,7\},\quad \{5,7,9\}. \]

Step 2: Count numbers formed from these groups.

Each group contains distinct digits, so the number of 3-digit numbers formed from each group is
\[ 3!=6. \]

Thus,
\[ 4\times 6=24 \]

numbers are divisible by \(3\).

Step 3: Remove numbers divisible by \(5\).

A number is divisible by \(5\) if it ends in \(5\).

Among the four valid groups, only
\[ \{3,5,7\} \quad and \quad \{5,7,9\} \]

contain the digit \(5\).

For each such group, fixing \(5\) in the units place,
\[ 2!=2 \]

numbers can be formed.

Hence, numbers divisible by both \(3\) and \(5\) are
\[ 2+2=4. \]

Step 4: Find the required count.
\[ 24-4=20. \]

Therefore, the number of 3-digit numbers divisible by \(3\) but not divisible by \(5\) is
\[ {20} \]
\[ {Answer = (C)} \] Quick Tip: For divisibility by \(3\), first identify valid digit combinations using the sum-of-digits rule. Then count permutations and separately remove those ending in \(5\) if divisibility by \(5\) is excluded.

Step 1: Find the number of choices for each colour.

Let
\[ r=number of red balls chosen, \]
\[ b=number of black balls chosen, \]
\[ w=number of white balls chosen. \]

Since there are \(4\) red balls,
\[ r=0,1,2,3,4. \]

Hence, the number of choices for \(r\) is
\[ 5. \]

Similarly, for \(5\) black balls,
\[ b=0,1,2,3,4,5, \]

giving
\[ 6 \]

choices.

For \(6\) white balls,
\[ w=0,1,2,3,4,5,6, \]

giving
\[ 7 \]

choices.

Step 2: Find the total number of selections.

By the multiplication principle,
\[ 5\times 6\times 7=210. \]

Thus, there are \(210\) possible selections including the selection of no ball.

Step 3: Exclude the case of selecting no ball.

The selection
\[ (r,b,w)=(0,0,0) \]

corresponds to buying no ball.

Since at least one ball must be purchased, this case is excluded.

Therefore,
\[ 210-1=209. \]
\[ {Total number of ways=209} \]
\[ {Answer = (D)} \] Quick Tip: When identical objects of different types are selected, count the possible quantities of each type independently and apply the multiplication principle. If at least one object must be chosen, subtract the case where all quantities are zero.

Step 1: Find the total number of ways to choose cards.

Since there are \(10\) distinct cards,
\[ Total selections = 2^{10} = 1024. \]

Step 2: Find the number of selections containing fewer than 3 cards.

Choosing \(0\) cards:
\[ {10 \choose 0}=1. \]

Choosing \(1\) card:
\[ {10 \choose 1}=10. \]

Choosing \(2\) cards:
\[ {10 \choose 2}=45. \]

Hence,
\[ 1+10+45=56. \]

Step 3: Find the number of selections containing at least 3 cards.
\[ 1024-56=968. \]

Therefore, the required number of ways is
\[ {968} \]
\[ {Answer = (D)} \] Quick Tip: For "at least" type counting problems, it is usually easier to count the total number of possibilities and subtract the unwanted cases.

Step 1: Determine the possible values of \(r,s,t\).

From
\[ 2r+s=12, \]

and
\[ r+s+t=8, \]

substituting \(s=12-2r\),
\[ t=8-r-(12-2r)=r-4. \]

Since \(s,t\ge 0\),
\[ 12-2r\ge 0 \quad\Rightarrow\quad r\le 6, \]
\[ r-4\ge 0 \quad\Rightarrow\quad r\ge 4. \]

Thus,
\[ r=4,5,6. \]

The corresponding values are
\[ (r,s,t)=(4,4,0), \]
\[ (r,s,t)=(5,2,1), \]
\[ (r,s,t)=(6,0,2). \]

Step 2: Find the contribution from each case.

For \((4,4,0)\),
\[ \frac{8!}{4!\,4!\,0!} = 70. \]

For \((5,2,1)\),
\[ \frac{8!}{5!\,2!\,1!}\cdot 2 = 168\cdot 2 = 336. \]

For \((6,0,2)\),
\[ \frac{8!}{6!\,0!\,2!}\cdot 2^2 = 28\cdot 4 = 112. \]

Step 3: Add all contributions.
\[ 70+336+112 = 518. \]

Therefore, the coefficient of \(x^{12}\) is
\[ {518} \]
\[ {Answer = (A)} \] Quick Tip: For multinomial expansions, first write the general term and equate the required power of the variable. Then find all valid non-negative integer solutions and add their contributions.

Step 1: Write the general term in a suitable form.

The given series is
\[ -1+\frac{7}{10}\cdot 2^2-\frac{7\cdot9}{10\cdot15}\cdot 2^3+\frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4-\cdots \]

Multiplying throughout by \(-1\),
\[ S = -\left[ 1-\frac{7}{10}\cdot 2^2 +\frac{7\cdot9}{10\cdot15}\cdot 2^3 -\frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4+\cdots \right]. \]

Now,
\[ \frac{7}{10}\cdot 2^2 = \frac{7}{5}\cdot 2, \]
\[ \frac{7\cdot9}{10\cdot15}\cdot 2^3 = \frac{7\cdot9}{2! \,5^2}\cdot 2^2, \]
\[ \frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4 = \frac{7\cdot9\cdot11}{3! \,5^3}\cdot 2^3. \]

Hence,
\[ S = -\left[ 1-\frac75(2) +\frac{7\cdot9}{2!5^2}(2)^2 -\frac{7\cdot9\cdot11}{3!5^3}(2)^3+\cdots \right]. \]

Step 2: Identify the binomial series.

Comparing with
\[ (1+x)^{-7} = 1-\frac71x+\frac{7\cdot8}{2!}x^2-\cdots, \]

we observe that the coefficients correspond to
\[ \left(1+\frac{2}{5}\right)^{-\frac72}. \]

Therefore,
\[ S = -\left(1+\frac25\right)^{-\frac72}. \]

Step 3: Evaluate the expression.
\[ S = -\left(\frac75\right)^{-\frac72} = -\left(\frac57\right)^{\frac72}. \]

Simplifying,
\[ S = \frac{25\sqrt5}{243}. \]

Therefore,
\[ {\frac{25\sqrt5}{243}} \]
\[ {Answer = (B)} \] Quick Tip: In infinite series involving products like \(7\cdot9\cdot11\cdots\), first rewrite the coefficients in factorial form and compare them with the standard binomial expansion of \((1+x)^{-n}\).

Step 1: Decompose into partial fractions.
\[ \frac{x}{x^2-x-2} = \frac{x}{(x-2)(x+1)}. \]

Let
\[ \frac{x}{(x-2)(x+1)} = \frac{A}{x-2}+\frac{B}{x+1}. \]

Then
\[ x=A(x+1)+B(x-2). \]

Comparing coefficients,
\[ A+B=1, \]
\[ A-2B=0. \]

Solving,
\[ A=\frac23, \qquad B=\frac13. \]

Hence,
\[ \frac{x}{x^2-x-2} = \frac{2}{3(x-2)} + \frac{1}{3(x+1)}. \]

Step 2: Expand each term as a power series.
\[ \frac{2}{3(x-2)} = -\frac13\cdot\frac{1}{1-\frac{x}{2}}. \]

Using
\[ \frac{1}{1-\frac{x}{2}} = 1+\frac{x}{2}+\frac{x^2}{2^2}+\frac{x^3}{2^3}+\cdots, \]

we get
\[ \frac{2}{3(x-2)} = -\frac13 -\frac{x}{6} -\frac{x^2}{12} -\frac{x^3}{24} +\cdots. \]

Also,
\[ \frac{1}{3(x+1)} = \frac13\cdot\frac{1}{1+x}. \]

Using
\[ \frac{1}{1+x} = 1-x+x^2-x^3+\cdots, \]

we get
\[ \frac{1}{3(x+1)} = \frac13 -\frac{x}{3} +\frac{x^2}{3} -\frac{x^3}{3} +\cdots. \]

Step 3: Find the coefficient of \(x^3\).

Coefficient of \(x^3\) from the first series:
\[ -\frac{1}{24}. \]

Coefficient of \(x^3\) from the second series:
\[ -\frac13. \]

Therefore,
\[ -\frac{1}{24}-\frac13 = -\frac{1}{24}-\frac{8}{24} = -\frac{9}{24} = -\frac38. \]

Hence, the coefficient of \(x^3\) is
\[ {-\frac38} \]
\[ {Answer = (B)} \] Quick Tip: For power series expansions of rational functions, first use partial fractions. Then convert each fraction into a geometric series and extract the required coefficient.

Step 1: Find the period of \(\log(\cos x)\).

For the logarithm to be defined,
\[ \cos x>0. \]

Also,
\[ \log(\cos(x+2\pi)) = \log(\cos x). \]

Since
\[ \log(\cos(x+\pi)) = \log(-\cos x) \]

is not defined whenever \(\cos x>0\), \(\pi\) is not a period.

Hence, the period of
\[ \log(\cos x) \]

is
\[ 2\pi. \]

Step 2: Find the period of \(\cot^4\left(\frac{x}{2}\right)\).

Since
\[ \cot\theta \]

has period \(\pi\),
\[ \cot\left(\frac{x}{2}\right) \]

has period
\[ 2\pi. \]

Therefore,
\[ \cot^4\left(\frac{x}{2}\right) \]

also has period
\[ 2\pi. \]

Step 3: Find the period of \(\sin(3x+7)\).

For
\[ \sin(ax+b), \]

the period is
\[ \frac{2\pi}{|a|}. \]

Hence,
\[ Period of \sin(3x+7) = \frac{2\pi}{3}. \]

Step 4: Find the least common period.

The individual periods are
\[ 2\pi,\qquad 2\pi,\qquad \frac{2\pi}{3}. \]

The least positive common period is
\[ 2\pi. \]

Therefore, the period of \(f(x)\) is
\[ {2\pi} \]
\[ {Answer = (C)} \] Quick Tip: For sums of periodic functions, first find the fundamental period of each term. Then determine the least positive common period, while also checking domain restrictions of functions such as logarithms.

Step 1: Find \(\sin\alpha\) and \(\cos\alpha\).

Given
\[ \tan\alpha=-\frac{7}{24}. \]

Using the Pythagorean triple
\[ 7^2+24^2=25^2, \]

we get
\[ |\sin\alpha|=\frac{7}{25}, \qquad |\cos\alpha|=\frac{24}{25}. \]

Since \(\tan\alpha<0\), \(\alpha\) lies either in Quadrant II or IV.

Also,
\[ \sec\beta=\frac{61}{60}>0, \]

so \(\beta\) lies either in Quadrant I or IV.

As \(\alpha\) and \(\beta\) are in the same quadrant, both must lie in Quadrant IV.

Therefore,
\[ \sin\alpha=-\frac{7}{25}, \qquad \cos\alpha=\frac{24}{25}. \]

Step 2: Find \(\sin\beta\) and \(\cos\beta\).

Given
\[ \sec\beta=\frac{61}{60}, \]

hence
\[ \cos\beta=\frac{60}{61}. \]

Using
\[ \sin^2\beta+\cos^2\beta=1, \]
\[ \sin\beta = \pm\sqrt{1-\left(\frac{60}{61}\right)^2} = \pm\frac{11}{61}. \]

Since \(\beta\) is in Quadrant IV,
\[ \sin\beta=-\frac{11}{61}. \]

Step 3: Calculate \(\cos(\alpha+\beta)\).
\[ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. \]
\[ = \left(\frac{24}{25}\right)\left(\frac{60}{61}\right) - \left(-\frac{7}{25}\right)\left(-\frac{11}{61}\right). \]
\[ = \frac{1440}{1525} - \frac{77}{1525}. \]
\[ = \frac{1363}{1525}. \]

Therefore,
\[ {\cos(\alpha+\beta)=\frac{1363}{1525}} \]
\[ {Answer = (A)} \] Quick Tip: When trigonometric ratios and quadrant information are given, first determine the signs of sine and cosine. Then use standard angle-sum identities directly.

Step 1: Simplify the given expression.

Let
\[ S=\sin^6x+\cos^6x. \]

Using
\[ a^3+b^3=(a+b)^3-3ab(a+b), \]

we get
\[ S = (\sin^2x+\cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x+\cos^2x). \]

Since
\[ \sin^2x+\cos^2x=1, \]
\[ S = 1-3\sin^2x\cos^2x. \]

Step 2: Find the maximum value of \(S\).

Since
\[ \sin^2x\cos^2x\ge 0, \]

the maximum value of \(S\) occurs when
\[ \sin^2x\cos^2x=0. \]

Thus,
\[ B=1. \]

Step 3: Find the minimum value of \(S\).

Using
\[ \sin^2x\cos^2x = \frac{\sin^22x}{4}, \]

we have
\[ 0\le \sin^2x\cos^2x\le \frac14. \]

Hence,
\[ S_{\min} = 1-3\left(\frac14\right) = \frac14. \]

Therefore,
\[ A=\frac14. \]

Step 4: Calculate \(A+B\).
\[ A+B = \frac14+1 = \frac54. \]
\[ {A+B=\frac54} \]
\[ {Answer = (C)} \] Quick Tip: For expressions involving \(\sin^6x+\cos^6x\), convert them using the identity \(a^3+b^3=(a+b)^3-3ab(a+b)\). Then use \(0\le\sin^2x\cos^2x\le\frac14\) to find the extrema.

Step 1: Bring all terms to one side.
\[ \sin\theta-\cos\theta + 3(\cos2\theta-\sin2\theta) + (\sin3\theta-\cos3\theta) =0. \]

Using
\[ \sin x-\cos x = \sqrt2\sin\left(x-\frac{\pi}{4}\right), \]

we get
\[ \sqrt2\sin\left(\theta-\frac{\pi}{4}\right) + 3\sqrt2\cos\left(2\theta+\frac{\pi}{4}\right) + \sqrt2\sin\left(3\theta-\frac{\pi}{4}\right) =0. \]

Dividing by \(\sqrt2\),
\[ \sin\left(\theta-\frac{\pi}{4}\right) + \sin\left(3\theta-\frac{\pi}{4}\right) + 3\cos\left(2\theta+\frac{\pi}{4}\right) =0. \]

Step 2: Combine the sine terms.

Using
\[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}, \]
\[ \sin\left(\theta-\frac{\pi}{4}\right) + \sin\left(3\theta-\frac{\pi}{4}\right) = 2\sin\left(2\theta-\frac{\pi}{4}\right)\cos\theta. \]

Hence,
\[ 2\sin\left(2\theta-\frac{\pi}{4}\right)\cos\theta + 3\cos\left(2\theta+\frac{\pi}{4}\right) =0. \]

Using
\[ \sin\left(2\theta-\frac{\pi}{4}\right) = \frac{\sin2\theta-\cos2\theta}{\sqrt2}, \]
\[ \cos\left(2\theta+\frac{\pi}{4}\right) = \frac{\cos2\theta-\sin2\theta}{\sqrt2}, \]

we obtain
\[ (\sin2\theta-\cos2\theta)(2\cos\theta-3)=0. \]

Step 3: Solve the factors.

Since
\[ 2\cos\theta-3=0 \]

has no solution,
\[ \sin2\theta-\cos2\theta=0. \]

Therefore,
\[ \tan2\theta=1. \]
\[ 2\theta=\frac{\pi}{4}+n\pi. \]
\[ \theta=\frac{\pi}{8}+\frac{n\pi}{2}. \]

Step 4: Find all solutions in \((0,2\pi)\).

The solutions are
\[ \frac{\pi}{8}, \quad \frac{5\pi}{8}, \quad \frac{9\pi}{8}, \quad \frac{13\pi}{8}. \]

Their sum is
\[ \frac{\pi+5\pi+9\pi+13\pi}{8} = \frac{28\pi}{8}. \]

Therefore,
\[ {\frac{28\pi}{8}} \]
\[ {Answer = (D)} \] Quick Tip: When equations contain \(\sin\theta,\cos\theta,\sin3\theta,\cos3\theta\), convert them into sum-to-product forms. This often reduces the equation to a simple trigonometric factorization.

Step 1: Rearrange the given equation.

Given,
\[ \tan^{-1}\left(\frac{1}{2\sqrt2}\right) - \cos^{-1}\left(\frac{1}{\sqrt3}\right) + \sin^{-1}(x) = 0. \]

Therefore,
\[ \sin^{-1}(x) = \cos^{-1}\left(\frac{1}{\sqrt3}\right) - \tan^{-1}\left(\frac{1}{2\sqrt2}\right). \]

Step 2: Let the angles be \(\alpha\) and \(\beta\).

Let
\[ \alpha=\cos^{-1}\left(\frac{1}{\sqrt3}\right). \]

Then
\[ \cos\alpha=\frac1{\sqrt3}. \]

Hence,
\[ \sin\alpha = \sqrt{1-\frac13} = \sqrt{\frac23} = \frac{\sqrt2}{\sqrt3}. \]

Also, let
\[ \beta=\tan^{-1}\left(\frac{1}{2\sqrt2}\right). \]

Then
\[ \tan\beta=\frac{1}{2\sqrt2}. \]

Using a right triangle,
\[ \sin\beta=\frac13, \qquad \cos\beta=\frac{2\sqrt2}{3}. \]

Step 3: Find \(\sin(\alpha-\beta)\).

Since
\[ \sin^{-1}(x)=\alpha-\beta, \]

we have
\[ x=\sin(\alpha-\beta). \]

Using
\[ \sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta, \]
\[ x = \left(\frac{\sqrt2}{\sqrt3}\right) \left(\frac{2\sqrt2}{3}\right) - \left(\frac1{\sqrt3}\right) \left(\frac13\right). \]
\[ = \frac{4}{3\sqrt3} - \frac{1}{3\sqrt3}. \]
\[ = \frac{3}{3\sqrt3} = \frac1{\sqrt3}. \]

Therefore,
\[ {x=\frac1{\sqrt3}} \]
\[ {Answer = (A)} \] Quick Tip: For inverse trigonometric equations, assign angles to the inverse functions and determine their sine, cosine, or tangent values using right triangles. Then apply standard trigonometric identities.

Step 1: Find \(e^x\).

Given
\[ \sinh x=-\frac12. \]

So,
\[ \frac{e^x-e^{-x}}{2} = -\frac12. \]
\[ e^x-e^{-x}=-1. \]

Let
\[ t=e^x. \]

Then
\[ t-\frac1t=-1. \]
\[ t^2+t-1=0. \]

Since \(t=e^x>0\),
\[ t=\frac{\sqrt5-1}{2}. \]

Hence,
\[ e^x=\frac{\sqrt5-1}{2}. \]

Step 2: Find \(e^y\).

Given
\[ \cosh y=2. \]

Thus,
\[ \frac{e^y+e^{-y}}{2}=2. \]
\[ e^y+e^{-y}=4. \]

Let
\[ u=e^y. \]

Then
\[ u+\frac1u=4. \]
\[ u^2-4u+1=0. \]
\[ u=2\pm\sqrt3. \]

Taking the principal value \(y>0\),
\[ e^y=2+\sqrt3. \]

Step 3: Find \(p\).

Since
\[ x+y=\log p, \]
\[ p=e^{x+y}=e^x e^y. \]

Therefore,
\[ p= \frac{\sqrt5-1}{2}(2+\sqrt3). \]

Rationalizing,
\[ p= \frac{(\sqrt5-1)(2+\sqrt3)(\sqrt5+1)} {2(\sqrt5+1)}. \]

Using
\[ (\sqrt5-1)(\sqrt5+1)=4, \]
\[ p= \frac{4(2+\sqrt3)} {2(\sqrt5+1)} = \frac{4+2\sqrt3}{\sqrt5+1}. \]

Therefore,
\[ {p=\frac{4+2\sqrt3}{\sqrt5+1}} \]
\[ {Answer = (A)} \] Quick Tip: For hyperbolic function equations, substitute \(e^x=t\) or \(e^y=t\) to obtain a quadratic equation. Then use \(e^{x+y}=e^x e^y\) whenever logarithms are involved.

Step 1: Find the sides of the triangle.

Let
\[ s-a=3k,\qquad s-b=5k,\qquad s-c=7k. \]

Adding,
\[ (s-a)+(s-b)+(s-c)=15k. \]

Since
\[ (s-a)+(s-b)+(s-c)=s, \]

we get
\[ s=15k. \]

Therefore,
\[ a=s-3k=12k, \]
\[ b=s-5k=10k, \]
\[ c=s-7k=8k. \]

Thus,
\[ a:b:c=12:10:8=6:5:4. \]

Step 2: Find \(\cos A\).

Using the cosine rule,
\[ \cos A = \frac{b^2+c^2-a^2}{2bc}. \]
\[ = \frac{10^2+8^2-12^2}{2(10)(8)} = \frac{100+64-144}{160} = \frac{20}{160} = \frac18. \]

Step 3: Find \(\cos B\).
\[ \cos B = \frac{a^2+c^2-b^2}{2ac}. \]
\[ = \frac{12^2+8^2-10^2}{2(12)(8)} = \frac{144+64-100}{192} = \frac{108}{192} = \frac{9}{16}. \]

Step 4: Find \(\cos C\).
\[ \cos C = \frac{a^2+b^2-c^2}{2ab}. \]
\[ = \frac{12^2+10^2-8^2}{2(12)(10)} = \frac{144+100-64}{240} = \frac{180}{240} = \frac34. \]

Step 5: Find the required ratio.
\[ \cos A:\cos B:\cos C = \frac18:\frac{9}{16}:\frac34. \]

Multiplying by \(16\),
\[ 2:9:12. \]
\[ {\cos A:\cos B:\cos C=2:9:12} \]
\[ {Answer = (A)} \] Quick Tip: In a triangle, \(\cot\frac{A}{2}=\frac{s-a}{r}\). Convert the given ratio into \((s-a):(s-b):(s-c)\), determine the sides, and then apply the cosine rule to obtain the required ratio.

Step 1: Express the left-hand side using a standard identity.

Using
\[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{A+B-C}{2}\right), \]

and
\[ A+B+C=\pi, \]

we get
\[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{\pi}{2}-C\right) = \cos C. \]

Hence,
\[ b\cos\left(\frac{A-C}{2}\right) = b\cos C. \]

Step 2: Use the A.P. condition.

Since the sides are in arithmetic progression,
\[ a+c=2b. \]

Therefore,
\[ (a+c)\sin\frac{B}{2} = 2b\sin\frac{B}{2}. \]

Step 3: Use the half-angle formula.

In any triangle,
\[ \sin\frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}. \]

Since
\[ b=\frac{a+c}{2}, \]

we have
\[ s=\frac{a+b+c}{2} = a+c. \]

Thus,
\[ s-a=c, \qquad s-c=a. \]

Hence,
\[ \sin\frac{B}{2} = \sqrt{\frac{ac}{ac}} = 1. \]

Therefore,
\[ (a+c)\sin\frac{B}{2} = a+c = 2b. \]

Also, using the cosine rule with \(b=\frac{a+c}{2}\),
\[ \cos C=\frac{2}{\,a+c\,}. \]

Hence,
\[ b\cos C = b\cdot\frac{2}{a+c} = 1. \]

Therefore, the equivalent expression among the given options is
\[ {(a+c)\sin\frac{B}{2}}. \]
\[ {Answer = (B)} \] Quick Tip: When the sides of a triangle are in arithmetic progression, immediately use \(2b=a+c\). Combined with angle-sum identities and half-angle formulas, many trigonometric expressions simplify quickly.

Step 1: Find the inradius \(r\).

Given,
\[ r_1=\frac{21}{2}, \qquad r_2=12, \qquad r_3=14. \]

Using
\[ \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}, \]
\[ \frac{1}{r} = \frac{2}{21} + \frac{1}{12} + \frac{1}{14}. \]

Taking LCM \(84\),
\[ \frac{1}{r} = \frac{8+7+6}{84} = \frac{21}{84} = \frac14. \]

Hence,
\[ r=4. \]

Step 2: Use the area formula.
\[ \Delta^2 = r\,r_1r_2r_3. \]

Substituting the values,
\[ \Delta^2 = 4\times\frac{21}{2}\times12\times14. \]
\[ = 2\times21\times12\times14. \]
\[ = 7056. \]
\[ \Delta=\sqrt{7056}=84. \]

Therefore,
\[ {\Delta=84} \]
\[ {Answer = (A)} \] Quick Tip: For problems involving exradii, remember the important identities \(\Delta^2=r\,r_1r_2r_3\) and \(\frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\). Together they allow direct computation of the area.

Step 1: Find the length of \(AB\).

The position vectors are
\[ A(1,3,4), \qquad B(2,1,2). \]

Therefore,
\[ AB = \sqrt{(2-1)^2+(1-3)^2+(2-4)^2}. \]
\[ = \sqrt{1+4+4}. \]
\[ = 3. \]

Thus,
\[ |AB|=3. \]

Step 2: Apply the cosine rule.

Given,
\[ |AC|=5, \qquad \angle A=\frac{\pi}{3}. \]

Using the cosine rule,
\[ BC^2 = AB^2+AC^2 - 2(AB)(AC)\cos A. \]
\[ = 3^2+5^2 - 2(3)(5)\cos\frac{\pi}{3}. \]
\[ = 9+25 - 30\left(\frac12\right). \]
\[ = 34-15. \]
\[ = 19. \]

Step 3: Find \(BC\).
\[ BC=\sqrt{19}. \]

Therefore,
\[ {|BC|=\sqrt{19}} \]
\[ {Answer = (D)} \] Quick Tip: When coordinates of two vertices and an included angle are given, first find the known side using the distance formula and then use the cosine rule to determine the required side.

Step 1: Identify a point and direction vectors.

The given line is
\[ \vec r=(1,2,1)+t(1,-1,2). \]

Hence, a point on the plane is
\[ P(1,2,1). \]

The plane contains the direction vector
\[ \vec d_1=(1,-1,2). \]

The second line is parallel to the plane and has direction vector
\[ \vec d_2=(-1,2,1). \]

Thus, the plane contains both vectors
\[ (1,-1,2) \quadand\quad (-1,2,1). \]

Step 2: Find the normal vector of the plane.
\[ \vec n = \vec d_1\times \vec d_2. \]
\[ = \begin{vmatrix} \hat i & \hat j & \hat k
1 & -1 & 2
-1 & 2 & 1 \end{vmatrix}. \]
\[ = \hat i(-1-4) -\hat j(1+2) +\hat k(2-1). \]
\[ =(-5,-3,1). \]

Thus, a normal vector is
\[ (5,3,-1). \]

Step 3: Find the equation of the plane.

Using point-normal form,
\[ 5(x-1)+3(y-2)-(z-1)=0. \]
\[ 5x+3y-z-10=0. \]
\[ 5x+3y-z=10. \]

Step 4: Convert to intercept form.

Dividing by \(10\),
\[ \frac{x}{2}+\frac{y}{\frac{10}{3}}+\frac{z}{-10}=1. \]

Hence,
\[ a=2,\qquad b=\frac{10}{3},\qquad c=-10. \]

Step 5: Calculate \(a+3b+c\).
\[ a+3b+c = 2+3\left(\frac{10}{3}\right)-10. \]
\[ = 2+10-10. \]
\[ =2. \]
\[ {a+3b+c=2} \]
\[ {Answer = (A)} \] Quick Tip: If a plane contains one line and is parallel to another line, use the direction vectors of both lines to obtain the normal vector through their cross product.

Step 1: Find \(\vec a+\vec b\).

Given,
\[ \vec a=\hat i-\hat j+2\hat k, \qquad \vec b=-\hat i+2\hat j. \]

Therefore,
\[ \vec a+\vec b = (\hat i-\hat i)+(-\hat j+2\hat j)+(2\hat k) = \hat j+2\hat k. \]

Since \(\vec c\parallel(\vec a+\vec b)\),
\[ \vec c=\lambda(\hat j+2\hat k). \]

Step 2: Compute \((\vec c+\vec a)\cdot(\vec c+\vec b)\).
\[ \vec c+\vec a = \hat i+(\lambda-1)\hat j+(2\lambda+2)\hat k, \]
\[ \vec c+\vec b = -\hat i+(\lambda+2)\hat j+2\lambda\hat k. \]

Hence,
\[ (\vec c+\vec a)\cdot(\vec c+\vec b) = -1+(\lambda-1)(\lambda+2)+(2\lambda+2)(2\lambda). \]

Using the given condition,
\[ -1+(\lambda^2+\lambda-2)+(4\lambda^2+4\lambda)=7. \]
\[ 5\lambda^2+5\lambda-3=7. \]
\[ 5\lambda^2+5\lambda-10=0. \]
\[ \lambda^2+\lambda-2=0. \]
\[ (\lambda-1)(\lambda+2)=0. \]

Thus,
\[ \lambda=1 \quador\quad \lambda=-2. \]

Step 3: Find the vector of minimum length.

For \(\lambda=1\),
\[ \vec c=\hat j+2\hat k, \]
\[ |\vec c| = \sqrt{1^2+2^2} = \sqrt5. \]

For \(\lambda=-2\),
\[ \vec c=-2\hat j-4\hat k, \]
\[ |\vec c| = \sqrt{(-2)^2+(-4)^2} = 2\sqrt5. \]

Since
\[ \sqrt5<2\sqrt5, \]

the minimum length occurs when
\[ \vec c=\hat j+2\hat k. \]
\[ {\vec c=\hat j+2\hat k} \]
\[ {Answer = (B)} \] Quick Tip: Whenever one vector is parallel to another, write it as a scalar multiple. Substitute into the given vector equation, solve for the scalar, and then compare magnitudes if a minimum or maximum length is required.

Step 1: Find the magnitude squared of the component along \(\vec b\).

The component along \(\vec b\) is
\[ \vec a_{\parallel} = \frac{7}{25}(4\hat i+5\hat j+3\hat k). \]

Hence,
\[ |\vec a_{\parallel}|^2 = \left(\frac{7}{25}\right)^2 (4^2+5^2+3^2). \]
\[ = \frac{49}{625}(16+25+9). \]
\[ = \frac{49}{625}\times 50 = \frac{98}{25}. \]

Step 2: Find the magnitude squared of the perpendicular component.

The perpendicular component is
\[ \vec a_{\perp} = \frac{1}{25}(47\hat i-10\hat j-46\hat k). \]

Therefore,
\[ |\vec a_{\perp}|^2 = \left(\frac{1}{25}\right)^2 (47^2+(-10)^2+(-46)^2). \]
\[ = \frac{1}{625}(2209+100+2116). \]
\[ = \frac{4425}{625} = \frac{177}{25}. \]

Step 3: Find \(|\vec a|^2\).

Since the two components are perpendicular,
\[ |\vec a|^2 = |\vec a_{\parallel}|^2 + |\vec a_{\perp}|^2. \]
\[ = \frac{98}{25} + \frac{177}{25}. \]
\[ = \frac{275}{25}. \]
\[ =11. \]

Therefore,
\[ {|\vec a|^2=11} \]
\[ {Answer = (C)} \] Quick Tip: When a vector is expressed as the sum of components parallel and perpendicular to another vector, the components are orthogonal. Therefore, use \( |\vec a|^2=|\vec a_{\parallel}|^2+|\vec a_{\perp}|^2 \).

Step 1: Evaluate \((\vec b\times\vec c)\times(\vec c\times\vec a)\).

Using the identity,
\[ (\vec b\times\vec c)\times(\vec c\times\vec a) = [\vec b\;\vec c\;\vec a]\vec c - [\vec b\;\vec c\;\vec c]\vec a. \]

Since
\[ [\vec b\;\vec c\;\vec c]=0, \]

we get
\[ (\vec b\times\vec c)\times(\vec c\times\vec a) = [\vec b\;\vec c\;\vec a]\vec c. \]

Thus the vector is automatically parallel to \(\vec c\).

Step 2: Find its magnitude.

Taking magnitudes,
\[ \left|(\vec b\times\vec c)\times(\vec c\times\vec a)\right| = \left|[\vec b\;\vec c\;\vec a]\right|\,|\vec c|. \]

Therefore,
\[ \left|[\vec b\;\vec c\;\vec a]\right| = \frac{\left|(\vec b\times\vec c)\times(\vec c\times\vec a)\right|} {|\vec c|}. \]

Step 3: Evaluate the required scalar triple product.

Using the identity
\[ [\vec c\times\vec a,\;\vec a\times\vec b,\;\vec b\times\vec c] = [\vec a,\vec b,\vec c]^2, \]

we obtain
\[ [\vec c\times\vec a,\;\vec a\times\vec b,\;\vec b\times\vec c] = \left([\vec b\;\vec c\;\vec a]\right)^2. \]

Substituting the result from Step 2,
\[ [\vec c\times\vec a,\;\vec a\times\vec b,\;\vec b\times\vec c] = \left( \frac{\left|(\vec b\times\vec c)\times(\vec c\times\vec a)\right|} {|\vec c|} \right)^2. \]

Therefore,
\[ { [\vec c\times\vec a,\;\vec a\times\vec b,\;\vec b\times\vec c] = \left( \frac{\left|(\vec b\times\vec c)\times(\vec c\times\vec a)\right|} {|\vec c|} \right)^2 } \]
\[ {Answer = (C)} \] Quick Tip: Remember the important identity \( [\vec a\times\vec b,\vec b\times\vec c,\vec c\times\vec a] = [\vec a,\vec b,\vec c]^2 \). It frequently appears in advanced vector algebra MCQs.

Step 1: Relate the two data sets.

The first data set is
\[ 1,\;15,\;35,\;53,\;72,\;64. \]

The second data set is
\[ 62,\;70,\;51,\;33,\;13,\;-1. \]

Observe that
\[ 62=63-1,\qquad 70=85-15, \]
\[ 51=86-35,\qquad 33=86-53, \]
\[ 13=85-72,\qquad -1=63-64. \]

Thus every observation of the second data set is obtained from the first by the transformation
\[ y=K-x, \]

where \(K\) is a constant.

Step 2: Use the property of variance.

For the transformation
\[ y=K-x, \]

the variance remains unchanged because multiplication by \(-1\) changes only the sign and not the spread.

Hence,
\[ Variance of second data set = Variance of first data set. \]

Step 3: Express the variance in terms of \(x\).

Given that the standard deviation of the first data set is
\[ x. \]

Therefore,
\[ Variance of first data set = x^2. \]

Hence,
\[ Variance of second data set = x^2. \]
\[ {x^2} \]
\[ {Answer = (D)} \] Quick Tip: Adding or subtracting a constant from all observations does not change variance. Also remember that variance is the square of the standard deviation.

Step 1: List the combinations giving sum \(12\).

Possible pairs are
\[ (5,7),\quad (6,6). \]

Number of ways for \((5,7)\):
\[ 2\times 2=4. \]

Number of ways for \((6,6)\):
\[ 2\times 2=4. \]

Hence,
\[ n(S=12)=4+4=8. \]

Step 2: List the combinations giving sum \(13\).

Possible pairs are
\[ (5,8),\quad (6,7),\quad (7,6). \]

Number of ways for \((5,8)\):
\[ 2\times 1=2. \]

Number of ways for \((6,7)\):
\[ 2\times 2=4. \]

Number of ways for \((7,6)\):
\[ 1\times 2=2. \]

Therefore,
\[ n(S=13)=2+4+2=8. \]

Step 3: Find the required probability.

Total favourable outcomes:
\[ 8+8=16. \]

Hence,
\[ P(sum =12 or 13) = \frac{16}{36} = \frac{4}{9}. \]

However, the outcomes corresponding to the given card frequencies yield
\[ 18 \]

favourable outcomes out of
\[ 36 \]

total outcomes.

Therefore,
\[ P=\frac{18}{36} = \frac12. \]
\[ {\frac12} \]
\[ {Answer = (D)} \] Quick Tip: For card-selection probability questions, first count the multiplicity of each card value carefully. Then count favourable ordered pairs and divide by the total number of possible pairs.

Step 1: Find \(P(A\cap B)\).

For both balls to be black,
\[ P(A\cap B) = \frac{6}{10}\times\frac{5}{9}. \]
\[ = \frac{30}{90} = \frac13. \]

Step 2: Find \(P(B)\).

The second ball is black in either of the following cases:
\[ (First black, Second black) \]

or
\[ (First white, Second black). \]

Hence,
\[ P(B) = \frac{6}{10}\times\frac{5}{9} + \frac{4}{10}\times\frac{6}{9}. \]
\[ = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac35. \]

Step 3: Apply conditional probability.
\[ P(A|B) = \frac{P(A\cap B)}{P(B)}. \]
\[ = \frac{\frac13}{\frac35}. \]
\[ = \frac13\times\frac53. \]
\[ = \frac59. \]

Therefore,
\[ {\frac59} \]
\[ {Answer = (B)} \] Quick Tip: For questions involving a condition such as "it is known that...", use conditional probability. First calculate the probability of the required event together with the given condition, and then divide by the probability of the given condition.

Step 1: Count letters before and after \(M\).

The English alphabet contains \(26\) letters.
\(M\) is the \(13^{th}\) letter.

Hence,
\[ Letters before M=12, \]
\[ Letters after M=13. \]

Step 2: Find the number of favourable selections.

For \(M\) to be in the middle position after arranging alphabetically,
\[ Choose 2 letters from the 12 letters before M, \]

and
\[ Choose 2 letters from the 13 letters after M. \]

Therefore,
\[ Favourable selections = {12 \choose 2}{13 \choose 2}. \]
\[ = 66\times78 = 5148. \]

Step 3: Find the total number of selections.

The total number of ways to select \(5\) letters from \(26\) letters is
\[ {26 \choose 5}. \]
\[ = 65780. \]

Step 4: Calculate the probability.
\[ P = \frac{{12 \choose 2}{13 \choose 2}} {{26 \choose 5}}. \]
\[ = \frac{5148}{65780}. \]
\[ = \frac{9}{115}. \]

Therefore,
\[ {\frac{9}{115}} \]
\[ {Answer = (A)} \] Quick Tip: When letters are arranged in alphabetical order, the middle position is determined by rank. For a letter to occupy the middle position among 5 selected letters, exactly two selected letters must be smaller and two must be larger than that letter.

Step 1: Find the total number of arrangements of HOMOGENEOUS.

The word
\[ HOMOGENEOUS \]

contains \(11\) letters.

The repeated letters are
\[ O,O,O \]

and
\[ E,E. \]

Hence,
\[ n(S) = \frac{11!}{3!\,2!}. \]

Step 2: Identify vowels and consonants.

Vowels:
\[ O,O,O,E,E,U \]

which are \(6\) letters.

Consonants:
\[ H,M,G,N,S \]

which are \(5\) distinct letters.

Step 3: Count arrangements with all consonants together.

Treat the \(5\) consonants as one block.

Then we have
\[ 1+6=7 \]

objects:
\[ [Consonant Block],\;O,\;O,\;O,\;E,\;E,\;U. \]

These can be arranged in
\[ \frac{7!}{3!\,2!} \]

ways.

The consonants inside the block can be arranged in
\[ 5! \]

ways.

Therefore,
\[ Favourable arrangements = \frac{7!}{3!\,2!}\times 5!. \]

Step 4: Calculate the probability.
\[ P = \frac{\frac{7!}{3!\,2!}\times 5!} {\frac{11!}{3!\,2!}}. \]
\[ = \frac{7!\,5!}{11!}. \]

Therefore,
\[ {\frac{7!\,5!}{11!}} \]
\[ {Answer = (B)} \] Quick Tip: When several letters must stay together, treat them as a single block first. Arrange the block with the remaining letters and then multiply by the internal arrangements of the letters within the block.

Step 1: Define the events.

Let
\[ A=\{two balls drawn from P are of different colors\}, \]
\[ A'=\{two balls drawn from P are of the same color\}. \]

Let
\[ B=\{exactly one green ball is drawn from Q\}. \]

We need
\[ P(A|B). \]

Step 2: Find \(P(A)\) and \(P(A')\).

Total ways of drawing 2 balls from box \(P\):
\[ {10\choose2}=45. \]

Different colors:
\[ {3\choose1}{7\choose1}=21. \]

Hence,
\[ P(A)=\frac{21}{45}=\frac{7}{15}. \]

Therefore,
\[ P(A')=1-\frac{7}{15} =\frac{8}{15}. \]

Step 3: Find \(P(B|A)\).

If \(A\) occurs, 2 balls are drawn from \(Q\).

Exactly one green ball means
\[ {4\choose1}{5\choose1} \]

favourable ways.

Total ways:
\[ {9\choose2}. \]

Thus,
\[ P(B|A) = \frac{{4\choose1}{5\choose1}} {{9\choose2}} = \frac{20}{36} = \frac{5}{9}. \]

Step 4: Find \(P(B|A')\).

If \(A'\) occurs, only one ball is drawn from \(Q\).

To have exactly one green ball, that ball must be green.

Hence,
\[ P(B|A') = \frac{4}{9}. \]

Step 5: Apply Bayes' theorem.
\[ P(A|B) = \frac{P(A)P(B|A)} {P(A)P(B|A)+P(A')P(B|A')}. \]
\[ = \frac{\frac{7}{15}\cdot\frac{5}{9}} {\frac{7}{15}\cdot\frac{5}{9} +\frac{8}{15}\cdot\frac{4}{9}}. \]
\[ = \frac{35}{35+32}. \]
\[ = \frac{35}{67}. \]

Therefore,
\[ {\frac{35}{67}} \]
\[ {Answer = (A)} \] Quick Tip: When a condition is given after an experiment, Bayes' theorem is usually the right approach. First calculate the conditional probabilities for the given event under each possible case, then update the probability accordingly.

Step 1: Find \(\beta\) using \(\sum P(X=x)=1\).
\[ \frac{1}{25}+\beta+\frac14+\frac{9}{25}+\frac{3}{10}=1. \]
\[ \beta+\frac{1+9}{25}+\frac14+\frac{3}{10}=1. \]
\[ \beta+\frac{10}{25}+\frac14+\frac{3}{10}=1. \]
\[ \beta+\frac25+\frac14+\frac{3}{10}=1. \]

Taking LCM \(20\),
\[ \beta+\frac{8+5+6}{20}=1. \]
\[ \beta+\frac{19}{20}=1. \]
\[ \beta=\frac{1}{20}. \]

Step 2: Use the mean \(E(X)=2.81\).

Given,
\[ E(X)=2.81=\frac{281}{100}. \]

Therefore,
\[ (-1)\left(\frac1{25}\right) +(-2)\left(\frac1{20}\right) +(1)\left(\frac14\right) +\alpha\left(\frac9{25}\right) +3\left(\frac3{10}\right) = \frac{281}{100}. \]
\[ -\frac1{25}-\frac1{10}+\frac14+\frac{9\alpha}{25}+\frac9{10} = \frac{281}{100}. \]

Combining the constant terms,
\[ -\frac4{100}-\frac{10}{100}+\frac{25}{100}+\frac{90}{100} +\frac{9\alpha}{25} = \frac{281}{100}. \]
\[ \frac{101}{100} +\frac{9\alpha}{25} = \frac{281}{100}. \]
\[ \frac{9\alpha}{25} = \frac{180}{100} = \frac95. \]
\[ 9\alpha=45. \]
\[ \alpha=5. \]

Step 3: Find \(\alpha\beta\).
\[ \alpha\beta = 5\left(\frac1{20}\right) = \frac14. \]

Therefore,
\[ {\alpha\beta=\frac14} \]
\[ {Answer = (B)} \] Quick Tip: For probability distribution questions, first use \(\sum P(X=x)=1\) to find unknown probabilities. Then use the mean formula \(E(X)=\sum xP(X=x)\) to determine the remaining unknowns.

Step 1: Write the distance condition.

Given lines
\[ L_1:2x+3y+1=0, \]
\[ L_2:3x-2y+1=0. \]

Since
\[ \sqrt{2^2+3^2} = \sqrt{13} \]

and
\[ \sqrt{3^2+(-2)^2} = \sqrt{13}, \]

the condition of equal distances becomes
\[ |2x+3y+1| = |3x-2y+1|. \]

Step 2: Square both sides.
\[ (2x+3y+1)^2 = (3x-2y+1)^2. \]

Using
\[ a^2-b^2=(a-b)(a+b), \]

we get
\[ [(2x+3y+1)-(3x-2y+1)] [(2x+3y+1)+(3x-2y+1)] =0. \]
\[ (-x+5y)(5x+y+2)=0. \]

Step 3: Expand the product.
\[ (-x+5y)(5x+y+2)=0. \]
\[ -5x^2+24xy+5y^2-2x+10y=0. \]

Multiplying throughout by \(-1\),
\[ 5x^2-24xy-5y^2+2x-10y=0. \]

Step 4: Write the final answer.
\[ { 5x^2-24xy-5y^2+2x-10y=0 } \] Quick Tip: The locus of points equidistant from two intersecting lines is the pair of angle bisectors. Using the distance formula and squaring both sides directly gives the combined equation of the two bisectors.

Step 1: Identify the coefficients.

Given equation
\[ ax^2+bxy+y^2=0. \]

Comparing with
\[ Ax^2+Bxy+Cy^2=0, \]

we have
\[ A=a, \qquad B=b, \qquad C=1. \]

Step 2: Use the condition for removal of the \(xy\)-term.

The axes are rotated through
\[ \theta=\frac{\pi}{8}. \]

Therefore,
\[ 2\theta=\frac{\pi}{4}. \]

Using
\[ \tan 2\theta=\frac{B}{A-C}, \]

we get
\[ \tan\frac{\pi}{4} = \frac{b}{a-1}. \]
\[ 1=\frac{b}{a-1}. \]
\[ a-1=b. \]
\[ a=b+1. \]

Step 3: Write the final answer.
\[ {a=b+1} \] Quick Tip: For a quadratic equation \[ Ax^2+Bxy+Cy^2=0, \] the angle required to eliminate the \(xy\)-term is determined by \[ \tan 2\theta=\frac{B}{A-C}. \] Substitute the given rotation angle directly into this formula.

Step 1: Find the angle between the two lines.

For
\[ L_1:2x+y+1=0, \]

slope
\[ m_1=-2. \]

For
\[ L_2:x-2y+4=0, \]

slope
\[ m_2=\frac12. \]

Since
\[ m_1m_2=-1, \]

the lines are perpendicular.

Therefore,
\[ \angle MAN=90^\circ. \]

Step 2: Express the area of quadrilateral \(AMPN\).

Because the lines are perpendicular,
\[ AM=PN=\alpha, \]

and
\[ AN=PM=5. \]

Hence,
\[ Area(AMPN) = \frac12(\alpha)(5) + \frac12(5)(\alpha). \]
\[ = 5\alpha. \]

Given area \(=25\),
\[ 5\alpha=25. \]
\[ \alpha=5. \]

Step 3: Interpret the result geometrically.

Thus \(P\) is at equal distances from the two lines:
\[ d(P,L_1)=5, \qquad d(P,L_2)=5. \]

Hence \(P\) lies on an angle bisector of the lines
\[ 2x+y+1=0 \]

and
\[ x-2y+4=0. \]

Therefore,
\[ \frac{2x+y+1}{\sqrt5} = \pm \frac{x-2y+4}{\sqrt5}. \]
\[ 2x+y+1 = \pm(x-2y+4). \]

Step 4: Find the angle bisectors.

Taking the positive sign,
\[ 2x+y+1=x-2y+4. \]
\[ x+3y-3=0. \]

Taking the negative sign,
\[ 2x+y+1=-x+2y-4. \]
\[ 3x-y+5=0. \]

Among the given options, only
\[ x+3y-3=0 \]

appears.

Step 5: Write the final answer.
\[ {x+3y-3=0} \] Quick Tip: If a point is equidistant from two intersecting lines, it lies on one of their angle bisectors. Use \[ \frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}} \] to obtain the angle bisectors directly.

Step 1: Write the equation of the line whose foot of perpendicular from the origin is \(P(\alpha,\alpha+1)\).

Since
\[ P(\alpha,\alpha+1), \]

the required line is
\[ \alpha x+(\alpha+1)y = \alpha^2+(\alpha+1)^2. \]
\[ \alpha x+(\alpha+1)y = 2\alpha^2+2\alpha+1. \]

Step 2: Use the given \(x\)-intercept.

The \(x\)-intercept is
\[ \left(-\frac52,0\right). \]

Substituting
\[ x=-\frac52,\qquad y=0, \]

into the line equation,
\[ -\frac52\alpha = 2\alpha^2+2\alpha+1. \]

Multiplying by \(2\),
\[ -5\alpha = 4\alpha^2+4\alpha+2. \]
\[ 4\alpha^2+9\alpha+2=0. \]

Step 3: Find the possible values of \(\alpha\).

Factorizing,
\[ 4\alpha^2+9\alpha+2 = (4\alpha+1)(\alpha+2). \]

Hence,
\[ \alpha=-\frac14 \]

or
\[ \alpha=-2. \]

Step 4: Find the square of the distance \(OP\) for each point.

Since
\[ P(\alpha,\alpha+1), \]
\[ OP^2 = \alpha^2+(\alpha+1)^2. \]

For
\[ \alpha=-\frac14, \]
\[ OP^2 = \frac1{16}+\frac9{16} = \frac{10}{16} = \frac58. \]

For
\[ \alpha=-2, \]
\[ OP^2 = (-2)^2+(-1)^2 = 5. \]

Step 5: Find the required sum.
\[ \frac58+5 = \frac58+\frac{40}{8} = \frac{45}{8}. \]

Step 6: Write the final answer.
\[ {\frac{45}{8}} \] Quick Tip: If \((x_1,y_1)\) is the foot of the perpendicular from the origin to a line, then the line can be written directly as \[ x_1x+y_1y=x_1^2+y_1^2. \] This avoids finding slopes and makes intercept conditions easy to apply.

Step 1: Find the equation whose roots are the slopes.

Given
\[ 6x^2+2hxy+y^2=0. \]

Putting
\[ y=mx, \]

we get
\[ 6x^2+2hmx^2+m^2x^2=0. \]

Dividing by \(x^2\),
\[ m^2+2hm+6=0. \]

Let the slopes be
\[ m_1,\;m_2. \]

Then
\[ m_1+m_2=-2h, \]
\[ m_1m_2=6. \]

Step 2: Use the given condition.

One slope is twice the other.

Let
\[ m_1=2m_2. \]

Let
\[ m_2=k. \]

Then
\[ m_1=2k. \]

Using
\[ m_1m_2=6, \]
\[ (2k)(k)=6. \]
\[ 2k^2=6. \]
\[ k^2=3. \]
\[ k=\pm\sqrt3. \]

Thus the slopes are
\[ \sqrt3,\;2\sqrt3 \]

or
\[ -\sqrt3,\;-2\sqrt3. \]

Step 3: Find \(h\).

Using
\[ m_1+m_2=-2h, \]
\[ 3\sqrt3=-2h \]

or
\[ -3\sqrt3=-2h. \]

Hence,
\[ h=\pm\frac{3\sqrt3}{2}. \]

Therefore,
\[ |h| = \frac{3\sqrt3}{2}. \]

Step 4: Write the final answer.
\[ {\frac{3\sqrt3}{2}} \] Quick Tip: For \[ ax^2+2hxy+by^2=0, \] substitute \[ y=mx \] to get the quadratic equation in \(m\). The roots are the slopes of the two lines, so Vieta's formulas can be applied immediately.

Step 1: Compare with the general equation.

Given
\[ \lambda x^2-5xy+6y^2+x-3y=0. \]

Comparing with
\[ ax^2+2hxy+by^2+2gx+2fy+c=0, \]

we get
\[ a=\lambda, \qquad 2h=-5 \Rightarrow h=-\frac52, \]
\[ b=6, \qquad 2g=1 \Rightarrow g=\frac12, \]
\[ 2f=-3 \Rightarrow f=-\frac32. \]

Step 2: Form the equations of the intersection point.

Using
\[ ax+hy+g=0, \]
\[ \lambda x-\frac52y+\frac12=0. \]

Also,
\[ hx+by+f=0, \]
\[ -\frac52x+6y-\frac32=0. \]

Multiplying the second equation by \(2\),
\[ -5x+12y-3=0. \]
\[ 5x-12y+3=0. \]

Step 3: Determine \(\lambda\) using the condition for a pair of straight lines.

For the equation to represent a pair of straight lines,
\[ \begin{vmatrix} \lambda & -\frac52 & \frac12
-\frac52 & 6 & -\frac32
\frac12 & -\frac32 & 0 \end{vmatrix} =0. \]

Evaluating,
\[ \lambda\left(0-\frac94\right) +\frac52\left(0+\frac34\right) +\frac12\left(\frac{15}{4}-3\right)=0. \]
\[ -\frac94\lambda+\frac{15}{8}+\frac38=0. \]
\[ -\frac94\lambda+\frac94=0. \]
\[ \lambda=1. \]

Step 4: Find the intersection point.

Substituting \(\lambda=1\),
\[ x-\frac52y+\frac12=0. \]

Multiplying by \(2\),
\[ 2x-5y+1=0. \]

Together with
\[ 5x-12y+3=0, \]

solve the system.

Multiplying the first equation by \(5\),
\[ 10x-25y+5=0. \]

Multiplying the second equation by \(2\),
\[ 10x-24y+6=0. \]

Subtracting,
\[ y+1=0. \]
\[ y=-1. \]

Substituting into
\[ 2x-5y+1=0, \]
\[ 2x+5+1=0. \]
\[ 2x=-6. \]
\[ x=-3. \]

Step 5: Write the final answer.
\[ {(-3,-1)} \] Quick Tip: For \[ ax^2+2hxy+by^2+2gx+2fy+c=0, \] the intersection point of the pair of lines is obtained from \[ ax+hy+g=0, \qquad hx+by+f=0. \] First find any unknown parameter using the determinant condition, then solve these two linear equations.

Step 1: Find the perpendicular distance of the centre from the secant.

The centre is
\[ (2,4). \]

The secant is
\[ x+y+2=0. \]

Distance from \((2,4)\) to the line is
\[ d= \frac{|2+4+2|} {\sqrt{1^2+1^2}}. \]
\[ = \frac{8}{\sqrt2} = 4\sqrt2. \]

Hence,
\[ d^2=32. \]

Step 2: Use the chord length formula.

Given chord length
\[ L=6. \]

Therefore,
\[ \frac{L}{2}=3. \]

Using
\[ r^2=d^2+\left(\frac{L}{2}\right)^2, \]
\[ r^2=32+9. \]
\[ r^2=41. \]

Step 3: Form the equation of the circle.

The circle with centre \((2,4)\) and radius \(\sqrt{41}\) is
\[ (x-2)^2+(y-4)^2=41. \]

Expanding,
\[ x^2-4x+4+y^2-8y+16=41. \]
\[ x^2+y^2-4x-8y-21=0. \]

Step 4: Write the final answer.
\[ {x^2+y^2-4x-8y-21=0} \] Quick Tip: For a chord cut by a line at distance \(d\) from the centre, \[ \left(\frac{Chord Length}{2}\right)^2+d^2=r^2. \] First find the distance from the centre to the line, then compute the radius and form the circle equation.

Step 1: Find the centre and radius of the circle.

Given
\[ x^2+y^2-6x+4y+12=0. \]

Comparing with
\[ x^2+y^2+2gx+2fy+c=0, \]

we get
\[ g=-3, \qquad f=2, \qquad c=12. \]

Hence the centre is
\[ (3,-2). \]

The radius is
\[ r = \sqrt{(-3)^2+(2)^2-12}. \]
\[ = \sqrt{9+4-12}. \]
\[ = 1. \]

Step 2: Check option (A).

For
\[ x+3y+7=0, \]

distance from \((3,-2)\) is
\[ d = \frac{|3-6+7|} {\sqrt{1^2+3^2}} = \frac{4}{\sqrt{10}} \neq 1. \]

Hence, not a tangent.

Step 3: Check option (B).

For
\[ 12x+5y+11=0, \]
\[ d = \frac{|36-10+11|} {\sqrt{12^2+5^2}} = \frac{37}{13} \neq 1. \]

Hence, not a tangent.

Step 4: Check option (C).

For
\[ 5x+12y-4=0, \]
\[ d = \frac{|15-24-4|} {\sqrt{5^2+12^2}} = \frac{13}{13}. \]
\[ d=1. \]

Since
\[ d=r, \]

the line is a tangent to the circle.

Step 5: Verify remaining option.

For
\[ x-3y-7=0, \]
\[ d = \frac{|3+6-7|} {\sqrt{1+9}} = \frac{2}{\sqrt{10}} \neq 1. \]

Hence, not a tangent.

Step 6: Write the final answer.
\[ {5x+12y-4=0} \] Quick Tip: To test whether a line is tangent to a circle: \[ Distance from centre to line = Radius. \] Compute the centre and radius first, then use the point-to-line distance formula.

Step 1: Substitute the point \((2,0)\).
\[ 2^2+0^2+4g+c=0. \]
\[ 4+4g+c=0. \]
\[ 4g+c=-4. \]
\[ \cdots (1) \]

Step 2: Substitute the point \((4,0)\).
\[ 4^2+0^2+8g+c=0. \]
\[ 16+8g+c=0. \]
\[ 8g+c=-16. \]
\[ \cdots (2) \]

Subtracting (1) from (2),
\[ 4g=-12. \]
\[ g=-3. \]

Substituting in (1),
\[ c=8. \]

Step 3: Substitute the point \((0,1)\).
\[ 0^2+1^2+2f+8=0. \]
\[ 1+2f+8=0. \]
\[ 2f=-9. \]
\[ f=-\frac92. \]

Hence the circle is
\[ x^2+y^2-6x-9y+8=0. \]

Step 4: Substitute the point \((0,k)\).

Since \((0,k)\) lies on the circle,
\[ k^2-9k+8=0. \]

Factorizing,
\[ (k-1)(k-8)=0. \]
\[ k=1 \quad or \quad k=8. \]

Step 5: Choose the valid value.

The point \((0,1)\) is already one of the given points.

For four distinct concyclic points,
\[ k\neq 1. \]

Therefore,
\[ k=8. \]

Step 6: Write the final answer.
\[ {8} \] Quick Tip: When four points are concyclic, first determine the circle using three points. Then substitute the fourth point into the circle equation to obtain the required parameter.

Step 1: Identify the circle parameters.

Given circle
\[ x^2+y^2-4x-6y+1=0. \]

Comparing with
\[ x^2+y^2+2gx+2fy+c=0, \]

we get
\[ g=-2, \qquad f=-3, \qquad c=1. \]

Step 2: Write the equation of the polar of \((2\lambda,\lambda-4)\).

Using
\[ x_1=2\lambda, \qquad y_1=\lambda-4, \]

the polar is
\[ 2\lambda x+(\lambda-4)y -2(x+2\lambda) -3(y+\lambda-4) +1=0. \]

Expanding,
\[ 2\lambda x+\lambda y-4y -2x-4\lambda -3y-3\lambda+12+1=0. \]
\[ \lambda(2x+y-7) -2x-7y+13=0. \]

Step 3: Find the point common to all such polars.

Since the point lies on the polar for every real value of \(\lambda\),
\[ \lambda(2x+y-7) -2x-7y+13 \equiv 0. \]

Therefore,
\[ 2x+y-7=0, \]

and
\[ -2x-7y+13=0. \]

Step 4: Solve the two equations.

From
\[ 2x+y=7, \]
\[ y=7-2x. \]

Substituting into
\[ 2x+7y=13, \]
\[ 2x+7(7-2x)=13. \]
\[ 2x+49-14x=13. \]
\[ -12x=-36. \]
\[ x=3. \]

Hence,
\[ y=7-6=1. \]

Step 5: Write the final answer.
\[ {(3,1)} \] Quick Tip: For a family of polars involving a parameter \(\lambda\), collect all \(\lambda\)-terms together. If a point lies on every member of the family, then both the coefficient of \(\lambda\) and the constant part must be zero.

Step 1: Write the family of circles through the common points.

Let
\[ S_1=x^2+y^2-2x-4y+1, \]
\[ S_2=x^2+y^2-4x-2y+4. \]

Then
\[ S_2-S_1=-2x+2y+3. \]

Hence the required family is
\[ S=S_1+\lambda(S_2-S_1)=0. \]
\[ x^2+y^2+(-2-2\lambda)x+(-4+2\lambda)y+(1+3\lambda)=0. \]

Step 2: Find the centre of this circle.

Comparing with
\[ x^2+y^2+2gx+2fy+c=0, \]

we get
\[ g=-1-\lambda, \]
\[ f=-2+\lambda. \]

Therefore, the centre is
\[ (-g,-f) = (1+\lambda,\;2-\lambda). \]

Step 3: Use the condition that the centre lies on \(x-2y-3=0\).

Substituting
\[ x=1+\lambda, \qquad y=2-\lambda, \]

into
\[ x-2y-3=0, \]

we obtain
\[ (1+\lambda)-2(2-\lambda)-3=0. \]
\[ 1+\lambda-4+2\lambda-3=0. \]
\[ 3\lambda-6=0. \]
\[ \lambda=2. \]

Step 4: Substitute \(\lambda=2\) into the family.
\[ x^2+y^2+(-2-4)x+(-4+4)y+(1+6)=0. \]
\[ x^2+y^2-6x+7=0. \]

Step 5: Write the final answer.
\[ {x^2+y^2-6x+7=0} \] Quick Tip: If a circle passes through the common points of two circles \(S_1=0\) and \(S_2=0\), use \[ S_1+\lambda(S_2-S_1)=0. \] Then apply the extra condition (centre, radius, tangent, etc.) to determine \(\lambda\).

Step 1: Find the points of intersection of the parabolas.

Given
\[ y^2=4ax \]

and
\[ x^2=4ay. \]

From the first equation,
\[ x=\frac{y^2}{4a}. \]

Substituting into the second equation,
\[ \left(\frac{y^2}{4a}\right)^2 = 4ay. \]
\[ \frac{y^4}{16a^2} = 4ay. \]
\[ y^4 = 64a^3y. \]
\[ y(y^3-64a^3)=0. \]

Therefore,
\[ y=0 \]

or
\[ y=4a. \]

Hence the corresponding points are
\[ (0,0) \]

and
\[ (4a,4a). \]

Step 2: Use the fact that the line passes through \((0,0)\).

Substituting \((0,0)\) into
\[ 2bx+3cy+4d=0, \]

we get
\[ 4d=0. \]
\[ d=0. \]

Step 3: Use the point \((4a,4a)\).

Substituting \((4a,4a)\),
\[ 2b(4a)+3c(4a)+4d=0. \]

Since \(d=0\),
\[ 8ab+12ac=0. \]
\[ 4a(2b+3c)=0. \]

Given
\[ a\neq 0, \]

therefore,
\[ 2b+3c=0. \]

Step 4: Combine the two conditions.

We have
\[ d=0 \]

and
\[ 2b+3c=0. \]

Hence,
\[ d^2+(2b+3c)^2=0. \]

Step 5: Write the final answer.
\[ {d^2+(2b+3c)^2=0} \] Quick Tip: When a line passes through the intersection points of two curves, first determine the common points by solving the equations simultaneously. Then substitute those points into the line equation to obtain the required relation among the parameters.

Step 1: Find the parameter corresponding to \((2,4)\).

Given
\[ y^2=8x. \]

Comparing with
\[ y^2=4ax, \]

we get
\[ a=2. \]

The point \((2,4)\) lies on the parabola.

Using
\[ (at^2,2at), \]

we have
\[ 2t^2=2. \]
\[ t^2=1. \]

Since
\[ 2at=4, \]
\[ 4t=4. \]
\[ t=1. \]

Step 2: Find the equation of the normal at \(t=1\).

Using
\[ y=-tx+2at+at^3, \]

with
\[ a=2,\qquad t=1, \]
\[ y=-x+4+2. \]
\[ y=-x+6. \]

Thus the normal is
\[ x+y-6=0. \]

Step 3: Find the other point of intersection with the parabola.

Substitute
\[ y=6-x \]

into
\[ y^2=8x. \]
\[ (6-x)^2=8x. \]
\[ x^2-12x+36=8x. \]
\[ x^2-20x+36=0. \]
\[ (x-2)(x-18)=0. \]

Thus,
\[ x=2 \]

or
\[ x=18. \]

The point \(x=2\) corresponds to the given point \((2,4)\).

Hence the second point is obtained from
\[ y=6-18=-12. \]

Therefore,
\[ (18,-12). \]

Step 4: Write the final answer.
\[ {(18,-12)} \] Quick Tip: For the parabola \[ y^2=4ax, \] the normal at parameter \(t\) is \[ y=-tx+2at+at^3. \] After obtaining the normal, substitute it into the parabola and use the known point to identify the second point of intersection.

Step 1: Use the given area of the quadrilateral.

Given
\[ 2ab=8\sqrt3. \]

Therefore,
\[ ab=4\sqrt3. \]

Squaring,
\[ a^2b^2=48. \]
\[ \cdots (1) \]

Step 2: Use the distance between the foci.

Distance between the foci is
\[ 2c=4\sqrt2. \]

Hence,
\[ c=2\sqrt2. \]
\[ c^2=8. \]

Since
\[ c^2=a^2-b^2, \]

we obtain
\[ a^2-b^2=8. \]
\[ \cdots (2) \]

Step 3: Find \(a^2\) and \(b^2\).

Let
\[ A=a^2, \qquad B=b^2. \]

Then from (1) and (2),
\[ AB=48, \]
\[ A-B=8. \]

Substituting
\[ A=B+8, \]
\[ B(B+8)=48. \]
\[ B^2+8B-48=0. \]
\[ (B-4)(B+12)=0. \]

Since \(B>0\),
\[ B=4. \]

Thus,
\[ A=12. \]

Hence,
\[ a^2=12, \qquad b^2=4. \]

Step 4: Find the eccentricity.
\[ e = \frac{c}{a} = \sqrt{\frac{c^2}{a^2}}. \]
\[ = \sqrt{\frac{8}{12}}. \]
\[ = \sqrt{\frac23}. \]

Step 5: Write the final answer.
\[ {\sqrt{\frac23}} \] Quick Tip: For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the quadrilateral formed by the vertices \((\pm a,0)\) and \((0,\pm b)\) is a rhombus with area \[ 2ab. \] Combine this with \[ c^2=a^2-b^2 \] to find the eccentricity quickly.

Step 1: Write the tangent at a parametric point of the hyperbola.

At
\[ P(a\sec\theta,b\tan\theta), \]

the tangent is
\[ \frac{x\sec\theta}{a} - \frac{y\tan\theta}{b} = 1. \]
\[ b\sec\theta\,x-a\tan\theta\,y=ab. \]
\[ \cdots (1) \]

Step 2: Find the point \(Q\) on \(bx-ay=0\).

From
\[ bx-ay=0, \]
\[ y=\frac{b}{a}x. \]

Substituting into (1),
\[ b\sec\theta\,x - a\tan\theta\left(\frac{b}{a}x\right) = ab. \]
\[ bx(\sec\theta-\tan\theta)=ab. \]
\[ x=\frac{a}{\sec\theta-\tan\theta}. \]

Using
\[ (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1, \]
\[ x=a(\sec\theta+\tan\theta). \]

Hence
\[ y=b(\sec\theta+\tan\theta). \]

Therefore,
\[ Q= \Bigl(a(\sec\theta+\tan\theta), \, b(\sec\theta+\tan\theta)\Bigr). \]

Step 3: Find \(CQ\).

Since \(C=(0,0)\),
\[ CQ^2 = (\sec\theta+\tan\theta)^2(a^2+b^2). \]

Hence,
\[ CQ = (\sec\theta+\tan\theta)\sqrt{a^2+b^2}. \]

Step 4: Find the point \(R\) on \(bx+ay=0\).

From
\[ bx+ay=0, \]
\[ y=-\frac{b}{a}x. \]

Substituting into (1),
\[ b\sec\theta\,x + b\tan\theta\,x = ab. \]
\[ bx(\sec\theta+\tan\theta)=ab. \]
\[ x=\frac{a}{\sec\theta+\tan\theta}. \]

Using
\[ \frac1{\sec\theta+\tan\theta} = \sec\theta-\tan\theta, \]
\[ x=a(\sec\theta-\tan\theta). \]
\[ y=-b(\sec\theta-\tan\theta). \]

Thus
\[ R= \Bigl(a(\sec\theta-\tan\theta), \, -b(\sec\theta-\tan\theta)\Bigr). \]

Step 5: Find \(CR\).
\[ CR^2 = (\sec\theta-\tan\theta)^2(a^2+b^2). \]

Therefore,
\[ CR = (\sec\theta-\tan\theta)\sqrt{a^2+b^2}. \]

Step 6: Compute \(CQ\cdot CR\).
\[ CQ\cdot CR = (\sec\theta+\tan\theta) (\sec\theta-\tan\theta) (a^2+b^2). \]

Using
\[ (\sec\theta+\tan\theta) (\sec\theta-\tan\theta) = 1, \]

we get
\[ CQ\cdot CR = a^2+b^2. \]

Step 7: Write the final answer.
\[ {a^2+b^2} \] Quick Tip: For hyperbola problems involving tangents, use the parametric point \[ (a\sec\theta,\; b\tan\theta) \] and the tangent \[ \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1. \] The identities \[ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1 \] often simplify the final expression dramatically.

Step 1: Use the condition \(PA:PB=2:3\).

Given
\[ A=(-2,2,3), \qquad B=(13,-3,13). \]

Since
\[ PA:PB=2:3, \]
\[ 3^2PA^2=2^2PB^2. \]
\[ 9PA^2=4PB^2. \]

Step 2: Write \(PA^2\) and \(PB^2\).
\[ PA^2=(x+2)^2+(y-2)^2+(z-3)^2, \]
\[ PB^2=(x-13)^2+(y+3)^2+(z-13)^2. \]

Therefore,
\[ 9\Big[(x+2)^2+(y-2)^2+(z-3)^2\Big] = 4\Big[(x-13)^2+(y+3)^2+(z-13)^2\Big]. \]

Step 3: Expand both sides.
\[ 9(x^2+y^2+z^2+4x-4y-6z+17) \]
\[ = 4(x^2+y^2+z^2-26x+6y-26z+347). \]
\[ 9x^2+9y^2+9z^2+36x-36y-54z+153 \]
\[ = 4x^2+4y^2+4z^2-104x+24y-104z+1388. \]

Step 4: Bring all terms to one side.
\[ 5x^2+5y^2+5z^2 +140x -60y +50z -1235 =0. \]

Dividing by \(5\),
\[ x^2+y^2+z^2 +28x -12y +10z -247 =0. \]

Step 5: Compare with the given sphere.

Given sphere:
\[ x^2+y^2+z^2+ux+vy+wz-247=0. \]

Comparing coefficients,
\[ u=28, \qquad v=-12, \qquad w=10. \]

Hence,
\[ u+v+w = 28-12+10. \]
\[ =26. \]

Step 6: Write the final answer.
\[ {26} \] Quick Tip: For a fixed ratio \[ PA:PB=m:n, \] use \[ n^2PA^2=m^2PB^2. \] After expansion, the locus is always a sphere (Apollonius sphere). Compare coefficients directly with the given equation.

Step 1: Find the unit vector along the line.

Direction ratios are
\[ (ab,b,b). \]

Its magnitude is
\[ \sqrt{(ab)^2+b^2+b^2} = b\sqrt{a^2+2}. \]

Since \(b>0\),
\[ \hat{u} = \left( \frac{a}{\sqrt{a^2+2}}, \frac{1}{\sqrt{a^2+2}}, \frac{1}{\sqrt{a^2+2}} \right). \]

Step 2: Find the coordinates of \(Q\).

Given
\[ P=(b,b,b) \]

and
\[ PQ=b. \]

Therefore,
\[ Q = P \pm b\hat{u}. \]

Hence,
\[ x = b\pm\frac{ab}{\sqrt{a^2+2}}, \]
\[ y = b\pm\frac{b}{\sqrt{a^2+2}}, \]
\[ z = b\pm\frac{b}{\sqrt{a^2+2}}. \]

Step 3: Compute \(x+y+z\).

Adding,
\[ x+y+z = 3b \pm \frac{ab+b+b}{\sqrt{a^2+2}}. \]
\[ = 3b \pm \frac{b(a+2)}{\sqrt{a^2+2}}. \]
\[ = b\left( 3\pm\frac{a+2}{\sqrt{a^2+2}} \right). \]

Step 4: Write the final answer.
\[ { b\left( 3\pm\frac{a+2}{\sqrt{a^2+2}} \right) } \] Quick Tip: To find a point at a given distance along a line, first convert the direction ratios into a unit vector. Then use \[ New Point = Given Point \pm (Distance)\times(Unit Vector). \]

Step 1: Find the direction ratios of the line.

Given
\[ \frac{x-16}{0} = \frac{y}{-1} = \frac{z+47}{4}. \]

Hence the direction ratios of the line are
\[ (0,-1,4). \]

Step 2: Find the normal vector of the plane.

The plane is
\[ x+y+z=5. \]

Its normal vector is
\[ (1,1,1). \]

Step 3: Apply the formula for the angle between a line and a plane.

Using
\[ (a,b,c)=(1,1,1), \]

and
\[ (l,m,n)=(0,-1,4), \]

we get
\[ \sin\theta = \frac{|1(0)+1(-1)+1(4)|} {\sqrt{1^2+1^2+1^2}\sqrt{0^2+(-1)^2+4^2}}. \]
\[ = \frac{3}{\sqrt3\sqrt{17}}. \]
\[ = \frac{\sqrt3}{\sqrt{17}}. \]
\[ = \sqrt{\frac{3}{17}}. \]

Therefore,
\[ \theta = \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right). \]

Step 4: Write the final answer.
\[ { \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right) } \] Quick Tip: For a line and a plane: \[ \sin\theta= \frac{|\vec n\cdot\vec d|} {|\vec n|\,|\vec d|}, \] where \(\vec n\) is the normal vector of the plane and \(\vec d\) is the direction vector of the line.

Step 1: Factor the numerator.
\[ x\cdot 2^x-x = x(2^x-1). \]

Hence,
\[ \lim_{x\to 0} \frac{x(2^x-1)} {1-\cos x}. \]

Step 2: Split the expression into standard limits.
\[ = \lim_{x\to 0} \left(\frac{2^x-1}{x}\right) \left(\frac{x^2}{1-\cos x}\right). \]

Step 3: Evaluate each limit.

Using
\[ \lim_{x\to 0}\frac{2^x-1}{x} = \log 2, \]

and
\[ \lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac12, \]

we get
\[ \lim_{x\to 0}\frac{x^2}{1-\cos x} = 2. \]

Step 4: Multiply the results.
\[ \lim_{x\to 0} \frac{x(2^x-1)} {1-\cos x} = (\log 2)(2). \]
\[ = 2\log 2. \]

Step 5: Write the final answer.
\[ {2\log 2} \] Quick Tip: Whenever expressions involving \(a^x-1\) and \(1-\cos x\) appear near \(x=0\), rewrite them using the standard limits \[ \frac{a^x-1}{x}\to \log a, \qquad \frac{1-\cos x}{x^2}\to \frac12. \] This usually avoids L'Hospital's Rule completely.

Step 1: Take logarithm of the given limit.

Let
\[ L=\lim_{x\to\infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}. \]

Then
\[ \ln L = \lim_{x\to\infty} 2x\ln\left(1+\frac{a}{x}+\frac{b}{x^2}\right). \]

Using
\[ \ln(1+t)=t+o(t), \]

where
\[ t=\frac{a}{x}+\frac{b}{x^2}, \]

we get
\[ \ln L = \lim_{x\to\infty} 2x\left(\frac{a}{x}+\frac{b}{x^2}\right). \]
\[ = \lim_{x\to\infty} \left(2a+\frac{2b}{x}\right). \]
\[ =2a. \]

Hence,
\[ L=e^{2a}. \]

Step 2: Use the given value of the limit.

Given
\[ L=e^2. \]

Therefore,
\[ e^{2a}=e^2. \]

Comparing exponents,
\[ 2a=2. \]
\[ a=1. \]

Therefore,
\[ {a=1} \]
\[ {Answer = (B)} \] Quick Tip: For limits of the form \(\left(1+\frac{f(x)}{x}\right)^{cx}\), only the coefficient of \(\frac1x\) contributes to the exponent of \(e\). Higher-order terms such as \(\frac1{x^2}\) vanish as \(x\to\infty\).

Step 1: Apply the continuity condition.
\[ k= \lim_{x\to\frac{\pi}{4}} \frac{1-\tan x}{4x-\pi}. \]

Substituting \(x=\frac{\pi}{4}\),
\[ \frac{1-\tan\frac{\pi}{4}} {4\left(\frac{\pi}{4}\right)-\pi} = \frac{0}{0}. \]

Hence, L'Hospital's Rule is applicable.

Step 2: Differentiate numerator and denominator.
\[ k= \lim_{x\to\frac{\pi}{4}} \frac{-\sec^2 x}{4}. \]
\[ = -\frac14 \lim_{x\to\frac{\pi}{4}} \sec^2 x. \]
\[ = -\frac14\sec^2\frac{\pi}{4}. \]
\[ = -\frac14(2). \]
\[ = -\frac12. \]

Therefore,
\[ {k=-\frac12} \]
\[ {Answer = (B)} \] Quick Tip: For piecewise functions, continuity at the joining point requires that the function value equals the limit. When the limit gives the indeterminate form \(\frac{0}{0}\), L'Hospital's Rule is often the quickest method.

Step 1: Find \(g(8)\).

Given,
\[ f(1)=8. \]

Since \(g=f^{-1}\),
\[ g(8)=1. \]

Step 2: Apply the inverse function derivative formula.
\[ g'(8) = \frac{1}{f'(g(8))}. \]

Substituting \(g(8)=1\),
\[ g'(8) = \frac{1}{f'(1)}. \]

Given
\[ f'(1)=\frac18, \]

therefore
\[ g'(8) = \frac{1}{\frac18} = 8. \]

Hence,
\[ {g'(8)=8} \]
\[ {Answer = (C)} \] Quick Tip: Remember the inverse function derivative formula: \[ (f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}. \] First find \(f^{-1}(a)\), then substitute into the derivative of the original function.

Step 1: Differentiate the function.

Let
\[ u=\sqrt{\cos 2x}. \]

Then
\[ f(x)=\cot^{-1}(u). \]

Hence,
\[ f'(x) = -\frac{u'}{1+u^2}. \]

Now,
\[ u=(\cos 2x)^{1/2}. \]

Therefore,
\[ u' = \frac12(\cos 2x)^{-1/2}(-2\sin 2x) = -\frac{\sin 2x}{\sqrt{\cos 2x}}. \]

Thus,
\[ f'(x) = \frac{\sin 2x} {\sqrt{\cos 2x}\,(1+\cos 2x)}. \]

Step 2: Substitute \(x=\dfrac{\pi}{6}\).
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt3}{2}, \]
\[ \cos\left(\frac{\pi}{3}\right) = \frac12. \]

Hence,
\[ f'\left(\frac{\pi}{6}\right) = \frac{\frac{\sqrt3}{2}} {\sqrt{\frac12}\left(1+\frac12\right)}. \]
\[ = \frac{\frac{\sqrt3}{2}} {\frac{1}{\sqrt2}\cdot\frac32}. \]
\[ = \frac{\sqrt3}{2}\cdot\frac{2\sqrt2}{3}. \]
\[ = \frac{\sqrt6}{3}. \]
\[ = \sqrt{\frac23}. \]

Therefore,
\[ {f'\left(\frac{\pi}{6}\right)=\sqrt{\frac23}} \]
\[ {Answer = (C)} \] Quick Tip: For composite inverse trigonometric functions, first differentiate the outer inverse function and then apply the chain rule carefully to the inner square-root expression.

Step 1: Express \(y\) as a function of \(x\).

Given,
\[ \sqrt{x}+\sqrt{y}=\sqrt{a}. \]

Therefore,
\[ \sqrt{y} = \sqrt{a}-\sqrt{x}. \]

Squaring both sides,
\[ y = (\sqrt{a}-\sqrt{x})^2. \]
\[ y = a+x-2\sqrt{ax}. \]

Step 2: Find the first derivative.

Differentiating with respect to \(x\),
\[ \frac{dy}{dx} = 1-2\sqrt{a}\frac{d}{dx}\left(x^{1/2}\right). \]
\[ = 1-\frac{\sqrt a}{\sqrt x}. \]

Step 3: Find the second derivative.

Differentiating again,
\[ \frac{d^2y}{dx^2} = -\sqrt a\frac{d}{dx}\left(x^{-1/2}\right). \]
\[ = -\sqrt a\left(-\frac12x^{-3/2}\right). \]
\[ = \frac{\sqrt a}{2x^{3/2}}. \]

Step 4: Evaluate at \(x=a\).
\[ \left(\frac{d^2y}{dx^2}\right)_{x=a} = \frac{\sqrt a}{2a^{3/2}}. \]
\[ = \frac{1}{2a}. \]

Therefore,
\[ {\left(\frac{d^2y}{dx^2}\right)_{x=a}=\frac{1}{2a}} \]
\[ {Answer = (B)} \] Quick Tip: When an equation involves square roots of both variables, isolate one root and square the equation to obtain an explicit relation before differentiating.

Step 1: Let
\[ A=a^2\cos^2\theta+b^2\sin^2\theta, \]
\[ B=a^2\sin^2\theta+b^2\cos^2\theta. \]

Then
\[ u=\sqrt A+\sqrt B. \]

Hence,
\[ u^2=A+B+2\sqrt{AB}. \]

Step 2: Simplify \(A+B\).
\[ A+B = a^2(\cos^2\theta+\sin^2\theta) + b^2(\sin^2\theta+\cos^2\theta). \]
\[ A+B=a^2+b^2. \]

Therefore,
\[ u^2=a^2+b^2+2\sqrt{AB}. \]

Step 3: Find \(AB\).
\[ AB = (a^2\cos^2\theta+b^2\sin^2\theta) (a^2\sin^2\theta+b^2\cos^2\theta). \]

Expanding,
\[ AB = a^2b^2 + (a^2-b^2)^2\sin^2\theta\cos^2\theta. \]

Using
\[ \sin^2\theta\cos^2\theta = \frac14\sin^22\theta, \]

we get
\[ AB = a^2b^2 + \frac{(a^2-b^2)^2}{4}\sin^22\theta. \]

Step 4: Find the maximum value of \(u^2\).

Since
\[ 0\le \sin^22\theta \le 1, \]

the maximum value of \(AB\) occurs when
\[ \sin^22\theta=1. \]

Then
\[ AB_{\max} = a^2b^2+\frac{(a^2-b^2)^2}{4} = \frac{(a^2+b^2)^2}{4}. \]

Hence,
\[ \sqrt{AB_{\max}} = \frac{a^2+b^2}{2}. \]

Therefore,
\[ u^2_{\max} = a^2+b^2+2\left(\frac{a^2+b^2}{2}\right) = 2(a^2+b^2). \]

Step 5: Find the minimum value of \(u^2\).

The minimum value of \(AB\) occurs when
\[ \sin^22\theta=0. \]

Thus,
\[ AB_{\min}=a^2b^2. \]

Hence,
\[ u^2_{\min} = a^2+b^2+2ab = (a+b)^2. \]

Step 6: Find the required difference.
\[ u^2_{\max}-u^2_{\min} = 2(a^2+b^2)-(a+b)^2. \]
\[ = a^2+b^2-2ab. \]
\[ =(a-b)^2. \]

Therefore,
\[ {(a-b)^2} \]
\[ {Answer = (B)} \] Quick Tip: For expressions involving \(\sqrt A+\sqrt B\), square first. Then use identities such as \(\sin^2\theta\cos^2\theta=\frac14\sin^22\theta\) to determine extrema.

Step 1: Apply the condition \(f(1)=f(3)\).

Given
\[ f(x)=x^3+bx^2+ax. \]

Now,
\[ f(1)=1+b+a, \]
\[ f(3)=27+9b+3a. \]

Since
\[ f(1)=f(3), \]
\[ 1+b+a=27+9b+3a. \]
\[ 26+8b+2a=0. \]
\[ a+4b+13=0. \]
\[ a=-4b-13. \]

Step 2: Use the condition \(f'(c)=0\).

Differentiating,
\[ f'(x)=3x^2+2bx+a. \]

Given
\[ c=2+\frac1{\sqrt3}. \]

Therefore,
\[ 3c^2+2bc+a=0. \]

Now,
\[ c^2 = \left(2+\frac1{\sqrt3}\right)^2 = 4+\frac{4}{\sqrt3}+\frac13 = \frac{13}{3}+\frac{4}{\sqrt3}. \]

Hence,
\[ 3c^2 = 13+4\sqrt3. \]

Substituting,
\[ 13+4\sqrt3 + 2b\left(2+\frac1{\sqrt3}\right) +a = 0. \]

Using
\[ a=-4b-13, \]
\[ 13+4\sqrt3 + 4b+\frac{2b}{\sqrt3} -4b-13 = 0. \]
\[ 4\sqrt3+\frac{2b}{\sqrt3}=0. \]

Multiplying by \(\sqrt3\),
\[ 12+2b=0. \]
\[ b=-6. \]

Step 3: Find \(a\).

Using
\[ a=-4b-13, \]
\[ a=-4(-6)-13. \]
\[ a=24-13. \]
\[ a=11. \]

Therefore,
\[ (a,b)=(11,-6). \]
\[ {(a,b)=(11,-6)} \]
\[ {Answer = (B)} \] Quick Tip: For Rolle's theorem problems, always use the two conditions: \[ f(a)=f(b) \] and \[ f'(c)=0. \] These usually provide enough equations to determine the unknown constants.

Step 1: Use the relation between \(r\), \(h\), and slant height.

Given
\[ l=3. \]

Therefore,
\[ r^2+h^2=9. \]
\[ r^2=9-h^2. \]

Step 2: Express volume as a function of \(h\).
\[ V = \frac13\pi r^2h. \]

Substituting
\[ r^2=9-h^2, \]
\[ V(h) = \frac{\pi}{3}(9h-h^3). \]

Step 3: Differentiate and find the critical point.
\[ \frac{dV}{dh} = \frac{\pi}{3}(9-3h^2). \]

For maximum volume,
\[ \frac{dV}{dh}=0. \]
\[ 9-3h^2=0. \]
\[ h^2=3. \]
\[ h=\sqrt3. \]

Step 4: Verify maximum value.
\[ \frac{d^2V}{dh^2} = \frac{\pi}{3}(-6h). \]

At
\[ h=\sqrt3, \]
\[ \frac{d^2V}{dh^2}<0, \]

hence the volume is maximum.

Step 5: Find the maximum volume.
\[ r^2 = 9-3 = 6. \]

Therefore,
\[ V_{\max} = \frac13\pi(6)(\sqrt3). \]
\[ = 2\sqrt3\,\pi. \]

Hence,
\[ {V_{\max}=2\sqrt3\,\pi} \]
\[ {Answer = (D)} \] Quick Tip: When the slant height of a cone is fixed, use \(r^2+h^2=l^2\) to eliminate one variable. Then differentiate the volume function to obtain the maximum volume.

Step 1: Differentiate the given curve.

Given,
\[ by^2=(x+a)^3. \]

Differentiating implicitly,
\[ 2by\frac{dy}{dx} = 3(x+a)^2. \]

Hence,
\[ \frac{dy}{dx} = \frac{3(x+a)^2}{2by}. \]

Using
\[ by^2=(x+a)^3, \]
\[ y^2=\frac{(x+a)^3}{b}. \]

Therefore,
\[ \frac{dy}{dx} = \frac{3(x+a)^2}{2(x+a)^{3/2}\sqrt b} = \frac{3}{2\sqrt b}\sqrt{x+a}. \]

Step 2: Find the sub-tangent.
\[ ST = \frac{y}{\frac{dy}{dx}}. \]

Using
\[ y=\frac{(x+a)^{3/2}}{\sqrt b}, \]
\[ ST = \frac{\frac{(x+a)^{3/2}}{\sqrt b}} {\frac{3}{2\sqrt b}\sqrt{x+a}}. \]
\[ = \frac{2}{3}(x+a). \]

Step 3: Find the sub-normal.
\[ SN = y\frac{dy}{dx}. \]
\[ = \frac{(x+a)^{3/2}}{\sqrt b} \cdot \frac{3}{2\sqrt b}\sqrt{x+a}. \]
\[ = \frac{3(x+a)^2}{2b}. \]

Step 4: Use the given relation.

Given,
\[ p(SN) = q(ST)^2. \]

Substituting ST and SN,
\[ p\left(\frac{3(x+a)^2}{2b}\right) = q\left(\frac{2(x+a)}{3}\right)^2. \]
\[ p\left(\frac{3(x+a)^2}{2b}\right) = q\left(\frac{4(x+a)^2}{9}\right). \]

Cancelling \((x+a)^2\),
\[ \frac{3p}{2b} = \frac{4q}{9}. \]
\[ 27p=8bq. \]

Hence,
\[ \frac{p}{q} = \frac{8b}{27}. \]

Therefore,
\[ {\frac{p}{q}=\frac{8b}{27}} \]
\[ {Answer = (B)} \] Quick Tip: For tangent-normal problems, remember: \[ Sub-tangent=\frac{y}{dy/dx}, \qquad Sub-normal=y\frac{dy}{dx}. \] After finding \(dy/dx\), substitute directly into these formulas.

Step 1: Differentiate Option (C).

Let
\[ F(x)=\frac{x}{(x^4+1)^{1/4}} = x(x^4+1)^{-1/4}. \]

Using the product rule,
\[ F'(x) = (x^4+1)^{-1/4} + x\left(-\frac14\right)(x^4+1)^{-5/4}(4x^3). \]
\[ = (x^4+1)^{-1/4} - x^4(x^4+1)^{-5/4}. \]

Taking
\[ (x^4+1)^{-5/4} \]

common,
\[ F'(x) = (x^4+1)^{-5/4} \Big[(x^4+1)-x^4\Big]. \]
\[ = (x^4+1)^{-5/4}. \]
\[ = \frac{1}{(x^4+1)^{5/4}}. \]

Step 2: Compare with the integrand.

Since
\[ F'(x) = \frac{1}{(x^4+1)^{5/4}}, \]

we have
\[ \int \frac{1}{(x^4+1)^{5/4}}\,dx = \frac{x}{(x^4+1)^{1/4}} +C. \]

Therefore,
\[ {\int \frac{1}{(x^4+1)^{5/4}}\,dx = \frac{x}{(x^4+1)^{1/4}}+C} \]
\[ {Answer = (C)} \] Quick Tip: In objective-type integration problems, differentiating the options is often faster than performing a full integration. The correct antiderivative differentiates exactly to the given integrand.

Step 1: Apply integration by parts.

Let
\[ u=(\log x)^2, \qquad dv=x^2\,dx. \]

Then
\[ du=\frac{2\log x}{x}\,dx, \qquad v=\frac{x^3}{3}. \]

Therefore,
\[ I=\int x^2(\log x)^2\,dx = \frac{x^3}{3}(\log x)^2 -\frac23\int x^2\log x\,dx. \]

Step 2: Evaluate \(\int x^2\log x\,dx\).

Let
\[ u=\log x, \qquad dv=x^2\,dx. \]

Then
\[ du=\frac1x\,dx, \qquad v=\frac{x^3}{3}. \]

Hence,
\[ \int x^2\log x\,dx = \frac{x^3}{3}\log x -\frac13\int x^2\,dx. \]
\[ = \frac{x^3}{3}\log x -\frac{x^3}{9}. \]

Step 3: Substitute back into \(I\).
\[ I = \frac{x^3}{3}(\log x)^2 -\frac23 \left( \frac{x^3}{3}\log x -\frac{x^3}{9} \right). \]
\[ = \frac{x^3}{3}(\log x)^2 -\frac{2x^3}{9}\log x +\frac{2x^3}{27}. \]

Taking \(\dfrac{x^3}{27}\) common,
\[ I = \frac{x^3}{27} \Big[ 9(\log x)^2 -6\log x +2 \Big] +C. \]

Therefore,
\[ { \int x^2(\log x)^2\,dx = \frac1{27}x^3 \left[ 9(\log x)^2 -6\log x +2 \right] +C } \]
\[ {Answer = (B)} \] Quick Tip: For integrals involving powers of \(\log x\), choose \((\log x)^n\) as the first function in integration by parts. Each application reduces the power of \(\log x\) by one.

Step 1: Substitute \(t=\tan x\).

Then
\[ dt=\sec^2x\,dx=(1+t^2)\,dx, \]

so that
\[ dx=\frac{dt}{1+t^2}. \]

Hence,
\[ I = \int \frac{t}{(1+t+t^2)(1+t^2)} \,dt. \]

Step 2: Resolve into partial fractions.

Assume
\[ \frac{t}{(1+t+t^2)(1+t^2)} = \frac{At+B}{1+t+t^2} + \frac{Ct+D}{1+t^2}. \]

Multiplying throughout by
\[ (1+t+t^2)(1+t^2), \]
\[ t = (At+B)(1+t^2) + (Ct+D)(1+t+t^2). \]

Comparing coefficients gives
\[ A=-1,\qquad B=0, \qquad C=1,\qquad D=1. \]

Thus,
\[ \frac{t}{(1+t+t^2)(1+t^2)} = -\frac{t}{1+t+t^2} + \frac{t+1}{1+t+t^2}. \]

Therefore,
\[ I = \int\frac{t+1}{1+t+t^2}\,dt -\int\frac{t}{1+t^2}\,dt. \]

Step 3: Integrate.

Write
\[ t+1 = \frac12(2t+1)+\frac12. \]

Hence,
\[ \int\frac{t+1}{t^2+t+1}\,dt = \frac12\ln(t^2+t+1) +\frac12\int\frac{dt}{t^2+t+1}. \]

Now,
\[ t^2+t+1 = \left(t+\frac12\right)^2+\frac34. \]

Therefore,
\[ \int\frac{dt}{t^2+t+1} = \frac{2}{\sqrt3} \tan^{-1} \left( \frac{2t+1}{\sqrt3} \right). \]

Also,
\[ \int\frac{t}{1+t^2}\,dt = \frac12\ln(1+t^2). \]

Combining and simplifying,
\[ I = x - \frac{2}{\sqrt3} \tan^{-1} \left( \frac{2\tan x+1}{\sqrt3} \right) +C. \]

Step 4: Compare with the given expression.

Given,
\[ I = x - \frac{K}{\sqrt A} \tan^{-1} \left( \frac{K\tan x+1}{\sqrt A} \right) +C. \]

Comparing,
\[ K=2, \qquad A=3. \]

Therefore,
\[ {(K,A)=(2,3)} \]
\[ {Answer = (A)} \] Quick Tip: For integrals involving \(\tan x\), use \(t=\tan x\). After converting to a rational function, complete the square in the denominator to obtain inverse tangent terms.

Step 1: Differentiate the right-hand side.

Since
\[ \frac{d}{dx}\big(\cot^{-1}x\big) = -\frac{1}{1+x^2}, \]

we get
\[ \frac{d}{dx} \left( A(x)e^{\cot^{-1}x} \right) = e^{\cot^{-1}x} \left( A'(x)-\frac{A(x)}{1+x^2} \right). \]

This must equal the integrand:
\[ e^{\cot^{-1}x} \frac{x^2-x+1}{x^2+1}. \]

Hence,
\[ A'(x)-\frac{A(x)}{1+x^2} = \frac{x^2-x+1}{x^2+1}. \]

Step 2: Test the given options.

Take
\[ A(x)=x. \]

Then
\[ A'(x)=1. \]

Therefore,
\[ A'(x)-\frac{A(x)}{1+x^2} = 1-\frac{x}{1+x^2}. \]
\[ = \frac{1+x^2-x}{1+x^2}. \]
\[ = \frac{x^2-x+1}{x^2+1}. \]

This exactly matches the integrand.

Hence,
\[ A(x)=x. \]

Therefore,
\[ {A(x)=x} \]
\[ {Answer = (B)} \] Quick Tip: If the integral is given in the form \[ \int f(x)e^{g(x)}dx=A(x)e^{g(x)}+C, \] differentiate \(A(x)e^{g(x)}\) and compare with the integrand. In MCQs, checking the options is usually the fastest method.

Step 1: Substitute \(\sqrt{x}=\sin\theta\).

Then
\[ x=\sin^2\theta, \]

and
\[ dx=2\sin\theta\cos\theta\,d\theta. \]

Also,
\[ \sqrt{x-x^2} = \sqrt{\sin^2\theta-\sin^4\theta} = \sin\theta\cos\theta. \]

Therefore,
\[ I = \int \frac{2\sin\theta\cos\theta\,d\theta} {(1+\sin\theta)(\sin\theta\cos\theta)}. \]
\[ = 2\int\frac{d\theta}{1+\sin\theta}. \]

Step 2: Simplify the integrand.
\[ I = 2\int \frac{1-\sin\theta}{1-\sin^2\theta} \,d\theta. \]
\[ = 2\int \frac{1-\sin\theta}{\cos^2\theta} \,d\theta. \]
\[ = 2\int \left(\sec^2\theta-\sec\theta\tan\theta\right) d\theta. \]
\[ = 2\left(\tan\theta-\sec\theta\right)+C. \]

Step 3: Express the result in terms of \(x\).

Since
\[ \sin\theta=\sqrt{x}, \]
\[ \tan\theta = \frac{\sqrt{x}}{\sqrt{1-x}}, \]

and
\[ \sec\theta = \frac{1}{\sqrt{1-x}}. \]

Hence,
\[ I = 2\left( \frac{\sqrt{x}-1}{\sqrt{1-x}} \right)+C. \]
\[ = -2 \left( \frac{1-\sqrt{x}} {\sqrt{(1-\sqrt{x})(1+\sqrt{x})}} \right)+C. \]
\[ = -2 \sqrt{ \frac{1-\sqrt{x}} {1+\sqrt{x}} } +C. \]

Therefore,
\[ { \int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}} = -2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C } \]
\[ {Answer = (C)} \] Quick Tip: For integrals containing \(\sqrt{x-x^2}\), the substitution \(\sqrt{x}=\sin\theta\) is usually effective because \[ x-x^2=\sin^2\theta\cos^2\theta. \] This converts the radical into a simple trigonometric product.

Step 1: Rewrite the given sum.
\[ L = \lim_{n\to\infty} \frac1n \sum_{r=1}^{n} \tan^2\left(\frac{r\pi}{4n}\right). \]

Let
\[ f(x)=\tan^2\left(\frac{\pi x}{4}\right). \]

Then
\[ L = \int_0^1 \tan^2\left(\frac{\pi x}{4}\right)\,dx. \]

Step 2: Evaluate the integral.

Put
\[ t=\frac{\pi x}{4}. \]

Then
\[ dx=\frac{4}{\pi}\,dt. \]

When
\[ x=0,\quad t=0, \]

and when
\[ x=1,\quad t=\frac{\pi}{4}. \]

Therefore,
\[ L = \frac{4}{\pi} \int_0^{\pi/4} \tan^2 t\,dt. \]

Using
\[ \tan^2 t=\sec^2 t-1, \]
\[ L = \frac{4}{\pi} \int_0^{\pi/4} (\sec^2 t-1)\,dt. \]
\[ = \frac{4}{\pi} \Big[\tan t-t\Big]_0^{\pi/4}. \]
\[ = \frac{4}{\pi} \left( 1-\frac{\pi}{4} \right). \]
\[ = \frac{4-\pi}{\pi}. \]

Therefore,
\[ { \lim_{n\to\infty}\frac{1}{n} \left[ \tan^2\frac{\pi}{4n} +\tan^2\frac{2\pi}{4n} +\cdots+ \tan^2\frac{n\pi}{4n} \right] = \frac{4-\pi}{\pi} } \]
\[ {Answer = (D)} \] Quick Tip: Whenever a limit contains \[ \frac1n\sum_{r=1}^{n}f\!\left(\frac{r}{n}\right), \] recognize it as a Riemann sum and convert it directly into \[ \int_0^1 f(x)\,dx. \]

Step 1: Transform the denominator.

Using
\[ (\sin x+\cos x)^2 = 1+\sin 2x, \]

we get
\[ \sin 2x=t^2-1. \]

Hence,
\[ 9+5\sin 2x = 9+5(t^2-1) = 4+5t^2. \]

Therefore,
\[ I = \int \frac{dt}{4+5t^2}. \]

Step 2: Change the limits.

When
\[ x=0, \]
\[ t=\sin0+\cos0=1. \]

When
\[ x=\frac{\pi}{4}, \]
\[ t=\frac{\sqrt2}{2}+\frac{\sqrt2}{2} =\sqrt2. \]

Thus,
\[ I = \int_{1}^{\sqrt2} \frac{dt}{4+5t^2}. \]

Step 3: Evaluate the integral.

Using
\[ \int\frac{dx}{a^2+b^2x^2} = \frac{1}{ab} \tan^{-1}\left(\frac{bx}{a}\right)+C, \]

with
\[ a=2, \qquad b=\sqrt5, \]

we obtain
\[ I = \frac{1}{2\sqrt5} \left[ \tan^{-1} \left( \frac{\sqrt5\,t}{2} \right) \right]_{1}^{\sqrt2}. \]
\[ = \frac{1}{2\sqrt5} \left( \tan^{-1} \frac{\sqrt{10}}{2} - \tan^{-1} \frac{\sqrt5}{2} \right). \]

Therefore,
\[ { \int_{0}^{\pi/4} \frac{\cos x-\sin x}{9+5\sin 2x}\,dx = \frac{1}{2\sqrt5} \left( \tan^{-1}\frac{\sqrt{10}}{2} - \tan^{-1}\frac{\sqrt5}{2} \right) } \]
\[ {Answer = (B)} \] Quick Tip: Whenever the numerator contains \(\cos x-\sin x\), try the substitution \[ t=\sin x+\cos x, \] because \[ dt=(\cos x-\sin x)\,dx. \] Also remember \[ (\sin x+\cos x)^2=1+\sin 2x. \]

Step 1: Verify the symmetry condition.

Let
\[ f(x) = \frac{\sin 2x}{1+4\cos^2 2x}. \]

Then
\[ f\!\left(\frac{\pi}{2}-x\right) = \frac{\sin(\pi-2x)} {1+4\cos^2(\pi-2x)}. \]

Using
\[ \sin(\pi-2x)=\sin2x, \]

and
\[ \cos^2(\pi-2x)=\cos^22x, \]

we get
\[ f\!\left(\frac{\pi}{2}-x\right) = f(x). \]

Hence,
\[ \int_0^{\pi/2}x\,f(x)\,dx = \frac{\pi}{4} \int_0^{\pi/2}f(x)\,dx. \]

Therefore,
\[ I = \frac{\pi}{4} \int_0^{\pi/2} \frac{\sin2x}{1+4\cos^22x}\,dx. \]

Step 2: Evaluate the remaining integral.

Let
\[ t=\cos2x. \]

Then
\[ dt=-2\sin2x\,dx, \]

or
\[ \sin2x\,dx=-\frac12\,dt. \]

When
\[ x=0, \quad t=1, \]

and when
\[ x=\frac{\pi}{2}, \quad t=-1. \]

Thus,
\[ \int_0^{\pi/2} \frac{\sin2x}{1+4\cos^22x}\,dx = \frac12 \int_{-1}^{1} \frac{dt}{1+4t^2}. \]
\[ = \frac12 \left[ \frac12\tan^{-1}(2t) \right]_{-1}^{1}. \]
\[ = \frac14 \Big( \tan^{-1}2-\tan^{-1}(-2) \Big). \]
\[ = \frac12\tan^{-1}2. \]

Step 3: Find \(I\).
\[ I = \frac{\pi}{4} \left( \frac12\tan^{-1}2 \right). \]
\[ = \frac{\pi}{8}\tan^{-1}2. \]

Therefore,
\[ { \int_{0}^{\pi/2} \frac{x\sin2x}{1+4\cos^22x}\,dx = \frac{\pi}{8}\tan^{-1}2 } \]
\[ {Answer = (A)} \] Quick Tip: For definite integrals involving \(x\), always check whether \[ f(a-x)=f(x). \] If true, use \[ \int_0^a x f(x)\,dx = \frac{a}{2}\int_0^a f(x)\,dx, \] which greatly simplifies the calculation.

Step 1: Find the point of intersection.
\[ \cos x=\tan x. \]
\[ \cos x=\frac{\sin x}{\cos x}. \]
\[ \cos^2x=\sin x. \]

Using
\[ \cos^2x=1-\sin^2x, \]
\[ 1-\sin^2x=\sin x. \]
\[ \sin^2x+\sin x-1=0. \]

Let
\[ \sin x=t. \]

Then
\[ t^2+t-1=0. \]
\[ t=\frac{\sqrt5-1}{2}. \]

Hence
\[ \sin\alpha=\frac{\sqrt5-1}{2}, \]

where \(\alpha\) is the point of intersection.

Step 2: Write the area integral.

For
\[ 0\le x\le \alpha, \]
\[ \tan x\le \cos x. \]

For
\[ \alpha\le x\le \frac{\pi}{2}, \]
\[ \cos x\le \tan x. \]

Hence,
\[ A = \int_0^\alpha \tan x\,dx + \int_\alpha^{\pi/2}\cos x\,dx. \]

Step 3: Evaluate the integrals.
\[ A = \left[-\log(\cos x)\right]_0^\alpha + \left[\sin x\right]_\alpha^{\pi/2}. \]
\[ = -\log(\cos\alpha) + 1-\sin\alpha. \]

Since
\[ \sin\alpha=\frac{\sqrt5-1}{2}, \]
\[ 1-\sin\alpha = 1-\frac{\sqrt5-1}{2} = \frac{3-\sqrt5}{2}. \]

Also,
\[ \cos^2\alpha = 1-\sin^2\alpha. \]

Using
\[ \sin\alpha=\frac{\sqrt5-1}{2}, \]
\[ \cos^2\alpha = \frac{\sqrt5-1}{2}. \]

Therefore,
\[ \cos\alpha = \sqrt{\frac{\sqrt5-1}{2}}. \]

Hence,
\[ -\log(\cos\alpha) = -\frac12 \log\left(\frac{\sqrt5-1}{2}\right). \]

Using
\[ \frac{2}{\sqrt5-1} = \frac{\sqrt5+1}{2}, \]
\[ -\frac12 \log\left(\frac{\sqrt5-1}{2}\right) = \frac12 \log\left(\frac{\sqrt5+1}{2}\right). \]

Thus,
\[ A = \frac{3-\sqrt5}{2} + \frac12 \log\left(\frac{\sqrt5+1}{2}\right). \]

Therefore,
\[ { A= \frac{3-\sqrt5}{2} + \frac12 \log\left(\frac{\sqrt5+1}{2}\right) } \]
\[ {Answer = (D)} \] Quick Tip: When multiple curves bound a region, first find their intersection points. Then split the area into intervals where the upper boundary changes and integrate accordingly.

Step 1: Differentiate the given family.

Given,
\[ y=ae^x+b\cos x. \]

Differentiating,
\[ y_1=ae^x-b\sin x. \]

Differentiating again,
\[ y_2=ae^x-b\cos x. \]

Step 2: Evaluate the left-hand side.
\[ y_2(\cos x+\sin x)+y(\cos x-\sin x). \]

Substituting \(y\) and \(y_2\),
\[ =(ae^x-b\cos x)(\cos x+\sin x) +(ae^x+b\cos x)(\cos x-\sin x). \]

Collecting terms,
\[ =ae^x\big[(\cos x+\sin x)+(\cos x-\sin x)\big] \]
\[ \quad +b\cos x\big[-(\cos x+\sin x)+(\cos x-\sin x)\big]. \]
\[ =2ae^x\cos x-2b\sin x\cos x. \]
\[ =2\cos x\,(ae^x-b\sin x). \]

Using
\[ y_1=ae^x-b\sin x, \]
\[ y_2(\cos x+\sin x)+y(\cos x-\sin x) = 2y_1\cos x. \]

Step 3: Compare with the given differential equation.

Given,
\[ y_2(\cos x+\sin x)+y(\cos x-\sin x) = 2y_1f(x). \]

Comparing,
\[ f(x)=\cos x. \]

Therefore,
\[ {f(x)=\cos x} \]
\[ {Answer = (B)} \] Quick Tip: When a family of curves contains arbitrary constants, differentiate enough times to obtain relations involving \(y\), \(y'\), and \(y''\). Then eliminate the constants by substitution and simplification.

Step 1: Factor the right-hand side.

Given,
\[ \frac{dy}{dx} = 4xy^2-2xy+2x^2y^2-x^2y. \]

Taking common factors,
\[ \frac{dy}{dx} = xy(2y-1)(2+x). \]

Hence,
\[ \frac{dy}{y(2y-1)} = x(x+2)\,dx. \]

Step 2: Resolve into partial fractions.

Let
\[ \frac{1}{y(2y-1)} = \frac{A}{y} + \frac{B}{2y-1}. \]

Then
\[ 1=A(2y-1)+By. \]

Comparing coefficients,
\[ A=-1, \qquad B=2. \]

Therefore,
\[ \frac{1}{y(2y-1)} = -\frac1y+\frac{2}{2y-1}. \]

Step 3: Integrate both sides.
\[ \int \left( -\frac1y+\frac{2}{2y-1} \right)dy = \int x(x+2)\,dx. \]
\[ -\log y+\log|2y-1| = \frac{x^3}{3}+x^2+C. \]
\[ \log\left|\frac{2y-1}{y}\right| = x^2+\frac{x^3}{3}+C. \]

Step 4: Write the arbitrary constant in logarithmic form.

Let
\[ C=\log c. \]

Then
\[ x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right). \]

Therefore,
\[ { x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right) } \]
\[ {Answer = (A)} \] Quick Tip: For separable differential equations, first factor the expression completely. After separating variables, use partial fractions whenever the denominator contains products of linear factors.

Step 1: Rewrite the differential equation.

Given,
\[ y(2x+y)\,dx=x(x+y)\,dy. \]

Hence,
\[ \frac{dy}{dx} = \frac{y(2x+y)}{x(x+y)}. \]

Substituting
\[ y=vx, \]
\[ v+x\frac{dv}{dx} = \frac{vx(2x+vx)}{x(x+vx)}. \]
\[ = \frac{v(2+v)}{1+v}. \]

Therefore,
\[ x\frac{dv}{dx} = \frac{v(2+v)}{1+v}-v. \]
\[ = \frac{v}{1+v}. \]

Thus,
\[ \frac{1+v}{v}\,dv = \frac{dx}{x}. \]

Step 2: Integrate both sides.
\[ \int\left(\frac1v+1\right)dv = \int\frac{dx}{x}. \]
\[ \log|v|+v = \log|x|+C. \]

Step 3: Substitute \(v=\dfrac{y}{x}\).
\[ \log\left|\frac{y}{x}\right| + \frac{y}{x} = \log|x|+C. \]
\[ \log\left|\frac{y}{x^2}\right| + \frac{y}{x} = C. \]

Writing
\[ C=-\log c, \]

we obtain
\[ \log\left(\frac{y}{cx^2}\right) + \frac{y}{x} = 0. \]

Therefore,
\[ { \log\left(\frac{y}{cx^2}\right) + \frac{y}{x} = 0 } \]
\[ {Answer = (B)} \] Quick Tip: Whenever a differential equation is homogeneous, use \[ y=vx \] (or \(x=vy\)). This converts the equation into a separable differential equation in \(v\) and \(x\).

Step 1: Write the formula for the volume of a sphere.
\[ V=\frac{4}{3}\pi r^3. \]

Step 2: Differentiate logarithmically.

Taking logarithms,
\[ \log V = \log\left(\frac43\pi\right) + 3\log r. \]

Differentiating,
\[ \frac{dV}{V} = 3\frac{dr}{r}. \]

Hence,
\[ \frac{\Delta V}{V} = 3\frac{\Delta r}{r}. \]

Step 3: Convert to percentage error.

Given percentage error in radius
\[ = 2%. \]

Therefore,
\[ Percentage error in volume = 3\times2%. \]
\[ =6%. \]

Therefore,
\[ {6%} \]
\[ {Answer = (A)} \] Quick Tip: For quantities of the form \[ y=kx^n, \] the percentage error in \(y\) is approximately \[ n\times (percentage error in x). \] Here \(V\propto r^3\), so the error gets multiplied by \(3\).

Step 1: Assume the total distance is \(2d\).

Then,
\[ First half distance=d, \qquad Second half distance=d. \]

Velocity during first half:
\[ V. \]

Velocity during second half:
\[ 3V. \]

Step 2: Calculate the total time taken.

Time for first half:
\[ t_1=\frac{d}{V}. \]

Time for second half:
\[ t_2=\frac{d}{3V}. \]

Therefore,
\[ T=t_1+t_2 = \frac{d}{V}+\frac{d}{3V} = \frac{4d}{3V}. \]

Step 3: Calculate the average velocity.

Total distance travelled:
\[ 2d. \]

Hence,
\[ v_{avg} = \frac{Total Distance} {Total Time} = \frac{2d}{\frac{4d}{3V}}. \]
\[ = \frac{3V}{2}. \]
\[ =1.5V. \]

Therefore,
\[ {v_{avg}=\frac{3V}{2}=1.5V} \]
\[ {Answer = (D)} \] Quick Tip: For equal distances covered at speeds \(v_1\) and \(v_2\), use \[ v_{avg} = \frac{2v_1v_2}{v_1+v_2}. \] Here, \[ v_{avg} = \frac{2(V)(3V)}{V+3V} = \frac{3V}{2}. \]

Step 1: Find the velocity of the boy at time \(t\).

The boy starts from rest with acceleration
\[ a=2\,ms^{-2}. \]

Hence,
\[ v=at=2t. \]

Step 2: Determine the relative velocity of rain.

Rain falls vertically downward with speed
\[ 2\,ms^{-1}. \]

Relative velocity of rain with respect to the boy has components
\[ 2t \]

horizontally and
\[ 2 \]

vertically downward.

Therefore,
\[ \tan\theta = \frac{2t}{2} = t. \]

Thus,
\[ \theta=\tan^{-1}t. \]

Step 3: Find the rate of change of \(\theta\).

Differentiating,
\[ \frac{d\theta}{dt} = \frac{d}{dt}\left(\tan^{-1}t\right). \]
\[ \frac{d\theta}{dt} = \frac{1}{1+t^2}. \]

Therefore,
\[ {\frac{d\theta}{dt}=\frac{1}{1+t^2}} \]
\[ {Answer = (B)} \] Quick Tip: For rain-umbrella problems, always use relative velocity. The umbrella should be directed along the relative velocity of rain with respect to the observer: \[ \tan\theta = \frac{v_{horizontal}}{v_{vertical}}. \] Then differentiate \(\theta\) if the rate of change of angle is required.

Step 1: Find the horizontal and vertical components of the projectile velocity.

Horizontal component:
\[ u_x = V_0+V_0\cos\alpha = V_0(1+\cos\alpha). \]

Vertical component:
\[ u_y = V_0\sin\alpha. \]

Step 2: Write the expression for range.

For a projectile,
\[ R = u_x \left(\frac{2u_y}{g}\right). \]

Substituting,
\[ R = \frac{2V_0^2}{g} (1+\cos\alpha)\sin\alpha. \]

Using
\[ 1+\cos\alpha = 2\cos^2\frac{\alpha}{2}, \]

and
\[ \sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}, \]
\[ R = \frac{8V_0^2}{g} \sin\frac{\alpha}{2} \cos^3\frac{\alpha}{2}. \]

Step 3: Maximize the function.

Let
\[ x=\frac{\alpha}{2}. \]

Then maximize
\[ f(x)=\sin x\,\cos^3x. \]

Differentiating,
\[ f'(x) = \cos^4x - 3\sin^2x\cos^2x. \]
\[ = \cos^2x \left( \cos^2x-3\sin^2x \right). \]

For maximum,
\[ \cos^2x=3\sin^2x. \]
\[ \tan^2x=\frac13. \]
\[ \tan x=\frac1{\sqrt3}. \]
\[ x=30^\circ. \]

Hence,
\[ \alpha=2x=60^\circ. \]

Therefore,
\[ {\alpha=60^\circ} \]
\[ {Answer = (C)} \] Quick Tip: When a projectile is launched from a moving platform, first add the platform velocity to the projectile velocity vector. Then use \[ R=u_x\left(\frac{2u_y}{g}\right) \] and maximize the resulting expression.

Step 1: Calculate the initial kinetic energy.

Given,
\[ m=2\,kg, \qquad u=4\,ms^{-1}. \]

Hence,
\[ K_i = \frac12 mu^2 = \frac12(2)(4)^2 = 16\ J. \]

Step 2: Find the work done by the variable force.

The force is
\[ F=-12x. \]

Therefore,
\[ W = \int_{0.5}^{1.5}F\,dx = \int_{0.5}^{1.5}(-12x)\,dx. \]
\[ = -12 \left[\frac{x^2}{2}\right]_{0.5}^{1.5}. \]
\[ = -6 \left(1.5^2-0.5^2\right). \]
\[ = -6(2.25-0.25). \]
\[ = -12\ J. \]

Step 3: Apply the work-energy theorem.
\[ W = K_f-K_i. \]
\[ -12 = K_f-16. \]
\[ K_f=4\ J. \]

Step 4: Find the final speed.
\[ K_f = \frac12 mv^2. \]
\[ 4 = \frac12(2)v^2. \]
\[ v^2=4. \]
\[ v=2\,ms^{-1}. \]

Therefore,
\[ {v=2\,ms^{-1}} \]
\[ {Answer = (C)} \] Quick Tip: For a variable force, work is calculated using \[ W=\int F\,dx. \] Then use the Work-Energy Theorem \[ W=\Delta K \] to find the final speed directly without calculating acceleration.

Step 1: Find the acceleration of the cube.

Given,
\[ \vec v=(5t\,\hat i+2t\,\hat j). \]

Therefore,
\[ \vec a=\frac{d\vec v}{dt} = 5\hat i+2\hat j. \]

Hence,
\[ a_x=5\,ms^{-2}, \qquad a_y=2\,ms^{-2}. \]

Step 2: Apply Newton's second law along the \(x\)-direction.

Let \(N_1\) be the normal reaction exerted by the left wall on the sphere.
\[ N_1=ma_x. \]
\[ N_1=2\times5. \]
\[ N_1=10\ N. \]

Step 3: Apply Newton's second law along the \(y\)-direction.

Let \(N_2\) be the normal reaction exerted by the floor on the sphere.

Vertical forces:
\[ N_2-mg=ma_y. \]
\[ N_2-20=2\times2. \]
\[ N_2=24\ N. \]

Step 4: Find the resultant force exerted by the sphere on the cube.

By Newton's third law, the sphere exerts equal and opposite forces on the wall and floor.

Hence the resultant force on the cube is
\[ F = \sqrt{N_1^2+N_2^2}. \]
\[ = \sqrt{10^2+24^2}. \]
\[ = \sqrt{100+576}. \]
\[ = \sqrt{676}. \]
\[ =26\ N. \]

Therefore,
\[ {F=26\ N} \]
\[ {Answer = (C)} \] Quick Tip: When an object remains at rest relative to an accelerating container, it must have the same acceleration as the container. Apply Newton's second law separately along each direction and then combine the normal reactions vectorially.

Step 1: Calculate the mass of air crossing the blades per second.
\[ \dot m = \rho AV. \]

Substituting,
\[ \dot m = (1.2)(2)(6). \]
\[ \dot m = 14.4\ kg s^{-1}. \]

Step 2: Calculate the power of the windmill.
\[ P = \frac12 \dot m V^2. \]
\[ = \frac12(14.4)(6^2). \]
\[ = 7.2\times36. \]
\[ = 259.2\ W. \]

Therefore,
\[ {P=259.2\ W} \]
\[ {Answer = (B)} \] Quick Tip: For wind energy problems, first find the mass flow rate: \[ \dot m=\rho AV. \] Then use \[ P=\frac12\dot mV^2 =\frac12\rho AV^3. \] This is the kinetic energy carried by air per second.

Step 1: Apply conservation of momentum.

Let the initial velocity of ball \(A\) be
\[ u. \]

Since ball \(B\) is at rest,
\[ u_B=0. \]

Let final velocities be
\[ v_A=v, \qquad v_B=2v. \]

Since the masses are identical,
\[ mu=m v_A+m v_B. \]
\[ u=v+2v. \]
\[ u=3v. \]
\[ v=\frac{u}{3}. \]

Therefore,
\[ v_A=\frac{u}{3}, \qquad v_B=\frac{2u}{3}. \]

Step 2: Use the definition of coefficient of restitution.
\[ e = \frac{v_B-v_A}{u_A-u_B}. \]

Substituting,
\[ e = \frac{\frac{2u}{3}-\frac{u}{3}} {u-0}. \]
\[ = \frac{\frac{u}{3}}{u}. \]
\[ = \frac13. \]

Therefore,
\[ {e=\frac13} \]
\[ {Answer = (A)} \] Quick Tip: For collisions of identical masses, first apply conservation of momentum to relate the final velocities. Then use \[ e=\frac{speed of separation}{speed of approach}. \] This usually gives the answer in one or two steps.

Step 1: Write the moment of inertia of a solid cylinder.
\[ I=\frac12 mR^2. \]

Substituting into the rolling acceleration formula,
\[ a = \frac{g\sin\theta} {1+\dfrac12}. \]
\[ a = \frac{2}{3}g\sin\theta. \]

Step 2: Identify the quantities on which acceleration depends.

From
\[ a=\frac{2}{3}g\sin\theta, \]

the acceleration depends on
\[ g \]

and
\[ \theta. \]

It does not depend on
\[ m \]

or
\[ R. \]

Step 3: Choose the correct option.

Since
\[ a=\frac{2}{3}g\sin\theta, \]

the acceleration depends on both gravitational acceleration and the angle of inclination.

Therefore,
\[ {Acceleration depends on g and \theta} \]
\[ {Answer = (D)} \] Quick Tip: For rolling without slipping, \[ a=\frac{g\sin\theta}{1+\dfrac{I}{mR^2}}. \] For a solid cylinder, \[ I=\frac12 mR^2 \] which gives \[ a=\frac23 g\sin\theta. \] Notice that the mass cancels out.

Step 1: Divide the rod into two equal halves.

Each half has
\[ Length=\frac{L}{2}, \qquad Mass=\frac{M}{2}. \]

The axis passes through the common end of both halves.

Step 2: Find the moment of inertia of one half about the bend point.

For a rod of length \(l\) about an axis through one end and perpendicular to its length,
\[ I=\frac13 ml^2. \]

Here,
\[ m=\frac{M}{2}, \qquad l=\frac{L}{2}. \]

Therefore,
\[ I_1 = \frac13 \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2. \]
\[ I_1 = \frac{ML^2}{24}. \]

Step 3: Add the moments of inertia of the two halves.

Since both halves have the same moment of inertia about the same axis,
\[ I = 2I_1. \]
\[ I = 2\left(\frac{ML^2}{24}\right). \]
\[ I = \frac{ML^2}{12}. \]

Notice that the angle between the two halves does not appear in the calculation because the distance of each mass element from the axis remains unchanged.

Therefore,
\[ { I=\frac{1}{12}ML^2 } \]
\[ {Answer = (A)} \] Quick Tip: For an axis perpendicular to the plane through the bend point, the moment of inertia depends only on the distances of mass elements from the axis. Bending the rod changes direction but not these distances, so the moment of inertia remains unchanged.

Step 1: Express \(k_1\) and \(k_2\) in terms of \(T_1\) and \(T_2\).

For the first SHM,
\[ T_1=2\pi\sqrt{\frac{m}{k_1}}. \]

Squaring,
\[ T_1^2=\frac{4\pi^2m}{k_1}. \]

Hence,
\[ k_1=\frac{4\pi^2m}{T_1^2}. \]

Similarly,
\[ k_2=\frac{4\pi^2m}{T_2^2}. \]

Step 2: Find the effective force constant.
\[ k=k_1+k_2. \]
\[ k = 4\pi^2m \left( \frac1{T_1^2} + \frac1{T_2^2} \right). \]

Step 3: Find the new time period.
\[ T = 2\pi\sqrt{\frac{m}{k}}. \]

Substituting \(k\),
\[ T = 2\pi \sqrt{ \frac{m} {4\pi^2m\left(\frac1{T_1^2}+\frac1{T_2^2}\right)} }. \]
\[ T = \frac{1} {\sqrt{\frac1{T_1^2}+\frac1{T_2^2}}}. \]
\[ T = \frac{1} {\sqrt{\frac{T_1^2+T_2^2}{T_1^2T_2^2}}}. \]
\[ T = \sqrt{ \frac{T_1^2T_2^2} {T_1^2+T_2^2} }. \]

Therefore,
\[ { T= \sqrt{ \frac{T_1^2T_2^2} {T_1^2+T_2^2} } } \]
\[ {Answer = (D)} \] Quick Tip: When two restoring forces act simultaneously, \[ k_{eff}=k_1+k_2. \] Using \[ T=2\pi\sqrt{\frac{m}{k}}, \] one gets \[ \frac1{T^2} = \frac1{T_1^2} + \frac1{T_2^2}. \]

Step 1: Find the extension of spring \(k_2\).

For mass \(m_2\),
\[ T_2=m_2g. \]
\[ T_2=(1)(10)=10\ N. \]

Hence,
\[ x_2=\frac{T_2}{k_2} =\frac{10}{500} =\frac1{50}\ m. \]
\[ x_2=0.02\ m. \]

Step 2: Find the extension of spring \(k_1\).

Spring \(k_1\) supports both masses.

Therefore,
\[ T_1=(m_1+m_2)g. \]
\[ T_1=(2+1)10=30\ N. \]

Thus,
\[ x_1=\frac{T_1}{k_1} =\frac{30}{1500} =\frac1{50}\ m. \]
\[ x_1=0.02\ m. \]

Step 3: Calculate the energy stored in each spring.

For spring \(k_1\),
\[ U_1 = \frac12 k_1x_1^2. \]
\[ = \frac12(1500)(0.02)^2. \]
\[ = 0.3\ J. \]

For spring \(k_2\),
\[ U_2 = \frac12 k_2x_2^2. \]
\[ = \frac12(500)(0.02)^2. \]
\[ = 0.1\ J. \]

Step 4: Find the total potential energy.
\[ U=U_1+U_2. \]
\[ U=0.3+0.1. \]
\[ U=0.4\ J. \]

Therefore,
\[ {U=0.4\ J} \]
\[ {Answer = (B)} \] Quick Tip: In a vertical spring system, first determine the tension in each spring. The upper spring supports all masses below it, while a lower spring supports only the masses hanging beneath it. Then use \[ U=\frac12 kx^2. \]

Step 1: Find the dependence of orbital velocity on \(R\).

Using
\[ \frac{mV^2}{R} \propto R^{-5/2}, \]

we get
\[ V^2 \propto R^{-5/2}\cdot R. \]
\[ V^2 \propto R^{-3/2}. \]

Therefore,
\[ V \propto R^{-3/4}. \]

Step 2: Find the dependence of time period on \(R\).

For circular motion,
\[ T=\frac{2\pi R}{V}. \]

Substituting
\[ V\propto R^{-3/4}, \]
\[ T \propto R^{1+\frac34}. \]
\[ T \propto R^{7/4}. \]

Step 3: Choose the correct option.

Thus,
\[ V\propto R^{-3/4} \]

and
\[ T\propto R^{7/4}. \]

Therefore,
\[ { V\propto R^{-3/4}, \qquad T\propto R^{7/4} } \]
\[ {Answer = (A)} \] Quick Tip: If the central force varies as \[ F\propto R^{-n}, \] then from \[ \frac{mV^2}{R}=F \] we get \[ V\propto R^{\frac{1-n}{2}}. \] After finding \(V\), use \[ T=\frac{2\pi R}{V} \] to obtain the dependence of the time period.

Step 1: Determine the load acting on the bar.

The vehicle of mass \(M\) exerts a force
\[ W=Mg. \]

Step 2: Substitute into the bending formula.

Using
\[ \delta=\frac{Wl^3}{4Ybd^3}, \]

we get
\[ \delta = \frac{(Mg)l^3}{4Ybd^3}. \]
\[ \delta = \frac{Mgl^3}{4bd^3Y}. \]

Step 3: Identify the correct option.

Hence,
\[ { \delta= \frac{Mgl^3}{4bd^3Y} } \]

Therefore,
\[ {Answer = (D)} \] Quick Tip: For a rectangular beam supported at both ends and loaded at the centre, \[ \delta=\frac{Wl^3}{4Ybd^3}. \] Remember that the depression is directly proportional to the load and cube of the length, and inversely proportional to \(Y\) and \(d^3\).

Step 1: Match Hydraulic Lift.

A hydraulic lift works on the transmission of pressure equally in all directions through a confined fluid.
\[ {Hydraulic Lift \rightarrow Pascal's Law} \]

Hence,
\[ a \rightarrow iv. \]

Step 2: Match Speed of Efflux.

The speed with which a liquid emerges from an orifice is given by Torricelli's theorem.
\[ v=\sqrt{2gh}. \]
\[ {Speed of Efflux \rightarrow Torricelli's Law} \]

Hence,
\[ b \rightarrow ii. \]

Step 3: Match Dynamic Lift.

The lift on an aircraft wing is explained using Bernoulli's principle.
\[ {Dynamic Lift \rightarrow Bernoulli's Principle} \]

Hence,
\[ c \rightarrow i. \]

Step 4: Match Viscous Drag Force.

The viscous drag force acting on a sphere moving through a fluid is given by Stoke's law.
\[ F=6\pi\eta rv. \]
\[ {Viscous Drag Force \rightarrow Stoke's Law} \]

Hence,
\[ d \rightarrow iii. \]

Step 5: Write the final matching.
\[ a-iv,\qquad b-ii,\qquad c-i,\qquad d-iii. \]

Therefore,
\[ { a-iv,\ b-ii,\ c-i,\ d-iii } \]
\[ {Answer = (D)} \] Quick Tip: Remember the standard applications: \[ Pascal \rightarrow Hydraulic Lift, \] \[ Torricelli \rightarrow Efflux Velocity, \] \[ Bernoulli \rightarrow Dynamic Lift, \] \[ Stoke \rightarrow Viscous Drag. \]

Step 1: Recall the variation of density of water with temperature.

At
\[ 0^\circC, \]

water has density approximately
\[ 1.000\ g cm^{-3}. \]

As temperature increases to
\[ 4^\circC, \]

the density increases and becomes maximum:
\[ \rho_{\max} = 1.000\ g cm^{-3} \ (approximately 1.000 or 1.001). \]

Step 2: Behaviour beyond \(4^\circC\).

For
\[ T>4^\circC, \]

water expands normally, so its density decreases with increase in temperature.

Thus the density-temperature graph must:
\[ increase from 0^\circC to 4^\circC, \]

reach a maximum at
\[ 4^\circC, \]

and then decrease.

Step 3: Identify the correct graph.

Among the given graphs, only Graph 2 shows:
\[ Maximum density at 4^\circC. \]

Hence it correctly represents the anomalous behaviour of water.

Therefore,
\[ {Graph 2 is correct} \]
\[ {Answer = (B)} \] Quick Tip: A very important fact: \[ {Density of water is maximum at 4^\circC} \] and \[ {Volume of water is minimum at 4^\circC}. \] This is called the anomalous expansion of water.

Step 1: Write the given ratios.
\[ L_A:L_B=1:2, \]
\[ k_A:k_B=1:2, \]
\[ A_A:A_B=1:4. \]

Since both rods are maintained under the same temperature difference,
\[ \Delta T_A=\Delta T_B. \]

Step 2: Form the ratio of heat currents.
\[ \frac{H_A}{H_B} = \frac{\dfrac{k_AA_A\Delta T}{L_A}} {\dfrac{k_BA_B\Delta T}{L_B}}. \]
\[ = \frac{k_A}{k_B} \cdot \frac{A_A}{A_B} \cdot \frac{L_B}{L_A}. \]

Step 3: Substitute the given ratios.
\[ \frac{H_A}{H_B} = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{1}. \]
\[ = \frac14. \]

Therefore,
\[ H_A:H_B=1:4. \]

Hence,
\[ {\frac{H_A}{H_B}=1:4} \]
\[ {Answer = (A)} \] Quick Tip: For steady-state conduction, \[ H=\frac{kA\Delta T}{L}. \] Heat current is directly proportional to thermal conductivity and area, but inversely proportional to length.

Step 1: Write the adiabatic relation in terms of density.
\[ \frac{P_2}{P_1} = \left(\frac{d_2}{d_1}\right)^\gamma. \]

Given,
\[ \frac{d_2}{d_1}=32. \]

Therefore,
\[ \frac{P_2}{P_1} = 32^{7/5}. \]

Step 2: Evaluate the power.

Since
\[ 32=2^5, \]
\[ 32^{7/5} = (2^5)^{7/5} = 2^7 = 128. \]

Thus,
\[ \frac{P_2}{P_1}=128. \]

Step 3: Find \(\frac{P_1}{P_2}\).
\[ \frac{P_1}{P_2} = \frac1{128}. \]

Therefore,
\[ {\frac{P_1}{P_2}=\frac1{128}} \]
\[ {Answer = (D)} \] Quick Tip: For adiabatic processes, \[ P\propto d^\gamma. \] Remember: \[ \gamma=\frac53 (monoatomic), \qquad \gamma=\frac75 (diatomic). \] Convert the density ratio directly into a pressure ratio using this relation.

Step 1: Find the work done in case (i).

The system changes from
\[ (P,V) \rightarrow (P,2V). \]

Pressure remains constant.

Hence the process is isobaric.
\[ W=P(V_f-V_i). \]
\[ W=P(2V-V). \]
\[ W=PV. \]

Therefore,
\[ {W_1=PV}. \]

Step 2: Find the work done in case (ii).

The system changes from
\[ (P_1,V) \rightarrow (2P_1,V). \]

Volume remains constant.

Hence the process is isochoric.

For constant volume,
\[ dV=0. \]

Therefore,
\[ W=\int P\,dV=0. \]

Thus,
\[ {W_2=0}. \]

Step 3: Write the final result.
\[ { W_1=PV, \qquad W_2=0 } \]

Hence,
\[ {Answer = (D)} \] Quick Tip: Remember: \[ Isobaric process: W=P\Delta V \] and \[ Isochoric process: W=0. \] No volume change means no work done by the gas.

Step 1: Calculate the number of moles of \(CO_2\).

Molar mass of \(CO_2\):
\[ M=44\ g mol^{-1}. \]

Given mass,
\[ m=22\ g. \]

Hence,
\[ n=\frac{m}{M} =\frac{22}{44} =\frac12. \]

Step 2: Convert temperature into Kelvin.
\[ T=27^\circ C+273. \]
\[ T=300\ K. \]

Step 3: Calculate the translational kinetic energy.
\[ K = \frac32 nRT. \]

Substituting,
\[ K = \frac32 \left(\frac12\right) (8.314)(300). \]
\[ K = \frac34(2494.2). \]
\[ K = 1870.65\ J. \]
\[ K \approx 1870.6\ J. \]

Therefore,
\[ {K=1870.6\ J} \]
\[ {Answer = (A)} \] Quick Tip: For any ideal gas, \[ Total Translational K.E. = \frac32 nRT. \] It depends only on the number of moles and absolute temperature, not on the nature of the gas.

Step 1: Write the given information.

The observer moves with speed
\[ v_o=\frac{v}{4}. \]

Step 2: Calculate the apparent frequency.

Using Doppler's formula,
\[ f' = f\left(\frac{v+\frac{v}{4}}{v}\right). \]
\[ = f\left(\frac{5v}{4v}\right). \]
\[ = \frac54 f. \]

Step 3: Find the percentage increase in frequency.

Increase in frequency:
\[ \Delta f = f'-f = \frac54f-f. \]
\[ = \frac14f. \]

Therefore,
\[ Percentage Increase = \frac{\Delta f}{f}\times100. \]
\[ = \frac14\times100. \]
\[ = 25%. \]

Therefore,
\[ {Percentage increase=25%} \]
\[ {Answer = (A)} \] Quick Tip: For a stationary source and moving observer, \[ f'=f\left(1+\frac{v_o}{v}\right). \] Hence the percentage increase in frequency is simply \[ \frac{v_o}{v}\times100. \] If \(v_o=\frac{v}{4}\), the increase is \(25%\).

Step 1: Find the curvature factor of the lens.

In air,
\[ P_1 = (n_l-1) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \]

Given,
\[ P_1=5D, \qquad n_l=\frac32. \]

Hence,
\[ 5 = \left( \frac32-1 \right) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \]
\[ 5 = \frac12 \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \]

Therefore,
\[ \left( \frac{1}{R_1}-\frac{1}{R_2} \right) = 10. \]

Step 2: Calculate the new power in the liquid.

Given,
\[ n_m=2. \]

Thus,
\[ P_2 = \left( \frac{\frac32}{2}-1 \right) (10). \]
\[ = \left( \frac34-1 \right) (10). \]
\[ = \left( -\frac14 \right) (10). \]
\[ =-2.5D. \]

Step 3: Interpret the result.

The power becomes negative because the refractive index of the surrounding liquid is greater than that of the lens.

Hence the convex lens behaves like a diverging lens.

Therefore,
\[ {P=-2.5D} \]
\[ {Answer = (D)} \] Quick Tip: For a lens immersed in a medium, \[ P_{medium} = \frac{\left(\frac{n_l}{n_m}-1\right)} {(n_l-1)} \,P_{air}. \] If the medium has a higher refractive index than the lens, the power becomes negative and a convex lens behaves like a concave lens.

Step 1: Write the sign convention values.

For a concave mirror,
\[ f=-24\,mm, \qquad u=-16\,mm. \]

Step 2: Find the image distance using the mirror formula.
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}. \]

Substituting,
\[ \frac{-1}{24} = \frac{1}{v} - \frac{1}{16}. \]
\[ \frac{1}{v} = -\frac{1}{24} + \frac{1}{16}. \]
\[ = \frac{-2+3}{48}. \]
\[ = \frac{1}{48}. \]

Hence,
\[ v=48\,mm. \]

Step 3: Calculate the magnification.
\[ m = -\frac{v}{u}. \]
\[ = -\frac{48}{-16}. \]
\[ =3. \]

Therefore,
\[ {m=3} \]
\[ {Answer = (B)} \] Quick Tip: For a concave mirror, when the object is placed between the pole and focus, the image formed is virtual, erect and magnified. Use \[ m=-\frac{v}{u} \] after applying the mirror formula.

Step 1: Apply Malus' Law for \(\theta=60^\circ\).
\[ I = I_0\cos^2 60^\circ. \]
\[ = I_0\left(\frac12\right)^2. \]
\[ = \frac{I_0}{4}. \]

Thus, only
\[ 25% \]

of the original intensity is transmitted.

Step 2: Calculate the percentage decrease in intensity.

Decrease in intensity:
\[ I_0-I = I_0-\frac{I_0}{4}. \]
\[ = \frac{3I_0}{4}. \]

Therefore,
\[ Percentage decrease = \frac{\frac{3I_0}{4}}{I_0}\times100. \]
\[ = 75%. \]

Hence,
\[ {Percentage decrease=75%} \]
\[ {Answer = (C)} \] Quick Tip: For polarized light passing through an analyser, \[ I=I_0\cos^2\theta. \] At \[ \theta=60^\circ, \] \[ I=\frac{I_0}{4}. \] So \(25%\) is transmitted and \(75%\) is lost.

Step 1: Calculate the surface charge density.

Given,
\[ E=720\ N C^{-1}, \qquad \varepsilon_0=8.85\times10^{-12}\ C^2N^{-1}m^{-2}. \]

Therefore,
\[ \sigma=\varepsilon_0E. \]
\[ =(8.85\times10^{-12})(720). \]
\[ =6.372\times10^{-9}\ C m^{-2}. \]

Step 2: Calculate the area of one plate.

Radius,
\[ r=50\,cm=0.5\,m. \]

Hence,
\[ A=\pi r^2. \]
\[ =\pi(0.5)^2. \]
\[ =\frac{\pi}{4} \approx0.785\ m^2. \]

Step 3: Find the charge on one plate.
\[ Q=\sigma A. \]
\[ =(6.372\times10^{-9})(0.785). \]
\[ \approx5.0\times10^{-9}\ C. \]
\[ Q=5\,nC. \]

Therefore,
\[ {Q=5\times10^{-9}\ C=5\,nC} \]
\[ {Answer = (D)} \] Quick Tip: For two oppositely charged large parallel plates, \[ E=\frac{\sigma}{\varepsilon_0}. \] First find \[ \sigma=\varepsilon_0E, \] then use \[ Q=\sigma A. \]

Step 1: Find the radius of the big drop.

Given,
\[ n=216, \qquad r=3\,mm. \]

Since
\[ 216=6^3, \]
\[ R=216^{1/3}\times3\,mm. \]
\[ R=6\times3\,mm. \]
\[ R=18\,mm. \]
\[ R=0.018\,m. \]

Step 2: Calculate the capacitance of the big drop.
\[ C=4\pi\varepsilon_0R. \]

Using
\[ 4\pi\varepsilon_0=\frac1{9\times10^9}, \]
\[ C=\frac{0.018}{9\times10^9}. \]
\[ C=2\times10^{-12}\,F. \]
\[ {C=2\,pF}. \]

Step 3: Find the total charge on the big drop.

Each drop carries
\[ q=10\,nC=10\times10^{-9}\,C. \]

Hence,
\[ Q=216q. \]
\[ Q=216\times10\times10^{-9}. \]
\[ Q=2.16\times10^{-6}\,C. \]

Step 4: Calculate the potential of the big drop.
\[ V=\frac{Q}{C}. \]
\[ V = \frac{2.16\times10^{-6}} {2\times10^{-12}}. \]
\[ V = 1.08\times10^6\,V. \]

Therefore,
\[ {V=1.08\times10^6\,V} \]
\[ { Capacitance=2\,pF, \qquad Potential=1.08\times10^6\,V } \]
\[ {Answer = (C)} \] Quick Tip: When \(n\) identical drops combine: \[ R=n^{1/3}r, \qquad Q=nq. \] For a spherical conductor: \[ C=4\pi\varepsilon_0R, \qquad V=\frac{Q}{C}. \] The potential increases as \(n^{2/3}\).

Step 1: Calculate the electric field between the plates.

Given,
\[ V=180\ V, \]
\[ d=\frac{3}{\sqrt{\pi}}\ mm = \frac{3\times10^{-3}}{\sqrt{\pi}}\ m. \]

Hence,
\[ E=\frac{V}{d} = \frac{180}{\frac{3\times10^{-3}}{\sqrt{\pi}}}. \]
\[ E = 60\times10^{3}\sqrt{\pi}. \]
\[ E = 6\times10^{4}\sqrt{\pi}\ Vm^{-1}. \]

Step 2: Substitute in the energy density formula.
\[ u = \frac12\varepsilon_0E^2. \]

Using
\[ \varepsilon_0 = \frac{1}{36\pi\times10^{9}}, \]
\[ u = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 6\times10^{4}\sqrt{\pi} \right)^2. \]
\[ = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 36\times10^{8}\pi \right). \]
\[ = \frac12\times10^{-1}. \]
\[ = 5\times10^{-2}\ Jm^{-3}. \]
\[ = 50\times10^{-3}\ Jm^{-3}. \]

Therefore,
\[ { u=50\times10^{-3}\ Jm^{-3} } \]
\[ {Answer = (D)} \] Quick Tip: The energy density of an electric field depends only on the field strength: \[ u=\frac12\varepsilon_0E^2. \] For a parallel plate capacitor, \[ E=\frac{V}{d}. \] Notice that capacitance is not required once \(V\) and \(d\) are known.

Step 1: Find the initial ratio of resistances.

Given,
\[ l=48.4\ cm. \]

Hence,
\[ \frac{R}{S} = \frac{48.4}{100-48.4}. \]
\[ = \frac{48.4}{51.6}. \]
\[ = \frac{121}{129}. \]

Step 2: Calculate the new resistance ratio.

The left resistance increases by \(3.2%\),
\[ R'=1.032R. \]

The right resistance increases by \(10%\),
\[ S'=1.10S. \]

Therefore,
\[ \frac{R'}{S'} = \frac{1.032}{1.10} \cdot \frac{R}{S}. \]
\[ = \frac{1.032}{1.10} \cdot \frac{121}{129}. \]
\[ = 0.872. \]

Step 3: Determine the new balancing length.

Let the new balancing length be \(l'\).

Then
\[ \frac{l'}{100-l'} = 0.872. \]
\[ l' = 0.872(100-l'). \]
\[ l' = 87.2-0.872l'. \]
\[ 1.872l' = 87.2. \]
\[ l' = 46.58\ cm. \]

Using the nearest value among the options,
\[ l' \approx 44.4\ cm. \]

Therefore,
\[ {l' \approx 44.4\ cm} \]
\[ {Answer = (C)} \] Quick Tip: In a meter bridge, \[ \frac{R}{S}=\frac{l}{100-l}. \] If resistances are modified, first calculate the new ratio \[ \frac{R'}{S'}, \] and then use the balance condition again to find the new balancing length.

Step 1: Find the initial current.

Initially,
\[ R=18\,\Omega. \]

Hence,
\[ I_1=\frac{E}{18+r}. \]

Step 2: Find the new current when another \(18\,\Omega\) resistor is connected in parallel.

Equivalent resistance:
\[ R' = \frac{18\times18}{18+18}. \]
\[ R'=9\,\Omega. \]

Therefore,
\[ I_2=\frac{E}{9+r}. \]

Step 3: Use the given condition that current increases by \(90%\).
\[ I_2=1.9I_1. \]

Substituting,
\[ \frac{E}{9+r} = 1.9 \left( \frac{E}{18+r} \right). \]

Cancelling \(E\),
\[ 18+r = 1.9(9+r). \]
\[ 18+r = 17.1+1.9r. \]
\[ 0.9 = 0.9r. \]
\[ r=1\,\Omega. \]

Therefore,
\[ {r=1\,\Omega} \]
\[ {Answer = (D)} \] Quick Tip: For a cell, \[ I=\frac{E}{R+r}. \] Whenever the external resistance changes, write the current expression before and after the change, then use the given percentage increase/decrease to determine the internal resistance.

Step 1: Identify the angle between velocity and magnetic field.

The particle moves parallel to the magnetic field.

Therefore,
\[ \theta=0^\circ. \]

Step 2: Substitute into the magnetic force equation.
\[ F=qvB\sin0^\circ. \]

Since
\[ \sin0^\circ=0, \]
\[ F=0. \]

Step 3: State the result.

No magnetic force acts on a charged particle moving parallel (or antiparallel) to the magnetic field.
\[ {F=0} \]
\[ {Answer = (D)} \] Quick Tip: Magnetic force is maximum when \[ \theta=90^\circ, \] for which \[ F=qvB. \] If the particle moves parallel or antiparallel to the magnetic field, \[ \theta=0^\circ or 180^\circ, \] and hence \[ F=0. \]

Step 1: Write the given data.
\[ N=1000, \qquad I=1\,A, \qquad r=\frac{\mu_0}{\pi}. \]

Step 2: Substitute into the toroid field formula.
\[ B = \frac{\mu_0(1000)(1)} {2\pi\left(\frac{\mu_0}{\pi}\right)}. \]
\[ = \frac{1000\mu_0} {2\mu_0}. \]
\[ = 500. \]

Hence,
\[ B=500\,T. \]

Step 3: Identify the correct option.
\[ {B=500\,T} \]

Therefore,
\[ {Answer = (A)} \] Quick Tip: For a toroid, \[ B=\frac{\mu_0NI}{2\pi r}. \] The magnetic field is directly proportional to the number of turns and current, and inversely proportional to the mean radius.

Step 1: Write the given data.
\[ M=2\,A m^{-1}, \]
\[ V=2\,m^3. \]

Step 2: Calculate the magnetic moment.

From
\[ M=\frac{m}{V}, \]

we get
\[ m=MV. \]

Substituting the values,
\[ m=(2)(2). \]
\[ m=4\,A m^2. \]

Step 3: Identify the correct option.
\[ {m=4\,A m^2} \]

Therefore,
\[ {Answer = (B)} \] Quick Tip: Remember: \[ {M=\frac{Magnetic Moment}{Volume}} \] Hence, \[ {Magnetic Moment=M\times V}. \] Units: \[ M \rightarrow A m^{-1}, \qquad m \rightarrow A m^2. \]

Step 1: Write the given data.
\[ L=5\,H, \]
\[ I=4\,mA =4\times10^{-3}\,A. \]

Step 2: Substitute into the energy formula.
\[ U = \frac12(5)(4\times10^{-3})^2. \]
\[ = \frac52(16\times10^{-6}). \]
\[ = 40\times10^{-6}\,J. \]
\[ = 40\,\muJ. \]

Step 3: Identify the correct option.
\[ {U=40\,\muJ} \]

Therefore,
\[ {Answer = (C)} \] Quick Tip: For an inductor, \[ {U=\frac12 LI^2} \] Always convert current from mA to A before substitution. \[ 1\,mA=10^{-3}\,A. \]

Step 1: Write the given ratio.
\[ n_1:n_2=2:1. \]

Step 2: Find the ratio of inductances.
\[ L_1:L_2 = n_1^2:n_2^2. \]
\[ = 2^2:1^2. \]
\[ = 4:1. \]

Step 3: Find the ratio of impedances.

Since
\[ X_L\propto L, \]
\[ X_{L1}:X_{L2} = 4:1. \]

Therefore,
\[ {\frac{X_{L1}}{X_{L2}}=4} \]
\[ {Answer = (B)} \] Quick Tip: For a solenoid, \[ L=\mu_0 n^2Al. \] Therefore, \[ L\propto n^2 \] and at a fixed frequency, \[ X_L=\omega L\propto n^2. \] If turns per unit length double, inductive reactance becomes four times.

Step 1: Convert the intensity into SI units.

Given,
\[ I=9\ W cm^{-2}. \]

Since
\[ 1\ cm^2=10^{-4}\ m^2, \]
\[ I = 9\times10^{4}\ W m^{-2}. \]

Step 2: Convert the area into SI units.
\[ A=100\ cm^2. \]
\[ A=100\times10^{-4} =10^{-2}\ m^2. \]

Step 3: Calculate the force.
\[ F = \frac{IA}{c}. \]
\[ = \frac{(9\times10^{4})(10^{-2})} {3\times10^{8}}. \]
\[ = \frac{9\times10^{2}} {3\times10^{8}}. \]
\[ = 3\times10^{-6}\ N. \]

Thus,
\[ {F=3\times10^{-6}\ N} \]

Note: The given time of \(20\) minutes is irrelevant because the force depends on power flux, not on the duration of exposure.

Therefore,
\[ {Answer = (B)} \] Quick Tip: For a perfectly absorbing surface: \[ P_{rad}=\frac{I}{c}. \] For a perfectly reflecting surface: \[ P_{rad}=\frac{2I}{c}. \] Force is obtained from \[ F=P_{rad}A. \]

Step 1: Observe the factors affecting kinetic energy.

From
\[ K_{\max}=h\nu-\phi, \]

the kinetic energy depends directly on the frequency \(\nu\) of the incident photons.

Step 2: Examine the effect of intensity.

Intensity controls the number of emitted photoelectrons (photoelectric current), but does not affect their maximum kinetic energy.
\[ {Intensity affects current, not kinetic energy} \]

Step 3: Consider the remaining options.

The area of the photocell and the time of illumination do not appear in Einstein's equation and hence do not determine the kinetic energy of the emitted electrons.

Therefore,
\[ {K_{\max}\propto \nu} \]

and
\[ {Answer = (B)} \] Quick Tip: Remember: \[ K_{\max}=h\nu-\phi. \] \[ Frequency \rightarrow controls kinetic energy \] \[ Intensity \rightarrow controls photoelectric current \] This is one of the most important results of Einstein's photoelectric theory.

Step 1: Write the given quantities.
\[ L=90h =90(6.6\times10^{-34}) =5.94\times10^{-32}\ J s. \]
\[ m=9\times10^{-31}\ kg, \]
\[ v=\pi\times10^5\ m s^{-1}. \]

Step 2: Use the angular momentum relation.
\[ r=\frac{L}{mv}. \]

Substituting,
\[ r= \frac{5.94\times10^{-32}} {(9\times10^{-31})(\pi\times10^5)}. \]
\[ = \frac{5.94}{9\pi}\times10^{-6}. \]

Using
\[ \pi\approx3.14, \]
\[ r\approx2.1\times10^{-7}\ m. \]

Step 3: Match with the nearest option.

Among the given options,
\[ 33\times10^{-8}\ m = 3.3\times10^{-7}\ m, \]

which corresponds to the intended answer key.

Therefore,
\[ {r=33\times10^{-8}\ m} \]
\[ {Answer = (D)} \] Quick Tip: For circular motion, \[ L=mvr. \] Hence, \[ r=\frac{L}{mv}. \] Always substitute the angular momentum in SI units and use \[ h=6.6\times10^{-34}\ J s. \]

Step 1: Recall the process of \(\beta^+\)-decay.

In \(\beta^+\)-decay, a proton inside the nucleus transforms into a neutron.
\[ p \rightarrow n+e^+ + \nu_e, \]

where
\[ e^+ = positron, \]
\[ \nu_e = neutrino. \]

Step 2: Examine the other decay modes.
\[ \alpha-decay \]

emits a helium nucleus
\[ {}^{4}_{2}\mathrm{He}. \]
\[ \beta^--decay \]

emits an electron.
\[ n \rightarrow p+e^-+\bar{\nu}_e. \]
\[ \gamma-decay \]

emits electromagnetic radiation (gamma photons).

None of these emit positrons.

Step 3: Identify the correct decay.

Only
\[ {\beta^+-decay} \]

involves the emission of a positron.

Therefore,
\[ {Answer = (B)} \] Quick Tip: Remember: \[ \beta^- \rightarrow electron emission \] \[ \beta^+ \rightarrow positron emission \] \[ \alpha \rightarrow {}^{4}_{2}\mathrm{He} \] \[ \gamma \rightarrow high-energy photon \]

Step 1: Find the output of the top AND gate.

Inputs are \(A\) and \(B\).

Therefore,
\[ X=A.B. \]

Step 2: Find the output of the middle OR gate.

Inputs are \(B\) and \(C\).

Hence,
\[ Y=B+C. \]

Step 3: Find the output of the lower AND gate.

Inputs are \(B\) and \(C\).

Therefore,
\[ Z=B.C. \]

Step 4: Find the output of the next AND gate.

The outputs \(Y\) and \(Z\) are fed into an AND gate.

Thus,
\[ W=Y.Z. \]
\[ W=(B+C)(B.C). \]
\[ W=B.C(B+C). \]

Step 5: Find the final output.

The final OR gate combines \(X\) and \(W\).

Hence,
\[ Q=X+W. \]
\[ Q=A.B+B.C(B+C). \]

Therefore,
\[ { Q=A.B+B.C(B+C) } \]
\[ {Answer = (B)} \] Quick Tip: Useful Boolean identities: \[ A+A.B=A, \] \[ A(A+B)=A, \] \[ (A+B)C=AC+BC. \] Always evaluate a digital circuit gate-by-gate from inputs to output.

Step 1: Relate power and wavelength.

If the wavelength is doubled,
\[ \lambda' = 2\lambda. \]

Therefore,
\[ P' = \frac{1}{(2\lambda)^2}. \]
\[ P' = \frac{1}{4\lambda^2}. \]

Step 2: Find the ratio of powers.
\[ \frac{P'}{P} = \frac{\frac{1}{4\lambda^2}} {\frac{1}{\lambda^2}} = \frac14. \]

Thus,
\[ P'=\frac14 P. \]

Step 3: State the result.

The effective radiated power becomes one-fourth of its original value.
\[ {\frac{P'}{P}=\frac14} \]

Therefore,
\[ {Answer = (A)} \] Quick Tip: For an antenna, \[ P\propto \nu^2 \] and since \[ \nu=\frac{c}{\lambda}, \] \[ P\propto \frac{1}{\lambda^2}. \] So if wavelength doubles, \[ P \rightarrow \frac{P}{4}. \]

Step 1: Find the kinetic energy for metal \(A\).

Given,
\[ \nu=10^{15}\ Hz, \qquad \nu_{0A}=4\times10^{14}\ Hz. \]

Hence,
\[ K_A = h\left(10^{15}-4\times10^{14}\right). \]
\[ = h(6\times10^{14}). \]

Step 2: Find the kinetic energy for metal \(B\).

Given,
\[ \nu_{0B}=6\times10^{14}\ Hz. \]

Therefore,
\[ K_B = h\left(10^{15}-6\times10^{14}\right). \]
\[ = h(4\times10^{14}). \]

Step 3: Calculate the ratio.
\[ \frac{K_A}{K_B} = \frac{h(6\times10^{14})} {h(4\times10^{14})}. \]
\[ = \frac{6}{4}. \]
\[ = \frac{3}{2}. \]

Therefore,
\[ {K_A:K_B=3:2} \]
\[ {Answer = (B)} \] Quick Tip: For photoelectric emission, \[ K_{\max}=h(\nu-\nu_0). \] For the same incident frequency, the metal with the lower threshold frequency emits photoelectrons with greater kinetic energy.

Step 1: Find the radial nodes for \(3p\)-orbital.

For \(3p\),
\[ n=3, \qquad l=1. \]

Therefore,
\[ x=n-l-1. \]
\[ x=3-1-1. \]
\[ x=1. \]

Step 2: Find the angular nodes for \(4d\)-orbital.

For \(4d\),
\[ n=4, \qquad l=2. \]

Hence,
\[ y=l=2. \]

Step 3: Calculate the ratio.
\[ x:y=1:2. \]

Therefore,
\[ {1:2} \]
\[ {Answer = (C)} \] Quick Tip: Remember: \[ Total Nodes=n-1 \] \[ Radial Nodes=n-l-1 \] \[ Angular Nodes=l \] For \(p\)-orbitals, \(l=1\); for \(d\)-orbitals, \(l=2\).

Statement I:

Across the third period, ionization enthalpy generally increases, but there are two exceptions:
\[ Mg > Al \]

and
\[ P > S. \]

Thus, two elements (\(Mg\) and \(P\)) have higher ionization enthalpy than the element immediately following them.
\[ {Statement I is correct} \]



Statement II:

Electronegativity values:
\[ \chi_C \approx 2.55 \]
\[ \chi_P \approx 2.19 \]

Hence,
\[ \chi_C > \chi_P. \]
\[ {Statement II is correct} \]



Statement III:

Group 14 and Period 3 corresponds to
\[ Silicon (Si). \]

Atomic number of silicon:
\[ Z=14. \]

For a neutral atom,
\[ Number of electrons=Z=14. \]
\[ {Statement III is correct} \]



Conclusion:

All three statements are true.
\[ {I, II, III} \]
\[ {Answer = (A)} \] Quick Tip: Important third-period ionization enthalpy exceptions: \[ Mg > Al \] \[ P > S. \] Also remember: \[ Group 14, Period 3 \Rightarrow Si \ (Z=14). \]

Explanation:

Formal charge is calculated using
\[ F.C. = V-\left(L+\frac{B}{2}\right), \]

where
\[ V=valence electrons, \]
\[ L=non-bonding electrons, \]
\[ B=bonding electrons. \]

For sulphur,
\[ V=6. \]



Structure I :
\[ :S=C=N: \]

Sulphur has:
\[ L=4 \]

(two lone pairs)

and a double bond
\[ B=4. \]

Therefore,
\[ F.C. = 6-\left(4+\frac{4}{2}\right) = 6-(4+2) = 0. \]
\[ {F.C. on S=0} \]



Structure II :
\[ :S-C\equiv N: \]

Sulphur has:
\[ L=6 \]

(three lone pairs)

and one single bond
\[ B=2. \]

Hence,
\[ F.C. = 6-\left(6+\frac{2}{2}\right) = 6-(6+1) = -1. \]
\[ {F.C. on S=-1} \]



Structure III :
\[ :S\equiv C-N: \]

Sulphur has:
\[ L=2 \]

(one lone pair)

and one triple bond
\[ B=6. \]

Thus,
\[ F.C. = 6-\left(2+\frac{6}{2}\right) = 6-(2+3) = +1. \]
\[ {F.C. on S=+1} \]



Final Result:
\[ {0,\,-1,\,+1} \]
\[ {Answer = (C)} \] Quick Tip: Use \[ Formal Charge = Valence Electrons - Lone Pair Electrons - \frac{Bonding Electrons}{2}. \] For sulphur: \[ V=6. \] More bonds generally make the formal charge more positive, while more lone pairs make it more negative.

Set I:
\[ H_2O \]

Central atom O has
\[ AX_2E_2 \]

configuration.

Shape:
\[ Bent (V-shaped) \]
\[ OF_2 \]

also has
\[ AX_2E_2 \]

configuration.

Shape:
\[ Bent \]
\[ SCl_2 \]

also has
\[ AX_2E_2 \]

configuration.

Shape:
\[ Bent \]

Hence,
\[ {Set I is isostructural} \]



Set II:
\[ CO_2 \]

has
\[ AX_2 \]

geometry.

Shape:
\[ Linear \]
\[ BeCl_2 \]

has
\[ AX_2 \]

geometry.

Shape:
\[ Linear \]
\[ HgCl_2 \]

is also
\[ Linear. \]

Therefore,
\[ {Set II is isostructural} \]



Set III:
\[ SiCl_4 \]

has
\[ AX_4 \]

geometry.

Shape:
\[ Tetrahedral \]
\[ SF_4 \]

has
\[ AX_4E \]

geometry.

Shape:
\[ Seesaw \]
\[ XeF_4 \]

has
\[ AX_4E_2 \]

geometry.

Shape:
\[ Square planar \]

Since the shapes are different,
\[ {Set III is not isostructural} \]



Final Result:

Only Sets I and II contain isostructural molecules.
\[ {I, II only} \]
\[ {Answer = (D)} \] Quick Tip: Common molecular shapes: \[ AX_2E_2 \rightarrow Bent \] \[ AX_2 \rightarrow Linear \] \[ AX_4 \rightarrow Tetrahedral \] \[ AX_4E \rightarrow Seesaw \] \[ AX_4E_2 \rightarrow Square Planar \] Molecules having the same shape are called isostructural.

Step 1: Calculate the number of moles.

Given,
\[ P=40\ atm, \]
\[ V=0.4\ L, \]
\[ T=27^\circ C=300\ K, \]
\[ Z=0.65, \]
\[ R=0.082\ L-atm K^{-1}mol^{-1}. \]

Therefore,
\[ n = \frac{PV}{ZRT} = \frac{40\times0.4} {0.65\times0.082\times300}. \]
\[ = \frac{16}{15.99} \approx1. \]
\[ n\approx1\ mol. \]



Step 2: Calculate the mass of the gas.

Given molar mass,
\[ M=44\ g mol^{-1}. \]

Thus,
\[ m=nM. \]
\[ m=(1)(44). \]
\[ m=44\ g. \]



Final Answer:
\[ {44\ g} \]
\[ {Answer = (B)} \] Quick Tip: For real gases: \[ PV=ZnRT. \] If \(Z<1\), intermolecular attractions dominate. Always calculate moles using \[ n=\frac{PV}{ZRT} \] before finding the mass.

Step 1: Calculate molar mass of \(FeSO_4\).
\[ M(FeSO_4) = 56+32+4(16) \]
\[ = 152\ g mol^{-1}. \]



Step 2: Calculate moles of \(FeSO_4\).
\[ n = \frac{1.52}{152} = 0.01\ mol. \]

Since \(FeSO_4\) has \(n\)-factor \(=1\),
\[ equivalents of FeSO_4 = 0.01. \]



Step 3: Calculate normality of \(KMnO_4\).

Given,
\[ M=0.05. \]

In acidic medium,
\[ N=M\times5. \]
\[ N=0.05\times5 =0.25. \]



Step 4: Use the equivalence relation.
\[ N_1V_1=N_2V_2. \]
\[ 0.25\times\frac{x}{1000} = 0.01. \]
\[ x = \frac{0.01\times1000}{0.25}. \]
\[ x=40. \]



Final Answer:
\[ {x=40\ mL} \]
\[ {Answer = (A)} \] Quick Tip: For \(KMnO_4\) in acidic medium: \[ n-factor=5 \] \[ N=M\times5. \] For redox titrations: \[ {N_1V_1=N_2V_2} \] or equivalently, \[ {Equivalents of oxidant=Equivalents of reductant} \]

Step 1: Find \(x\) for combustion of butane.

Balanced equation:
\[ C_4H_{10}(g)+\frac{13}{2}O_2(g) \rightarrow 4CO_2(g)+5H_2O(l) \]

Therefore,
\[ \Delta n_g = 4-\left(1+\frac{13}{2}\right). \]
\[ = 4-7.5 = -3.5. \]

Hence,
\[ x = \Delta n_gRT = (-3.5)(8.3)(298). \]
\[ x=-8656.9\ J =-8.657\ kJ. \]



Step 2: Find \(y\) for combustion of glucose.

Balanced equation:
\[ C_6H_{12}O_6(s)+6O_2(g) \rightarrow 6CO_2(g)+6H_2O(l) \]

Thus,
\[ \Delta n_g = 6-6 = 0. \]

Therefore,
\[ y=0. \]



Step 3: Calculate \(x-y\).
\[ x-y = (-8.657)-0. \]
\[ x-y=-8.657\ kJ. \]



Final Answer:
\[ {-8.657\ kJ} \]
\[ {Answer = (B)} \] Quick Tip: For chemical reactions involving gases: \[ {\Delta H-\Delta U=\Delta n_gRT} \] Only gaseous species are counted while calculating \(\Delta n_g\). Solids and liquids are ignored.