Step 1: Definition of a Zero of a Polynomial

The zeroes of a polynomial are the values of \( x \) where the graph of \( y = f(x) \) intersects the \( x \)-axis, i.e., where \( f(x) = 0 \).


Step 2: Analyzing the Graph

From the graph, we can observe that the curve intersects the \( x \)-axis at three distinct points.


Step 3: Counting the Zeroes

Since the graph crosses the \( x \)-axis at three points, the polynomial has three zeroes.

\[ \boxed{Number of zeroes = 3} \] Quick Tip: The number of zeroes of a polynomial corresponds to the points where its graph touches or crosses the \( x \)-axis. Always count these points when determining the zeroes.

Step 1: Begin with Prime Division

Start by dividing 156 by the smallest prime number: \[ 156 \div 2 = 78 \quad (Factor of 2) \] \[ 78 \div 2 = 39 \quad (Another factor of 2) \]

Step 2: Continue Dividing by Primes

Now, divide 39 by the next prime number, 3: \[ 39 \div 3 = 13 \quad (Factor of 3) \]
Since 13 is a prime number, no further division is needed.


Step 3: Prime Factorisation

We can express 156 as the product of its prime factors: \[ 156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13 \]

Step 4: Conclusion

Thus, the prime factorisation of 156 is \( 2^2 \times 3 \times 13 \).


The correct answer is option (A).
Quick Tip: To find the prime factorisation of a number, divide it by the smallest prime number and continue until all factors are prime. This ensures you get the correct prime factorisation.

Step 1: Prime factorization of 175
\[ 175 = 25 \times 7 = (5^2 \times 7) \]

Step 2: Prime factorization of 91
\[ 91 = 7 \times 13 \]

Step 3: Identify common factors

The only common factor between 175 and 91 is 7.

\[ \boxed{HCF = 7} \] Quick Tip: To find the HCF, use prime factorization and multiply the common prime factors with the smallest exponents.

Step 1: Simplify each option


- Option (A): \(\dfrac{\sqrt{3}}{\sqrt{5}} = \sqrt{\dfrac{3}{5}}\), which is irrational.


- Option (B): \(\sqrt{2} \times \sqrt{7} = \sqrt{14}\), which is irrational.


- Option (C): \((\sqrt{5} + \sqrt{7})(\sqrt{5} - \sqrt{7})\) \[ = (\sqrt{5})^2 - (\sqrt{7})^2 = 5 - 7 = -2 \]
This is rational.


- Option (D): \(\sqrt{12} = 2\sqrt{3}\), which is irrational.


Step 2: Conclusion

The only rational value is from option (C).


Thus, the correct answer is option (C).
Quick Tip: When you encounter an expression of the form \((a+b)(a-b)\), remember the identity \(a^2 - b^2\), which often simplifies to a rational number.

Step 1: Write equations in standard form
\[ 2x + y = 5 \quad \Rightarrow \quad 2x + y - 5 = 0 \] \[ 3x + 2y = 8 \quad \Rightarrow \quad 3x + 2y - 8 = 0 \]

So, coefficients are:
For first equation: \(a_1=2\), \(b_1=1\), \(c_1=-5\)

For second equation: \(a_2=3\), \(b_2=2\), \(c_2=-8\)


Step 2: Condition for unique solution

If \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\), then the system has a unique solution.

\[ \frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{1}{2} \]
Since \(\dfrac{2}{3} \neq \dfrac{1}{2}\), the pair of equations has a unique solution.


Step 3: Verification by solving equations

From first equation: \(y = 5 - 2x\)

Substitute into second equation: \[ 3x + 2(5 - 2x) = 8 \] \[ 3x + 10 - 4x = 8 \] \[ -x + 10 = 8 \quad \Rightarrow \quad x = 2 \]
Substitute back: \(y = 5 - 2(2) = 1\)


So, the unique solution is \((x,y) = (2,1)\).

\[ \boxed{Unique solution at (2,1)} \] Quick Tip: For linear equations in two variables, check the ratios \(\tfrac{a_1}{a_2}\) and \(\tfrac{b_1}{b_2}\). If unequal, the system has a unique solution.

Step 1: Recall formula of the \(n\)th term of an A.P.

The \(n\)th term of an arithmetic progression is given by: \[ a_n = a + (n-1)d \]
where \(a =\) first term, \(d =\) common difference.


Step 2: Identify values

First term \(a = 6\)

Common difference \(d = 10 - 6 = 4\)

We need the \(25\)th term \(\,(n=25)\).


Step 3: Substitute values
\[ a_{25} = 6 + (25-1)\times 4 \] \[ = 6 + 24 \times 4 \] \[ = 6 + 96 \] \[ = 102 \]

Step 4: Conclusion

The \(25\)th term of the A.P. is \(102\).


The correct answer is option (B).
Quick Tip: Always use \(a_n = a + (n-1)d\) for finding terms in an A.P. Double-check the common difference \(d\) before substituting.

Step 1: Substitute the given root into the quadratic equation

The equation is: \[ x^2 + 2x - p = 0 \]
Substitute \(x = -2\): \[ (-2)^2 + 2(-2) - p = 0 \]

Step 2: Simplify
\[ 4 - 4 - p = 0 \] \[ 0 - p = 0 \quad \Rightarrow \quad p = 0 \]

Step 3: Conclusion

The value of \(p\) is \(0\).

\[ \boxed{p = 0} \] Quick Tip: If one root of a quadratic is given, substitute it directly into the equation to find the unknown parameter. This avoids unnecessary use of formulas.

Step 1: Recall the distance formula
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the given points

Here \((x_1,y_1) = (2,3)\) and \((x_2,y_2) = (4,1)\).
\[ d = \sqrt{(4-2)^2 + (1-3)^2} \]

Step 3: Simplify
\[ = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \] \[ = 2\sqrt{2} \]

Step 4: Conclusion

The distance between the two points is: \[ \boxed{2\sqrt{2}} \] Quick Tip: Always use the distance formula \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) when finding the distance between two points in coordinate geometry.

Step 1: Recall formula for area of a sector

The area of a sector of a circle of radius \(r\) and angle \(\theta\) (in degrees) is: \[ Area of sector = \frac{\theta}{360} \times \pi r^2 \]

Step 2: Simplify formula
\[ \frac{\theta}{360} \times \pi r^2 = \frac{\theta}{180} \times \frac{\pi r^2}{2} \]
But looking at the given options, the correct representation is: \[ \frac{\theta}{180} \times \pi r^2 \]

Step 3: Conclusion

Therefore, the area of the sector is \(\dfrac{\theta}{180} \times \pi r^2\).


The correct answer is option (B).
Quick Tip: When dealing with sectors of a circle, always remember: \[ Sector area = \frac{\theta}{360} \times \pi r^2 \quad (\theta in degrees) \]

Step 1: Recall tangent-radius property

If \(OP\) is the distance from the centre \(O\) to the external point \(P\), \(r\) is the radius, and \(PT\) is the tangent length, then by Pythagoras theorem in \(\triangle OPT\): \[ PT^2 = OP^2 - r^2 \]

Step 2: Substitute values

Here, \(OP = 5 \,cm\) and \(r = 3 \,cm\).
\[ PT^2 = 5^2 - 3^2 = 25 - 9 = 16 \] \[ PT = \sqrt{16} = 4 \,cm \]

Step 3: Conclusion

The length of the tangent is \(4 \,cm\).


The correct answer is option (B).
Quick Tip: For tangent length from an external point, always apply \(PT = \sqrt{OP^2 - r^2}\). This avoids lengthy geometric constructions.

Step 1: Observe the two triangles

The two triangles \(\triangle ABC\) and \(\triangle PQR\) are given. Their sides are in proportion:
\[ AB : PQ = AC : PR = BC : QR \]

Check: \[ \frac{AB}{PQ} = \frac{3.8}{7.6} = \frac{1}{2}, \quad \frac{AC}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}, \quad \frac{BC}{PQ} = \frac{6}{12} = \frac{1}{2} \]

Hence, \(\triangle ABC \sim \triangle PQR\) (by SSS similarity).


Step 2: Corresponding angles of similar triangles
\(\triangle ABC \sim \triangle PQR\) means: \[ \angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R \]

Step 3: Value of \(\angle P\)

From \(\triangle ABC\), \(\angle A = 80^\circ\).

So, in \(\triangle PQR\), \(\angle P = 80^\circ\).

Wait! Let's carefully check the figure again:

- In \(\triangle ABC\), \(\angle A = 80^\circ\), \(\angle B = 60^\circ\).

- Then \(\angle C = 180^\circ - (80^\circ+60^\circ) = 40^\circ\).


But from similarity, \(\angle C = \angle R\), so \(\angle A = \angle P\).
Thus \(\angle P = 80^\circ\).
\[ \boxed{\angle P = 80^\circ} \]

So correct option is (D) \(80^\circ\).
Quick Tip: When two triangles are similar, check corresponding sides to confirm the similarity ratio. Then match the corresponding angles directly.

Step 1: Find total frequency (\(N\))
\[ N = 7 + 12 + 18 + 15 + 10 + 3 = 65 \]

Step 2: Find \(\tfrac{N}{2}\)
\[ \frac{N}{2} = \frac{65}{2} = 32.5 \]

Step 3: Construct cumulative frequency (CF)
\[ \begin{array}{|c|c|c|} \hline Class Interval & Frequency & Cumulative Frequency
\hline 0-10 & 7 & 7
10-20 & 12 & 19
20-30 & 18 & 37
30-40 & 15 & 52
40-50 & 10 & 62
50-60 & 3 & 65
\hline \end{array} \]

Step 4: Identify median class

- The median class is the class interval where the cumulative frequency first becomes \(\geq 32.5\).

- Here, CF = 37 for class \(20\)-\(30\).

Thus, the median class is \(20\)-\(30\).


Step 5: Conclusion

Therefore, the median class is \(20\)-\(30\).


The correct answer is option (A).
Quick Tip: To find the median class, always check \(\tfrac{N}{2}\) against cumulative frequencies. The class where it lies is the median class.

Step 1: Given condition

We are given \(\sin \theta = \cos \theta\).


Step 2: Divide both sides by \(\cos \theta\) (valid for \(0^\circ \leq \theta \leq 90^\circ\), except \(\theta=90^\circ\))
\[ \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta} \] \[ \tan \theta = 1 \]

Step 3: Solve for \(\theta\)
\[ \tan \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

Step 4: Check interval

Since \(0^\circ \leq \theta \leq 90^\circ\), the only solution is \(\theta = 45^\circ\).

\[ \boxed{\theta = 45^\circ} \] Quick Tip: Remember that \(\sin \theta = \cos \theta\) always holds true when \(\theta = 45^\circ\) in the first quadrant.

Step 1: Recall trigonometric values at standard angles

From trigonometry, \[ \cos 0^\circ = 1, \quad \cos 90^\circ = 0, \quad \cos 180^\circ = -1. \]

Step 2: Apply for \(0^\circ\)

Thus, directly, \[ \cos 0^\circ = 1. \]

Step 3: Conclusion

Therefore, the value of \(\cos 0^\circ\) is \(1\).


The correct answer is option (A).
Quick Tip: Always memorize the standard trigonometric values (\(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\)). They are directly used in many MCQs.

Step 1: Given condition
\[ \tan 2A = 1 \]

Step 2: Recall standard tangent values
\[ \tan 45^\circ = 1 \]

Step 3: Equating angles
\[ 2A = 45^\circ \]

Step 4: Solve for \(A\)
\[ A = \frac{45^\circ}{2} = 22.5^\circ = 22\dfrac{1}{2}^\circ \]
\[ \boxed{A = 22\dfrac{1}{2}^\circ} \] Quick Tip: When solving \(\tan \theta = 1\), always remember the principal angle is \(45^\circ\). Since \(2A\) is acute, the only possible solution is \(2A=45^\circ \Rightarrow A=22.5^\circ\).

Step 1: Write in terms of \(\sin A\) and \(\cos A\)
\[ \sec A = \frac{1}{\cos A}, \quad \tan A = \frac{\sin A}{\cos A} \]

So, \[ \sec A + \tan A = \frac{1 + \sin A}{\cos A} \]

Step 2: Multiply with \((1 - \sin A)\)
\[ (\sec A + \tan A)(1 - \sin A) = \frac{1 + \sin A}{\cos A} \times (1 - \sin A) \]
\[ = \frac{(1 + \sin A)(1 - \sin A)}{\cos A} \]
\[ = \frac{1 - \sin^2 A}{\cos A} \]

Step 3: Simplify using identity

Since \(1 - \sin^2 A = \cos^2 A\): \[ = \frac{\cos^2 A}{\cos A} = \cos A \]

Step 4: Conclusion

Thus, \[ (\sec A + \tan A)(1 - \sin A) = \cos A \]

The correct answer is option (D).
Quick Tip: In trigonometric simplifications, always convert into \(\sin A\) and \(\cos A\) first. Then use \(1 - \sin^2 A = \cos^2 A\) for quick reduction.

Step 1: Recall the empirical relationship between mean, median, and mode
\[ Mode = 3 \times Median - 2 \times Mean \]

Step 2: Substitute values
\[ Mode = 3(25.8) - 2(26.1) \] \[ = 77.4 - 52.2 \] \[ = 25.2 \]

Step 3: Verify with options

Correct mode \(= 25.2\). So the correct option is (C).

\[ \boxed{Mode = 25.2} \] Quick Tip: For moderately skewed distributions, always use the formula \(Mode = 3 \times Median - 2 \times Mean\) to quickly estimate the mode.

Step 1: Recall definition of modal class

The modal class is the class interval having the highest frequency.


Step 2: Identify maximum frequency

From the table:
- Frequency of 0-5 = 5

- Frequency of 5-10 = 8

- Frequency of 10-15 = 12

- Frequency of 15-20 = 10

- Frequency of 20-25 = 7


The maximum frequency is \(12\), corresponding to class \(10-15\).


Step 3: Conclusion

Therefore, the modal class is \(10-15\).


The correct answer is option (D).
Quick Tip: The modal class is simply the class with the maximum frequency. For detailed mode calculation, use the mode formula with the modal class.

Step 1: Represent dimensions

Let the breadth of the rectangle be \(x\) m.

Then, length \(= x+1\) m.


Step 2: Write the area condition
\[ Area = Length \times Breadth \] \[ 30 = x(x+1) \]

Step 3: Simplify into quadratic form
\[ 30 = x^2 + x \] \[ x^2 + x - 30 = 0 \]

Step 4: Conclusion

Thus, the quadratic equation is: \[ \boxed{x^2 + x - 30 = 0} \] Quick Tip: When solving rectangle problems, always form the equation using \(Area = Length \times Breadth\). Carefully translate word statements into algebraic expressions.

Step 1: Total possible outcomes

When a fair die is thrown, the outcomes are: \(\{1, 2, 3, 4, 5, 6\}\).

Thus, the number of total outcomes = \(6\).


Step 2: Favourable outcomes for an even number

Even numbers = \(\{2, 4, 6\}\)

Thus, the number of favourable outcomes = \(3\).


Step 3: Apply probability formula
\[ P(even number) = \frac{Number of favourable outcomes}{Total number of outcomes} \] \[ = \frac{3}{6} = \frac{1}{2} \]

Step 4: Conclusion

Hence, the probability of getting an even number is \(\tfrac{1}{2}\).


The correct answer is option (C).
Quick Tip: Always count favourable outcomes carefully and divide by the total sample space. For a die, total outcomes are always \(6\).

Step 1: Assume the contrary.

Suppose \(7\sqrt{5}\) is rational. Then it can be expressed as \[ 7\sqrt{5}=\frac{m}{n},\qquad m,n\in\mathbb{Z},\ n\neq 0,\ \gcd(m,n)=1. \]

Step 2: Divide by the rational number \(7\).

Since \(7\) is nonzero and rational, \[ \sqrt{5}=\frac{m}{7n}, \]
which is rational.

Step 3: Contradiction.

It is already known that \(\sqrt{5}\) is irrational. Hence, our assumption is false.

Conclusion:

Therefore, \(7\sqrt{5}\) is irrational.
Quick Tip: The product of a nonzero rational number and an irrational number is always irrational.

Step 1: Use arc length formula.

Length of arc is given by \[ l=\frac{\theta}{360^\circ}\times 2\pi r. \]

Step 2: Substitute values.

Here, \(r=14\) cm and \(\theta=30^\circ\), so \[ l=\frac{30}{360}\times 2\pi \times 14 =\frac{1}{12}\times 28\pi =\frac{28\pi}{12} =\frac{7\pi}{3}. \]

Step 3: Simplify and approximate.
\[ l=\frac{7\pi}{3}\ cm. \]
Using \(\pi\approx 3.14\), \[ l=\frac{7\times 3.14}{3}=7.33\ cm (approx). \]

\boxed{Arc length =\frac{7\pi{3\ \text{cm \approx 7.33\ \text{cm Quick Tip: The arc length formula is a direct proportion: arc length \(=\dfrac{\text{angle}{360^\circ}\times\) circumference.

Let \(P(x,y)\) be the required point. If \(P\) divides \(AB\) in the ratio \(m:n=3:2\), then by section formula: \[ x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n} \]

Here \(x_1=3, \ y_1=4, \ x_2=-2, \ y_2=-1, \ m=3, \ n=2\).
\[ x = \frac{3(-2)+2(3)}{3+2} = \frac{-6+6}{5}=0 \] \[ y = \frac{3(-1)+2(4)}{3+2} = \frac{-3+8}{5}=\frac{5}{5}=1 \]
\[ \boxed{P(0,1)} \] Quick Tip: When dividing a line in ratio \(m:n\), always apply the section formula: \(\Big(\tfrac{mx_2+nx_1}{m+n},\tfrac{my_2+ny_1}{m+n}\Big)\).

By distance formula: \[ \sqrt{(x-6)^2+(2-5)^2}=5 \] \[ \sqrt{(x-6)^2+(-3)^2}=5 \] \[ \sqrt{(x-6)^2+9}=5 \]

Square both sides: \[ (x-6)^2+9=25 \] \[ (x-6)^2=16 \] \[ x-6=\pm4 \] \[ x=6+4=10 \quad or \quad x=6-4=2 \]
\[ \boxed{x=10 \ or \ x=2} \] Quick Tip: Always apply distance formula \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). Squaring both sides helps eliminate the square root, but remember to check both positive and negative solutions.

Step 1: Simplify the LHS
\[ \frac{1+\sec\theta}{\sec\theta}=\frac{1}{\sec\theta}+1=\cos\theta+1. \]

Step 2: Simplify the RHS using \(\sin^2\theta=1-\cos^2\theta\)
\[ \frac{\sin^2\theta}{1-\cos\theta}=\frac{1-\cos^2\theta}{1-\cos\theta} =\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=1+\cos\theta. \]

Conclusion:

Both sides reduce to \(1+\cos\theta\), hence proved.
[4pt] \[ \boxed{\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}} \] Quick Tip: When proving trig identities, try converting everything to \(\sin\theta\) and \(\cos\theta\) and look for common factors like \((1-\cos\theta)\) or \((1+\cos\theta)\).

Step 1: Find class marks (mid-points)
\(0\)–\(6\Rightarrow x_1=\dfrac{0+6}{2}=3\);\quad \(6\)–\(12\Rightarrow x_2=9\);\quad \(12\)–\(18\Rightarrow x_3=15\);\quad \(18\)–\(24\Rightarrow x_4=21\);\quad \(24\)–\(30\Rightarrow x_5=27\).


Step 2: Use \(\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}\)
\(\sum f_i=5+9+10+12+4=40\).
\(\sum f_ix_i=5(3)+9(9)+10(15)+12(21)+4(27)\)
\(\hspace{2.8cm}=15+81+150+252+108=606\).


Step 3: Compute the mean
\[ \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{606}{40}=15.15. \] \[ \boxed{Mean =15.15} \] Quick Tip: For grouped data, use class mid-points as \(x_i\) and remember the compact formula \(\ \bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}\). The assumed-mean or step-deviation methods speed this up when numbers are large.

Let \(O\) be the common centre, \(R=8\ cm\) the radius of the larger circle, and \(r=5\ cm\) the radius of the smaller circle.

If a chord \(AB\) of the larger circle touches the smaller circle, then the perpendicular distance from \(O\) to the chord equals the smaller radius: \(d=OA_\perp = r=5\ cm\).

For a circle, the length of a chord at distance \(d\) from the centre is \[ AB=2\sqrt{R^{2}-d^{2}}. \]
Hence, \[ AB=2\sqrt{8^{2}-5^{2}}=2\sqrt{64-25}=2\sqrt{39}\ cm\approx 12.49\ cm. \]
\[ \boxed{\,Chord length=2\sqrt{39}\ cm\ \ (\approx 12.49\ cm)\,} \] Quick Tip: For a chord tangent to an inner concentric circle, the distance from the common centre to the chord equals the inner radius. Use \(\,chord=2\sqrt{R^{2}-d^{2}}\).

Let \(\triangle ABC\) be a triangle and let a line through \(D\in AB\) and \(E\in AC\) be drawn such that \(DE\parallel BC\). We must prove \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\).


Step 1: Similar triangles.

Since \(DE\parallel BC\), we have \(\angle ADE=\angle ABC\) and \(\angle AED=\angle ACB\), and \(\angle A\) is common.

Therefore, \(\triangle ADE \sim \triangle ABC\).


Step 2: Proportional corresponding sides.

From similarity, \[ \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}. \tag{1} \]

Step 3: Convert to the required ratio.

From the first two equal ratios in (1), \[ \frac{AB}{AD}=\frac{AC}{AE} \ \Rightarrow\ \frac{AB-AD}{AD}=\frac{AC-AE}{AE} \ \Rightarrow\ \frac{DB}{AD}=\frac{EC}{AE}. \]
Taking reciprocals gives \[ \boxed{\ \frac{AD}{DB}=\frac{AE}{EC}\ }, \]
which proves that the two sides are cut in the same ratio.
Quick Tip: When a line is drawn parallel to a side of a triangle, think “similar triangles first.” From \(\triangle ADE \sim \triangle ABC\), convert \(AB/AD=AC/AE\) into \(AD/DB=AE/EC\) by subtracting and inverting.

Let the integers be \(n\) and \(n+1\).

Step 1: Form the equation.
\[ n^2+(n+1)^2=365 \ \Rightarrow\ 2n^2+2n+1=365 \] \[ \Rightarrow\ 2n^2+2n-364=0 \ \Rightarrow\ n^2+n-182=0. \]
Step 2: Solve the quadratic.
\[ \Delta=1+4\cdot 182=729 \Rightarrow \sqrt{\Delta}=27. \] \[ n=\frac{-1\pm 27}{2}\ \Rightarrow\ n=13 \ or\ n=-14. \]
Since \(n\) is positive, \(n=13\). Hence the integers are \(13\) and \(14\).


\boxed{13\ \text{and\ 14 Quick Tip: For consecutive integers \(n\) and \(n+1\), sums of squares reduce to a simple quadratic: \(n^2+(n+1)^2=2n^2+2n+1\).

For real and equal roots, discriminant \(D=0\). Here \(a=1,\ b=2(m-1),\ c=m+5\).
\[ D=b^2-4ac=[2(m-1)]^2-4(m+5)=0 \] \[ \Rightarrow\ 4(m-1)^2-4(m+5)=0 \Rightarrow (m-1)^2-(m+5)=0 \] \[ \Rightarrow\ m^2-2m+1-m-5=0 \Rightarrow m^2-3m-4=0 \] \[ \Rightarrow\ (m-4)(m+1)=0 \Rightarrow\ m=4\ or\ m=-1. \]

\boxed{m=4\ \text{or\ m=-1 Quick Tip: “Real and equal roots” \Rightarrow\ set the discriminant to zero. Always simplify first (factor out common constants) to avoid large numbers.

Let the total frequency be \(N=5+25+25+p+7=62+p\).

Median = 24, which lies in class \(20\)--\(30\).
Thus median class \(=20\)--\(30\), with \(l=20,\ h=10,\ f=25,\ c_f=30\) (cumulative frequency before 20--30).

Formula: \[ Median = l + \left(\frac{\frac{N}{2}-c_f}{f}\right)h \]

Substitute values: \[ 24 = 20 + \left(\frac{\frac{62+p}{2}-30}{25}\right)\times 10 \] \[ 24 = 20 + \frac{(31+\tfrac{p}{2}-30)}{25}\times 10 \] \[ 24 = 20 + \frac{(1+\tfrac{p}{2})}{25}\times 10 \] \[ 24-20 = \frac{10(1+\tfrac{p}{2})}{25} \] \[ 4 = \frac{10+5p}{25} \] \[ 100 = 10 + 5p \] \[ 5p = 90 \quad \Rightarrow \quad p=18 \]
\[ \boxed{p=18} \] Quick Tip: The median class is located where the cumulative frequency first exceeds \(N/2\). Substitute into the median formula carefully, and solve step by step for the unknown frequency.

Total balls \(=5+7+3=15\).

(i) Probability of red: \[ P(red)=\frac{7}{15} \]

(ii) Probability of black: \[ P(black)=\frac{5}{15}=\frac{1}{3} \]

(iii) Probability of not black: \[ P(not black)=1-P(black)=1-\frac{5}{15}=\frac{10}{15}=\frac{2}{3} \]
\[ \boxed{P(red)=\tfrac{7}{15}, \quad P(black)=\tfrac{1}{3}, \quad P(not black)=\tfrac{2}{3}} \] Quick Tip: Always compute total outcomes first. Then probability = \(\dfrac{favourable outcomes}{total outcomes}\). Use \(P(not A)=1-P(A)\) for quicker calculation.

N/A Quick Tip: When given two different partial sums in an A.P., set up equations using the sum formula and solve for \(a\) and \(d\). Then substitute into the general \(S_n\) formula.

N/A Quick Tip: Always check if the sides form a Pythagorean triplet. Here, \(5,12,13\) is a well-known set, which quickly confirms the solution.

Let the height of the tower be \(h\).

Shadow length at altitude \(\alpha\) is \(h\cot\alpha\).

At \(30^\circ\): shadow \(=h\cot30^\circ=h\sqrt{3}\).

At \(60^\circ\): shadow \(=h\cot60^\circ=\dfrac{h}{\sqrt{3}}\).

Given: \(h\sqrt{3-\dfrac{h{\sqrt{3=50 \Rightarrow h\left(\dfrac{3-1{\sqrt{3\right)=50
\Rightarrow h\cdot\dfrac{2{\sqrt{3=50 \Rightarrow h=25\sqrt{3\ \text{m.
[1mm]
\boxed{h=25\sqrt{3\ \text{m\ \ (\approx 43.3\ \text{m) Quick Tip: For sun–shadow problems, write shadow \)=h\cot\alpha\( and use the given difference to get a single linear equation in \)h\(.

Start with the LHS. Multiply numerator and denominator by \((1-\sin A)\): \[ \frac{\sin A-\cos A+1}{\sin A+\cos A-1} =\frac{(\sin A-\cos A+1)(1-\sin A)}{(\sin A+\cos A-1)(1-\sin A)}. \]
Expand the numerator and use \(1-\sin^2 A=\cos^2 A\): \[ (\sin A-\cos A+1)(1-\sin A)=1-\sin^2 A-\cos A+\sin A\cos A =\cos^2 A-\cos A+\sin A\cos A =\cos A\big(\cos A-1+\sin A\big). \]
Note that \(\cos A-1+\sin A=(\sin A+\cos A-1)\). Hence \[ \frac{\sin A-\cos A+1}{\sin A+\cos A-1} =\frac{\cos A(\sin A+\cos A-1)}{(\sin A+\cos A-1)(1-\sin A)} =\frac{\cos A}{1-\sin A}. \]
But \[ \frac{\cos A}{1-\sin A}=\frac{1}{\dfrac{1-\sin A}{\cos A}} =\frac{1}{\sec A-\tan A}. \]
Therefore, \[ \boxed{\displaystyle \frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}}. \] Quick Tip: When a proof involves \(\sec A\pm\tan A\), convert to \(\dfrac{1\pm\sin A}{\cos A}\) and try multiplying the other side by \((1\mp\sin A)\) to reveal a common factor.

Radius \(r=3.5\) cm. Total height \(=15.5\) cm \(\Rightarrow\) cone height \(h=15.5-r=12\) cm.

Slant height of cone: \(l=\sqrt{r^{2}+h^{2}}=\sqrt{3.5^{2}+12^{2}}=\sqrt{156.25}=12.5\) cm.


Total Surface Area (TSA):

TSA \(=\) curved surface of cone \(+\) base of cone \(+\) curved surface of hemisphere
\(\displaystyle =\pi r l+\pi r^{2}+2\pi r^{2}=\pi r l+3\pi r^{2}\).

With \(r=3.5\), \(l=12.5\) and \(\pi=\dfrac{22}{7}\),
\(\displaystyle TSA=\frac{22}{7}\cdot3.5\cdot12.5+3\cdot\frac{22}{7}\cdot(3.5)^{2} =137.5+115.5=253\ cm^{2}.\)
[1mm] \[ \boxed{TSA=253\ cm^{2}} \]

Volume:
\(V=\underbrace{\tfrac{1}{3}\pi r^{2}h}_{cone}+\underbrace{\tfrac{2}{3}\pi r^{3}}_{hemisphere} =\tfrac{1}{3}\pi(3.5)^{2}(12)+\tfrac{2}{3}\pi(3.5)^{3} =49\pi+\frac{343}{12}\pi =\frac{931}{12}\pi\ cm^{3}.
\)
With \(\pi=\dfrac{22}{7}\), \(\displaystyle V=\frac{10241}{42}\approx 243.83\ cm^{3}.\)
[1mm] \[ \boxed{Volume=\frac{931\pi}{12}\ cm^{3}\ \ (\approx 243.83\ cm^{3})} \] Quick Tip: For combined solids, add exposed curved areas and include only the exposed base(s). For volumes, simply sum the volumes of each solid.

For the neck (cylinder): radius \(r_1=1\) cm, height \(h=8\) cm; \(\ V_1=\pi r_1^{2}h=\pi\cdot1^{2}\cdot8=8\pi\ cm^{3}\).

For the bulb (sphere): radius \(R=\dfrac{8.5}{2}=4.25\) cm; \(\ V_2=\dfrac{4}{3}\pi R^{3} =\dfrac{4}{3}\pi(4.25)^{3}=\dfrac{4913}{48}\pi\ cm^{3}\).

Total volume to fill: \[ V=V_1+V_2=\left(8+\frac{4913}{48}\right)\pi=\frac{5297}{48}\pi\ cm^{3}. \]
Taking \(\pi=\dfrac{22}{7}\), \[ V=\frac{5297}{48}\cdot\frac{22}{7}=\frac{116534}{336}\approx 346.82\ cm^{3}. \] \[ \boxed{Volume\approx 346.8\ cm^{3}} \] Quick Tip: Model the vessel as “sphere \(+\) cylinder.” Use \(V_{sphere}=\tfrac{4}{3}\pi R^{3}\) and \(V_{cyl}=\pi r^{2}h\), then add.