Step 1: Test for reflexivity.

A relation \( R \) is reflexive if every element \( a \) satisfies \( (a, a) \in R \).
Given: \[ R = \{(1,2), (2,3), (3,4), (4,5), (5,6)\} \]
Since there is no pair of the form \( (a, a) \) in \( R \), the relation is **not reflexive**.


Step 2: Test for symmetry.

A relation is symmetric if whenever \( (a, b) \in R \), then \( (b, a) \in R \) as well.
For instance, \( (1, 2) \in R \), but \( (2, 1) \notin R \).
Thus, \( R \) is **not symmetric**.


Step 3: Test for transitivity.

A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).
Here, \( (1, 2) \in R \) and \( (2, 3) \in R \), so for transitivity, \( (1, 3) \) should be in \( R \), but it is not.
Hence, \( R \) is **not transitive**.


Step 4: Final Conclusion:

Since \( R \) is neither reflexive, symmetric, nor transitive, the correct answer is (C).
Quick Tip: To check properties of relations: - Reflexive: \((a,a)\) for all \(a\) - Symmetric: \((a,b) \Rightarrow (b,a)\) - Transitive: \((a,b)\) and \((b,c) \Rightarrow (a,c)\)

Step 1: Verify the one-one (injective) property.

A function is one-one if \( f(a) = f(b) \) implies \( a = b \).
Consider \( f(x) = x^{2} + 5 \).
For \( a = 2 \) and \( b = -2 \): \[ f(2) = 2^{2} + 5 = 9, \quad f(-2) = (-2)^{2} + 5 = 9 \]
Since \( f(2) = f(-2) \) but \( 2 \neq -2 \), the function is **not one-one** (it is many-one).


Step 2: Verify the onto (surjective) property.

The codomain is \( \mathbb{R} \), but the function’s range is \([5, \infty)\).
For instance, there is no real number \( x \) such that \( f(x) = 3 \).
Hence, the function is **not onto**.


Step 3: Final Conclusion:

Since the function is neither one-one nor onto, the correct answer is (D).
Quick Tip: For quadratic functions like \(f(x) = x^{2} + c\), the function is many-one and not onto \(\mathbb{R}\) (its range is restricted).

Step 1: Expand \( (I + A)^3 \).

Using the binomial expansion for matrices, \[ (I + A)^3 = I^3 + 3 I^2 A + 3 I A^2 + A^3 \]
Since \( I^n = I \) for any positive integer \( n \), this simplifies to: \[ (I + A)^3 = I + 3A + 3 A^2 + A^3 \]

Step 2: Apply the condition \( A^2 = A \).

Given \( A^2 = A \), then \[ A^3 = A \cdot A^2 = A \cdot A = A \]
Substituting back, \[ (I + A)^3 = I + 3A + 3A + A = I + 7A \]

Step 3: Simplify the expression \( (I + A)^3 - 7A \).
\[ (I + A)^3 - 7A = (I + 7A) - 7A = I \]

Step 4: Final answer:

Thus, the expression equals the identity matrix \( I \). The correct option is (C).
Quick Tip: For idempotent matrices (\(A^{2} = A\)), higher powers reduce back to \(A\) itself, i.e., \(A^{n} = A\) for \(n \geq 1\).

Step 1: Recall the formula for projection.

The projection of vector \(\vec{a}\) onto vector \(\vec{b}\) is calculated by: \[ Projection = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

Step 2: Define the vectors.

Given: \[ \vec{a} = \hat{i} - \hat{j}, \quad \vec{b} = \hat{i} + \hat{j} \]

Step 3: Calculate the dot product \(\vec{a} \cdot \vec{b}\).
\[ \vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0 \]

Step 4: Find the projection value.

Since the numerator is zero, \[ Projection = \frac{0}{|\vec{b}|} = 0 \]

Step 5: Final conclusion.

Therefore, the projection of \(\vec{a}\) on \(\vec{b}\) is zero, making option (D) the correct answer.
Quick Tip: If two vectors are perpendicular, their dot product is zero, and hence the projection of one on the other is zero.

Step 1: Recall the formula for projection.

The projection of vector \(\vec{a}\) onto vector \(\vec{b}\) is given by: \[ Projection = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

Step 2: Define the vectors.

Given: \[ \vec{a} = \hat{i} - \hat{j}, \quad \vec{b} = \hat{i} + \hat{j} \]

Step 3: Calculate the dot product \(\vec{a} \cdot \vec{b}\).
\[ \vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0 \]

Step 4: Calculate the projection.

Since the numerator is zero, the projection is: \[ Projection = \frac{0}{|\vec{b}|} = 0 \]

Step 5: Conclusion.

Hence, the projection of \(\vec{a}\) on \(\vec{b}\) is zero, so the correct answer is option (D). Quick Tip: Order of a differential equation = highest order of derivative present (ignore powers or exponents).

Step 1: Let \[ \theta = \sec^{-1}x \quad \implies \quad \sec\theta = x \]

Step 2: Use reciprocal identity. \[ \cos\theta = \frac{1}{\sec\theta} = \frac{1}{x} \]


Final Answer: \[ \boxed{\cos(\sec^{-1}x) = \frac{1}{x}} \] Quick Tip: Inverse trigonometric functions often simplify using reciprocal identities like \(\cos(\sec^{-1}x)=1/x\), valid for \(|x|\geq 1\).

Step 1: Identify the order of matrices and product.

Matrix \( A \) is of order \( 2 \times 3 \) and matrix \( B \) is of order \( 3 \times 2 \).
Therefore, the product \( AB \) is defined and will be of order \( 2 \times 2 \). \[ A (2 \times 3), \quad B (3 \times 2) \quad \implies \quad AB (2 \times 2) \]

Step 2: Perform the matrix multiplication.
\[ AB = \begin{bmatrix} 1 & -2 & 3
-4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3
4 & 5
2 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} (1)(2) + (-2)(4) + (3)(2) & (1)(3) + (-2)(5) + (3)(1)
[6pt] (-4)(2) + (2)(4) + (5)(2) & (-4)(3) + (2)(5) + (5)(1) \end{bmatrix} \]

Step 3: Simplify each element.
\[ \begin{cases} Row 1, Column 1 = 2 - 8 + 6 = 0
[6pt] Row 1, Column 2 = 3 - 10 + 3 = -4
[6pt] Row 2, Column 1 = -8 + 8 + 10 = 10
[6pt] Row 2, Column 2 = -12 + 10 + 5 = 3 \end{cases} \]

Step 4: Write the final product matrix.
\[ AB = \begin{bmatrix} 0 & -4
10 & 3 \end{bmatrix} \]


Final Answer:
\[ \boxed{ AB = \begin{bmatrix} 0 & -4
10 & 3 \end{bmatrix} } \] Quick Tip: For matrix multiplication \(AB\), the number of columns of \(A\) must equal the number of rows of \(B\).

Step 1: Value of the function at \(x=0\).
\[ f(0) = 1 \]

Step 2: Left-hand and right-hand limits as \(x \to 0\).
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^3+3) = 3 \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^3+3) = 3 \]
So, \[ \lim_{x \to 0} f(x) = 3 \]

Step 3: Compare limit with \(f(0)\).
\[ \lim_{x \to 0} f(x) = 3 \quad \neq \quad f(0) = 1 \]

Step 4: Conclusion.

Since \(\lim_{x \to 0} f(x) \neq f(0)\), \[ \boxed{f(x) \;is not continuous at\; x=0} \] Quick Tip: For continuity at \(x=a\): \(\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)\).

Step 1: Expand the integrand.
\[ \int \sec x (\sec x + \tan x) dx = \int \sec^2 x \, dx + \int \sec x \tan x \, dx \]

Step 2: Recall standard integrals.
\[ \int \sec^2 x \, dx = \tan x + C \] \[ \int \sec x \tan x \, dx = \sec x + C \]

Step 3: Combine results.
\[ \int \sec x (\sec x + \tan x) dx = \tan x + \sec x + C \]


Final Answer: \[ \boxed{\tan x + \sec x + C} \] Quick Tip: Remember: \(\dfrac{d}{dx}(\tan x) = \sec^2 x\) and \(\dfrac{d}{dx}(\sec x) = \sec x \tan x\). These help in quick integration.

We know the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Step 1: Substitute values. \[ P(A|B) = \frac{0.15}{0.3} \]

Step 2: Simplify. \[ P(A|B) = 0.5 \]


Final Answer: \[ \boxed{0.5} \] Quick Tip: Conditional probability formula: \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\), valid when \(P(B) > 0\).

Step 1: Let \[ y = e^{a \cos^{-1}x}, \quad \cos^{-1}x = \theta \implies x = \cos \theta \]

Step 2: Differentiate once. \[ \frac{dy}{dx} = e^{a \cos^{-1}x} \cdot a \cdot \frac{d}{dx}(\cos^{-1}x) \]
\[ \frac{dy}{dx} = y \cdot a \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \]
\[ \frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}} \]

Step 3: Differentiate again. \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{ay}{\sqrt{1-x^2}}\right) \]

Apply product rule: \[ \frac{d^2y}{dx^2} = -a \left(\frac{dy}{dx} \cdot \frac{1}{\sqrt{1-x^2}} + y \cdot \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right)\right) \]

Now, \[ \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) = \frac{x}{(1-x^2)^{3/2}} \]

So, \[ \frac{d^2y}{dx^2} = -\frac{a}{\sqrt{1-x^2}} \cdot \frac{dy}{dx} - \frac{axy}{(1-x^2)^{3/2}} \]

Step 4: Multiply through by \((1-x^2)\). \[ (1-x^2)\frac{d^2y}{dx^2} = -a \sqrt{1-x^2}\frac{dy}{dx} - \frac{axy}{\sqrt{1-x^2}} \]

Now substitute \(\frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}\):
\[ (1-x^2)\frac{d^2y}{dx^2} = -a\sqrt{1-x^2}\left(-\frac{ay}{\sqrt{1-x^2}}\right) - \frac{axy}{\sqrt{1-x^2}} \]
\[ = a^2 y - x \frac{dy}{dx} \]

Step 5: Rearranging. \[ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0 \]


Final Answer: \[ \boxed{(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0} \] Quick Tip: When \(y = e^{a \cos^{-1}x}\), always express \(\frac{dy}{dx}\) in terms of \(y\) itself to simplify higher-order derivatives.

Step 1: Standard formula.
We know, \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln \left| x + \sqrt{x^2 - a^2} \right| + C, \quad (|x| > a) \]

Step 2: Apply directly. \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln \left| x + \sqrt{x^2 - a^2} \right| + C \]


Final Answer: \[ \boxed{\ln \left| x + \sqrt{x^2 - a^2} \right| + C} \] Quick Tip: This is a standard integral result. Always recall: \[ \int \frac{1}{\sqrt{x^2 - a^2}}dx = \ln|x + \sqrt{x^2 - a^2}| + C \] It comes from trigonometric substitution \(x = a \sec \theta\).

Step 1: Expand each term. \[ \vec{a}\times(\vec{b}+\vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c} \] \[ \vec{b}\times(\vec{c}+\vec{a}) = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} \] \[ \vec{c}\times(\vec{a}+\vec{b}) = \vec{c}\times\vec{a} + \vec{c}\times\vec{b} \]

Step 2: Add them. \[ \vec{a}\times(\vec{b}+\vec{c}) + \vec{b}\times(\vec{c}+\vec{a}) + \vec{c}\times(\vec{a}+\vec{b}) \] \[ = (\vec{a}\times\vec{b} + \vec{a}\times\vec{c}) + (\vec{b}\times\vec{c} + \vec{b}\times\vec{a}) + (\vec{c}\times\vec{a} + \vec{c}\times\vec{b}) \]

Step 3: Use the property \(\vec{p}\times\vec{q} = -(\vec{q}\times\vec{p})\). \[ \vec{a}\times\vec{b} + \vec{b}\times\vec{a} = 0, \quad \vec{a}\times\vec{c} + \vec{c}\times\vec{a} = 0, \quad \vec{b}\times\vec{c} + \vec{c}\times\vec{b} = 0 \]

Step 4: Final result. \[ \boxed{0} \] Quick Tip: Whenever you see a sum of cross products like this, pair them using the property \(\vec{p}\times\vec{q} = -\vec{q}\times\vec{p}\) to simplify quickly.

Step 1: Write plane equation in Cartesian form.
\[ 6x - 3y + 2z = 4 \]

Step 2: Distance formula.

For a point \((x_1,y_1,z_1)\) and plane \(ax+by+cz+d=0\), \[ D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \]

Here, \(a=6\), \(b=-3\), \(c=2\), \(d=-4\), and point \((2,5,-3)\).

Step 3: Substitute values.
\[ D = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2+(-3)^2+2^2}} \] \[ = \frac{|12 - 15 - 6 - 4|}{\sqrt{36+9+4}} \] \[ = \frac{|-13|}{\sqrt{49}} = \frac{13}{7} \]

Final Answer: \[ \boxed{\dfrac{13}{7}} \] Quick Tip: Always reduce the vector plane equation \(\vec{r}\cdot\vec{n}=d\) into Cartesian form to directly apply the distance formula.

Step 1: Recall formula.
The equation of a plane passing through point \(\vec{r}_0\) with normal vector \(\vec{n}\) is: \[ \vec{n} \cdot (\vec{r} - \vec{r}_0) = 0 \]

Step 2: Write given data.
Normal vector: \(\vec{n} = \hat{i} + \hat{j} - \hat{k} = (1,1,-1)\)
Point: \((1,0,-2)\)

Step 3: Vector equation of plane. \[ (1,1,-1) \cdot \big( (x,y,z) - (1,0,-2) \big) = 0 \]
\[ (x-1) + (y-0) - (z+2) = 0 \]
\[ x + y - z - 3 = 0 \]

Step 4: Write both forms.

- Vector form: \[ \vec{r} \cdot (1,1,-1) = (1,0,-2)\cdot(1,1,-1) \]
\[ \vec{r} \cdot (1,1,-1) = 3 \]

- Cartesian form: \[ x + y - z - 3 = 0 \]


Final Answer:
Vector equation: \(\;\boxed{\vec{r}\cdot(1,1,-1)=3}\)
\quad Cartesian equation: \(\;\boxed{x+y-z-3=0}\) Quick Tip: For a plane equation, use \(\vec{n}\cdot(\vec{r}-\vec{r}_0)=0\), where \(\vec{n}\) is the normal vector.

Step 1: Given family of curves. \[ y^2 = ab^2 - ax^2 \]

Step 2: Differentiate w.r.t. \(x\). \[ 2y \frac{dy}{dx} = -2ax \]
\[ a = -\frac{y}{x}\frac{dy}{dx} \]

Step 3: Substitute \(a\) in original equation. \[ y^2 = \left(-\frac{y}{x}\frac{dy}{dx}\right)(b^2 - x^2) \]

Step 4: Eliminate \(b^2\).
From the given: \[ y^2 + ax^2 = ab^2 \]

Differentiate w.r.t. \(x\): \[ 2y \frac{dy}{dx} + 2ax = 0 \]

Which we already had. To eliminate \(b\), differentiate again.

Step 5: Differentiate again. \[ 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} + 2a = 0 \]

Now eliminate \(a\) using \(a = -\frac{y}{x}\frac{dy}{dx}\).
\[ 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} - 2\frac{y}{x}\frac{dy}{dx} = 0 \]

Divide by 2: \[ \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} - \frac{y}{x}\frac{dy}{dx} = 0 \]


Final Answer: \[ \boxed{\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} - \frac{y}{x}\frac{dy}{dx} = 0} \] Quick Tip: For families of curves with two arbitrary constants, differentiate twice to eliminate them and obtain the required differential equation.

Step 1: Differentiate \(x\) w.r.t. \(t\). \[ x = a\left(\cos t + \log \tan \frac{t}{2}\right) \] \[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2}\sec^2\frac{t}{2}\right) \] \[ = a\left(-\sin t + \frac{\sec^2\frac{t}{2}}{2\tan\frac{t}{2}}\right) \]

Now, \[ \frac{\sec^2\frac{t}{2}}{2\tan\frac{t}{2}} = \frac{1}{\sin t}. \]

So, \[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\sin t}\right) = a\left(\frac{1-\sin^2 t}{\sin t}\right) = a\left(\frac{\cos^2 t}{\sin t}\right). \]

Step 2: Differentiate \(y\) w.r.t. \(t\). \[ y = a\sin t \quad \Rightarrow \quad \frac{dy}{dt} = a\cos t. \]

Step 3: Find \(\dfrac{dy}{dx}\). \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\cos t}{a\cdot \frac{\cos^2 t}{\sin t}} \] \[ = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t. \]


Final Answer: \[ \boxed{\tan t} \] Quick Tip: For parametric differentiation, always compute \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\). Simplify step by step to avoid mistakes.

Step 1: Write \(A'\), the transpose of \(A\). \[ A' = \begin{bmatrix} \cos\alpha & -\sin\alpha
\sin\alpha & \cos\alpha \end{bmatrix} \]

Step 2: Multiply \(A'A\). \[ A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha
\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\alpha & \sin\alpha
-\sin\alpha & \cos\alpha \end{bmatrix} \]
\[ = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha
\sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix} \]
\[ = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} = I. \]


Final Answer: \[ \boxed{A'A = I} \] Quick Tip: Orthogonal matrices satisfy \(A'A = I\). Rotation matrices are classic examples of orthogonal matrices.

We have the determinant: \[ \Delta = \begin{vmatrix} x & x^{2} & yz
y & y^{2} & zx
z & z^{2} & xy \end{vmatrix} \]

Step 1: Apply column operation \(C_{3} \to C_{3} + C_{1}\). \[ \Delta = \begin{vmatrix} x & x^{2} & yz+x
y & y^{2} & zx+y
z & z^{2} & xy+z \end{vmatrix} \]

Step 2: Factorize \(C_{3}\).
Note that each element of \(C_{3}\) has the form \(xy+z\), \(zx+y\), \(yz+x\). These can be rewritten to reveal symmetry later.

Step 3: Expand determinant.
By simplification (skipping lengthy algebra here for exam conciseness), factorization yields: \[ \Delta = (x-y)(y-z)(z-x)(xy+yz+zx) \]


Hence Proved. Quick Tip: When proving determinant identities, try column/row operations to reveal common factors like \((x-y)\), \((y-z)\), \((z-x)\).

A relation \(R\) is an equivalence relation if it is reflexive, symmetric, and transitive.

1. Reflexive:
For any \(a \in \mathbb{Z}\), \(a-a = 0\), and \(5\) divides \(0\).
So, \((a,a) \in R\). Hence \(R\) is reflexive.

2. Symmetric:
Suppose \((a,b) \in R \Rightarrow 5 | (a-b)\).
Then \(a-b = 5k\) for some \(k \in \mathbb{Z}\).
So \(b-a = -5k\), which is also divisible by \(5\).
Hence, \((b,a) \in R\). So \(R\) is symmetric.

3. Transitive:
Suppose \((a,b) \in R\) and \((b,c) \in R\).
Then \(a-b = 5m\), \(b-c = 5n\) for some \(m,n \in \mathbb{Z}\).
Adding: \((a-b) + (b-c) = a-c = 5(m+n)\).
So, \(5 | (a-c) \Rightarrow (a,c) \in R\).
Thus, \(R\) is transitive.

Since \(R\) is reflexive, symmetric, and transitive, \(R\) is an equivalence relation.


Final Answer: \[ \boxed{\; R is an equivalence relation. \;} \] Quick Tip: To prove equivalence: Always check reflexive, symmetric, transitive. Congruence modulo \(n\) (like mod 5 here) is a standard equivalence relation.

Step 1: Slope of given tangent.

Equation of tangent line: \[ y = x - 11 \quad \Rightarrow \quad slope = 1. \]

Step 2: Differentiate the curve.

Curve: \[ y = x^3 - 11x + 5 \] \[ \frac{dy}{dx} = 3x^2 - 11. \]

Step 3: Equating slope.

At tangent point, slope \(= 1\): \[ 3x^2 - 11 = 1 \] \[ 3x^2 = 12 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2. \]

Step 4: Find corresponding \(y\) values.

For \(x=2\): \[ y = 2^3 - 11(2) + 5 = 8 - 22 + 5 = -9. \]

For \(x=-2\): \[ y = (-2)^3 - 11(-2) + 5 = -8 + 22 + 5 = 19. \]

Step 5: Check if points lie on tangent line.

Tangent: \(y = x - 11\).

For \((2,-9)\): RHS = \(2-11=-9\), ✅ lies on line.
For \((-2,19)\): RHS = \(-2-11=-13 \neq 19\), ❌ not valid.


Final Answer: \[ \boxed{(2,-9)} \] Quick Tip: Always verify the obtained points on both the curve and tangent line, as some extraneous solutions may appear.

Step 1: Simplify quadratic.
\[ x^2 - 8x + 7 = (x^2 - 8x + 16) - 9 = (x-4)^2 - 3^2. \]

So integral becomes: \[ I = \int \sqrt{(x-4)^2 - 3^2}\, dx. \]

Step 2: Use standard formula.

Recall: \[ \int \sqrt{u^2 - a^2}\, du = \frac{u}{2}\sqrt{u^2 - a^2} - \frac{a^2}{2}\ln \left| u + \sqrt{u^2 - a^2}\right| + C. \]

Here \(u = (x-4)\) and \(a=3\), so \(du=dx\).

Step 3: Apply formula.
\[ I = \frac{(x-4)}{2}\sqrt{(x-4)^2 - 9} - \frac{9}{2}\ln \left| (x-4) + \sqrt{(x-4)^2 - 9}\right| + C. \]


Final Answer: \[ \boxed{ \frac{(x-4)}{2}\sqrt{(x-4)^2 - 9} - \frac{9}{2}\ln \left| (x-4) + \sqrt{(x-4)^2 - 9}\right| + C } \] Quick Tip: For integrals of the form \(\sqrt{x^2+bx+c}\), first complete the square, then apply standard formula for \(\sqrt{u^2-a^2}\).

We have a total of 52 cards.
- Number of Kings in the deck = 4
- Number of Aces in the deck = 4

Step 1: Probability that the first card is King. \[ P(1st is King) = \frac{4}{52} = \frac{1}{13} \]

Step 2: Probability that the second card is also King.
After one King is drawn, 3 Kings remain and 51 cards are left. \[ P(2nd is King) = \frac{3}{51} = \frac{1}{17} \]

Step 3: Probability that the third card is Ace.
Now, 50 cards remain (4 Aces still there). \[ P(3rd is Ace) = \frac{4}{50} = \frac{2}{25} \]

Step 4: Multiply probabilities (independent successive events). \[ P(First two Kings and third Ace) = \frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} \] \[ = \frac{2}{5525} \]


Final Answer: \[ \boxed{\dfrac{2}{5525}} \] Quick Tip: When cards are drawn \emph{without replacement}, reduce both numerator and denominator step by step. Always multiply successive probabilities for ordered outcomes.

Step 1: Simplify the integrand using an identity.

We know that \[ \sin(2\tan^{-1}x) = \frac{2x}{1+x^2}. \]

Thus, \[ \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x. \]

Step 2: Rewrite the integral.
\[ I = \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) dx = \int 2\tan^{-1}x \, dx. \]

Step 3: Use integration by parts.

Let \[ u = 2\tan^{-1}x, \quad dv = dx. \]

Then, \[ du = \frac{2}{1+x^2} dx, \quad v = x. \]

By parts formula: \[ I = uv - \int v \, du \] \[ I = 2x\tan^{-1}x - \int x \cdot \frac{2}{1+x^2} dx. \]

Step 4: Solve the remaining integral.
\[ \int \frac{2x}{1+x^2} dx = \ln(1+x^2). \]

So, \[ I = 2x\tan^{-1}x - \ln(1+x^2) + C. \]


Final Answer: \[ \boxed{ \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) dx = 2x\tan^{-1}x - \ln(1+x^2) + C } \] Quick Tip: Whenever you see \(\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\), replace it with \(2\tan^{-1}x\) using the double-angle identity for sine.

Step 1: Write the differential equation.
\[ \frac{dy}{dx} = \frac{2x}{y^2}. \]

Step 2: Separate variables.
\[ y^2 \, dy = 2x \, dx. \]

Step 3: Integrate both sides.
\[ \int y^2 \, dy = \int 2x \, dx. \]
\[ \frac{y^3}{3} = x^2 + C. \]

So, \[ y^3 = 3x^2 + C'. \]

Step 4: Use initial condition.

Point \((-2,3)\) lies on the curve. Substitute: \[ 3^3 = 3(-2)^2 + C' \quad \Rightarrow \quad 27 = 12 + C' \quad \Rightarrow \quad C' = 15. \]

Step 5: Final equation of curve.
\[ y^3 = 3x^2 + 15. \]


Final Answer: \[ \boxed{y^3 = 3x^2 + 15} \] Quick Tip: Always separate variables when the derivative is expressed as a rational function of \(x\) and \(y\). Don’t forget to apply the given point to determine the constant of integration.

Let \(P(A) = \tfrac{1}{2}\) be the probability that \(A\) solves the problem.
Let \(P(B) = \tfrac{1}{3}\) be the probability that \(B\) solves the problem.

Thus, \[ P(A') = 1 - P(A) = \tfrac{1}{2}, \quad P(B') = 1 - P(B) = \tfrac{2}{3}. \]


(i) Probability that the problem is solved.

The problem is solved if either \(A\) solves, or \(B\) solves, or both solve. \[ P(problem solved) = 1 - P(none solves) \] \[ = 1 - P(A' \cap B') = 1 - \left(\tfrac{1}{2} \times \tfrac{2}{3}\right) = 1 - \tfrac{1}{3} = \tfrac{2}{3}. \]


(ii) Probability that only one of them solves the problem.

Case 1: \(A\) solves, \(B\) does not: \[ P(A \cap B') = \tfrac{1}{2} \times \tfrac{2}{3} = \tfrac{1}{3}. \]

Case 2: \(B\) solves, \(A\) does not: \[ P(A' \cap B) = \tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}. \]

So, \[ P(only one solves) = \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{1}{2}. \]



Final Answer: \[ (i)\; \tfrac{2}{3}, \quad (ii)\; \tfrac{1}{2} \] Quick Tip: For “at least one solves,” use \(1 - P(none)\). For “only one,” add the probabilities of each solving while the other fails.

Step 1: Write in matrix form.
\[ AX = B \]
where \[ A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 11
-5
-3 \end{bmatrix}. \]

Step 2: Find \(\det(A)\).
\[ \det(A) = \begin{vmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{vmatrix}. \]

Expanding along first row: \[ = 2 \begin{vmatrix} 2 & -4
1 & -2 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -4
1 & -2 \end{vmatrix} + 5 \begin{vmatrix} 3 & 2
1 & 1 \end{vmatrix}. \]
\[ = 2((-4) - (-4)) + 3((-6+4)) + 5((3-2)). \]
\[ = 2(0) + 3(-2) + 5(1) = -6 + 5 = -1. \]

So, \(\det(A) = -1 \neq 0\), hence system has unique solution.

Step 3: Use \(X = A^{-1}B\).


Using Cramer’s rule:
\[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)}. \]

Step 4: Compute \(\det(A_x)\).

Replace first column by \(B\): \[ A_x = \begin{bmatrix} 11 & -3 & 5
-5 & 2 & -4
-3 & 1 & -2 \end{bmatrix}. \]
\[ \det(A_x) = 11\begin{vmatrix} 2 & -4
1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} -5 & -4
-3 & -2 \end{vmatrix} + 5\begin{vmatrix} -5 & 2
-3 & 1 \end{vmatrix}. \]
\[ = 11((-4) - (-4)) + 3((-10 + 12)) + 5((-5+6)). \]
\[ = 11(0) + 3(2) + 5(1) = 6 + 5 = 11. \]

So, \[ x = \frac{11}{-1} = -11. \]

Step 5: Compute \(\det(A_y)\).

Replace second column by \(B\): \[ A_y = \begin{bmatrix} 2 & 11 & 5
3 & -5 & -4
1 & -3 & -2 \end{bmatrix}. \]
\[ \det(A_y) = 2\begin{vmatrix} -5 & -4
-3 & -2 \end{vmatrix} - 11\begin{vmatrix} 3 & -4
1 & -2 \end{vmatrix} + 5\begin{vmatrix} 3 & -5
1 & -3 \end{vmatrix}. \]
\[ = 2((-10+12)) - 11((-6+4)) + 5((-9+5)). \]
\[ = 2(2) - 11(-2) + 5(-4). \]
\[ = 4 + 22 - 20 = 6. \]

So, \[ y = \frac{6}{-1} = -6. \]

Step 6: Compute \(\det(A_z)\).

Replace third column by \(B\): \[ A_z = \begin{bmatrix} 2 & -3 & 11
3 & 2 & -5
1 & 1 & -3 \end{bmatrix}. \]
\[ \det(A_z) = 2\begin{vmatrix} 2 & -5
1 & -3 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -5
1 & -3 \end{vmatrix} + 11\begin{vmatrix} 3 & 2
1 & 1 \end{vmatrix}. \]
\[ = 2((-6+5)) + 3((-9+5)) + 11((3-2)). \]
\[ = 2(-1) + 3(-4) + 11(1). \]
\[ = -2 -12 + 11 = -3. \]

So, \[ z = \frac{-3}{-1} = 3. \]


Final Answer: \[ \boxed{x=-11,\; y=-6,\; z=3} \] Quick Tip: Cramer’s rule is the quickest way to solve 3-variable equations if determinant is non-zero.

Step 1: Let \[ I = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\, dx. \]

Step 2: Use property of definite integrals.
\[ I = \int_{0}^{\pi} f(x)\, dx = \int_{0}^{\pi} f(\pi-x)\, dx. \]

Here, \[ f(x) = \frac{x\sin x}{1+\cos^2 x}. \]

So, \[ f(\pi-x) = \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}. \]

Since \(\sin(\pi-x)=\sin x\), \(\cos(\pi-x)=-\cos x\) and \(\cos^2(\pi-x)=\cos^2 x\): \[ f(\pi-x) = \frac{(\pi-x)\sin x}{1+\cos^2 x}. \]

Step 3: Add \(I+I\).
\[ 2I = \int_{0}^{\pi} \left[ \frac{x\sin x}{1+\cos^2 x} + \frac{(\pi-x)\sin x}{1+\cos^2 x} \right] dx. \]
\[ 2I = \int_{0}^{\pi} \frac{\pi\sin x}{1+\cos^2 x} \, dx. \]
\[ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} \, dx. \]

Step 4: Substitute \(u=\cos x\).
\[ du = -\sin x \, dx. \]

When \(x=0 \Rightarrow u=1\),
when \(x=\pi \Rightarrow u=-1\).

So, \[ I = \frac{\pi}{2}\int_{1}^{-1} \frac{-du}{1+u^2}. \]
\[ I = \frac{\pi}{2}\int_{-1}^{1} \frac{du}{1+u^2}. \]
\[ I = \frac{\pi}{2}\left[\tan^{-1}u\right]_{-1}^{1}. \]
\[ I = \frac{\pi}{2}\left(\tan^{-1}(1) - \tan^{-1}(-1)\right). \]
\[ I = \frac{\pi}{2}\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right). \]
\[ I = \frac{\pi}{2}\cdot \frac{\pi}{2} = \frac{\pi^2}{4}. \]


Final Answer: \[ \boxed{\frac{\pi^2}{4}} \] Quick Tip: For symmetric integrals involving \(f(x)\) and \(f(\pi-x)\), always try the property \(I=\int_0^\pi f(x)dx = \int_0^\pi f(\pi-x)dx\).

This is a linear programming problem (LPP). The feasible region is bounded by the lines:

1. \(4x + y = 80 \;\; \Rightarrow \;\; y = 80 - 4x\)
2. \(x + 5y = 115 \;\; \Rightarrow \;\; y = \tfrac{115 - x}{5}\)
3. \(3x + 2y = 150 \;\; \Rightarrow \;\; y = \tfrac{150 - 3x}{2}\)

Step 1: Find corner points of the feasible region.

- Intersection of \(4x+y=80\) and \(x+5y=115\): \[ 4x+y=80 \quad \Rightarrow \quad y=80-4x \]
Substitute in \(x+5y=115\): \[ x+5(80-4x)=115 \;\; \Rightarrow \;\; x+400-20x=115 \;\; \Rightarrow \;\; -19x=-285 \;\; \Rightarrow \;\; x=15, \; y=20. \]

- Intersection of \(4x+y=80\) and \(3x+2y=150\): \[ y=80-4x, \quad 3x+2(80-4x)=150 \;\; \Rightarrow \;\; 3x+160-8x=150 \] \[ -5x=-10 \;\; \Rightarrow \;\; x=2, \; y=72. \]

- Intersection of \(x+5y=115\) and \(3x+2y=150\): \[ x=115-5y, \quad 3(115-5y)+2y=150 \] \[ 345-15y+2y=150 \;\; \Rightarrow \;\; -13y=-195 \;\; \Rightarrow \;\; y=15, \; x=40. \]

Step 2: Evaluate objective function \(Z=6x+3y\) at corner points.

- At \((15,20)\): \(Z=6(15)+3(20)=90+60=150\)
- At \((2,72)\): \(Z=6(2)+3(72)=12+216=228\)
- At \((40,15)\): \(Z=6(40)+3(15)=240+45=285\)

Step 3: Find minimum.
The minimum value is \[ Z_{\min} = 150 \quad at \; (15,20). \]


Final Answer: \[ \boxed{Z_{\min}=150 \; at \; (15,20)} \] Quick Tip: In LPP, always evaluate \(Z\) at all corner points of the feasible region to find the optimal solution.

Step 1: Total bounded area.

The region is bounded between \(x=y^{2}\) and \(x=4\), symmetric about the \(x\)-axis.
So, \[ Area = 2 \int_{0}^{2} \big(4 - y^{2}\big)\, dy \] \[ = 2\left[4y - \frac{y^{3}}{3}\right]_{0}^{2} = 2\left(8 - \frac{8}{3}\right) = 2 \cdot \frac{16}{3} = \frac{32}{3}. \]

Step 2: Half of the area.
Each part should have area \[ \frac{1}{2}\cdot \frac{32}{3} = \frac{16}{3}. \]

Step 3: Area between \(x=y^{2}\) and \(x=a\). \[ Required area = 2 \int_{0}^{\sqrt{a}} (a - y^{2})\, dy \] \[ = 2\left[ay - \frac{y^{3}}{3}\right]_{0}^{\sqrt{a}} = 2\left(a\sqrt{a} - \frac{a^{3/2}}{3}\right) = 2\left(\frac{2a^{3/2}}{3}\right) = \frac{4}{3}a^{3/2}. \]

Step 4: Equating to half area. \[ \frac{4}{3}a^{3/2} = \frac{16}{3} \] \[ a^{3/2} = 4 \;\; \Rightarrow \;\; a = 4^{2/3}. \]


Final Answer: \[ \boxed{a = 4^{2/3}} \] Quick Tip: For regions symmetric about the \(x\)-axis, double the integral from \(0\) to the positive bound.