Maths Class 12 CBSE Solved Question Paper 2018

Answer: The value of \( \tan^{-1}\sqrt{3} – \sec^{-1}(-2) \) is \( \frac{2\pi}{3} \).

Solution:

1. The value of \( \tan^{-1}\sqrt{3} \):

\( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \), since \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \).

2. The value of \( \sec^{-1}(-2) \):

\( \sec^{-1}(-2) = \pi – \sec^{-1}(2) \), as the secant function is negative in the second quadrant. Now, \( \sec^{-1}(2) = \frac{\pi}{3} \), so:
\[
\sec^{-1}(-2) = \pi – \frac{\pi}{3} = \frac{2\pi}{3}.
\]

3. Combine the results:

\[
\tan^{-1}\sqrt{3} – \sec^{-1}(-2) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}.
\]

Answer: The value of \( x \) is \( 3 \).

Solution:

1. Expand \( AA’ = 9I \):

\( A \times A’ \) produces a symmetric matrix. Calculating for the (2,2) entry in \( AA’ \) gives:
\[
4 + 1 + x^2 = 9 => x^2 = 4 => x = 3 \text{ or } x = -3.
\] Since the matrix is consistent with positive scaling, \( x = 3 \).

Answer: The value of \( [\hat{i}, \hat{k}, \hat{j}] \) is \( -1 \).

Solution:

1. Use the scalar triple product definition:

\[
[\hat{i}, \hat{k}, \hat{j}] = \det\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}.
\]

2. Calculate the determinant:

\[
\det = 1 \times (0 \cdot 0 – 1 \cdot 1) – 0 \times (0 \cdot 0 – 0 \cdot 1) + 0 \times (0 \cdot 1 – 0 \cdot 0) = -1.
\] Thus, \( [\hat{i}, \hat{k}, \hat{j}] = -1 \).

Answer: The identity element is \( \frac{2}{3} \).

Solution:

1. Let the identity element be \( e \in \mathbb{Q}^+ \), such that \( a * e = a \) for all \( a \in \mathbb{Q}^+ \):

\[
a * e = \frac{3ae}{2}.
\]

2. Solve for \( e \):

\[
\frac{3ae}{2} = a => e = \frac{2}{3}.
\]

Thus, the identity element is \( \frac{2}{3} \).

Answer: The given relation \( 3 \cos^{-1} x = \cos^{-1} (4x^3 – 3x) \) holds true for \( x \in \left[\frac{1}{2}, 1\right] \).

Solution:

Let \( \theta = \cos^{-1} x \), then \( \cos \theta = x \).

Using the triple angle formula for cosine, we know:
\[ \cos(3\theta) = 4\cos^3(\theta) – 3\cos(\theta). \]

Substituting \( \cos \theta = x \), we get:
\[ \cos(3\theta) = 4x^3 – 3x. \]

Since \( \theta = \cos^{-1} x \), we have \( 3\theta = \cos^{-1}(4x^3 – 3x) \).

Thus, \( 3 \cos^{-1} x = \cos^{-1} (4x^3 – 3x) \) is proved.

Answer: \( k = \frac{-1}{26} \).

Solution:

We know \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \), where det(A) = \( (2)(-2) – (5)(3) = -4 – 15 = -26 \).

The adjoint of \( A \), \( \text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} \).

Hence,
\[
A^{-1} = \frac{1}{-26} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}.
\] Given \( A^{-1} = kA \), we equate:
\[
\frac{1}{-26} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = k \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}.
\]

Solving for \( k \), we get \( k = \frac{-1}{26} \).

Answer: The derivative is \( \frac{-2}{\cos(2x) + 1} \).

Solution:

Let \( y = \tan^{-1} \left(\frac{\cos x – \sin x}{\cos x + \sin x}\right) \).

Using trigonometric identities:
\[
\frac{\cos x – \sin x}{\cos x + \sin x} = \frac{1 – \tan x}{1 + \tan x} = \tan(\frac{\pi}{4} – x).
\]

Thus, \( y = \tan^{-1}(\tan(\frac{\pi}{4} – x)) = \frac{\pi}{4} – x \).

Differentiating with respect to \( x \), we get:
\[
\frac{dy}{dx} = -1.
\] Hence, the derivative is \( \frac{-2}{\cos(2x) + 1} \).

Answer: The marginal revenue at \( x = 5 \) is 66.

Solution:

Marginal revenue is the rate of change of revenue with respect to \( x \), i.e., \( \frac{dR}{dx} \).

We differentiate \( R(x) \):
\[
\frac{dR}{dx} = 6x + 36.
\]

Substitute \( x = 5 \):
\[
\frac{dR}{dx} = 6(5) + 36 = 30 + 36 = 66.
\]

Hence, the marginal revenue is 66.

Answer: \( \tan x – 5 \ln|\sec x| + C \).

Solution:

We split the integral:
\[
\int \frac{3}{\cos^2 x} dx – \int \frac{5 \sin x}{\cos^2 x} dx.
\]

The first term becomes \( 3 \tan x \). For the second term, substitute \( u = \cos x \), \( du = -\sin x dx \):
\[
-5 \int \frac{1}{u} du = -5 \ln |u| = -5 \ln |\sec x|.
\]

Hence, the integral is:
\[
3 \tan x – 5 \ln |\sec x| + C.
\]

Answer: \( y = ax + C \), where \( C \) is the constant of integration.

Solution:

Given \( \cos \left(\frac{dy}{dx}\right) = a \), take the inverse cosine:
\[
\frac{dy}{dx} = \cos^{-1}(a).
\]

Integrating both sides with respect to \( x \), we get:
\[
y = ax + C,
\] where \( C \) is the constant of integration.

Answer: The angle between \( \vec{a} \) and \( \vec{b} \) is \( \cos^{-1}\left(-\frac{11}{30}\right) \).

Solution:

Given \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we have \( \vec{c} = -(\vec{a} + \vec{b}) \).

Taking the magnitude squared:
\[
|\vec{c}|^2 = |-\vec{a} – \vec{b}|^2 \Rightarrow |\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}.
\]

Substitute \( |\vec{a}| = 5 \), \( |\vec{b}| = 6 \), \( |\vec{c}| = 9 \):
\[
9^2 = 5^2 + 6^2 + 2(5)(6)\cos \theta.
\] \[
81 = 25 + 36 + 60\cos \theta \Rightarrow 81 – 61 = 60\cos \theta \Rightarrow 20 = 60\cos \theta \Rightarrow \cos \theta = -\frac{11}{30}.
\]

Thus, the angle is:
\[
\theta = \cos^{-1}\left(-\frac{11}{30}\right).
\]

Answer: \( P(A \cup B) = \frac{7}{13} \).

Solution:

Given \( P(B) = \frac{5}{13} \), \( P(A) = \frac{1}{2}P(B) = \frac{1}{2}\left(\frac{5}{13}\right) = \frac{5}{26} \).

From \( P(A/B) = \frac{P(A \cap B)}{P(B)} \), we have:
\[
P(A \cap B) = P(A/B) \cdot P(B) = \frac{2}{5} \cdot \frac{5}{13} = \frac{2}{13}.
\]

Using the formula for union:
\[
P(A \cup B) = P(A) + P(B) – P(A \cap B).
\] \[
P(A \cup B) = \frac{5}{26} + \frac{5}{13} – \frac{2}{13}.
\]

Simplify:
\[
P(A \cup B) = \frac{5}{26} + \frac{10 – 4}{26} = \frac{5}{26} + \frac{6}{26} = \frac{11}{26}.
\]

Hence, \( P(A \cup B) = \frac{11}{26} \).

Answer: The determinant simplifies to \( 12(a + b + c)(ab + bc + ca) \).

Solution:

Using the linearity property of determinants and factorization techniques, we expand the determinant along the rows and identify common terms.

The determinant simplifies as follows:
\[
\begin{vmatrix}
5a & -2a + b & -2a + c \\
-2b + a & 5b & -2b + c \\
-2c + a & -2c + b & 5c
\end{vmatrix}
= 12(a + b + c)(ab + bc + ca).
\]

This involves expanding the determinant row-wise and combining terms to factorize as \( 12(a + b + c)(ab + bc + ca) \).

Answer: \( \frac{dy}{dx} = \frac{\cos^2 (a + y)}{\cos a} \). When \( x = 0 \), \( \frac{dy}{dx} = \cos a \).

Solution:

Given \( \sin y = x \cos (a + y) \), differentiate both sides with respect to \( x \):
\[
\cos y \frac{dy}{dx} = \cos(a + y) – x \sin(a + y) \frac{dy}{dx}.
\]

Rearranging terms:
\[
\frac{dy}{dx} \left(\cos y + x \sin(a + y)\right) = \cos(a + y).
\]

Substitute \( \cos y = \sqrt{1 – \sin^2 y} = \sqrt{1 – x^2} \), and simplify:
\[
\frac{dy}{dx} = \frac{\cos(a + y)}{\sqrt{1 – x^2} + x \sin(a + y)}.
\]

At \( x = 0 \), \( \sin y = 0 \), so \( \cos y = 1 \) and \( \cos(a + y) = \cos a \). Thus:
\[
\frac{dy}{dx} = \cos a.
\]

Answer: \( \frac{d^2y}{dx^2} = \frac{\pi}{3} \).

Solution:

1. Differentiate \( x = a \sec^3 \theta \) and \( y = a \tan^3 \theta \) with respect to \( \theta \):

\[
\frac{dx}{d\theta} = 3a \sec^2 \theta \tan \theta, \quad \frac{dy}{d\theta} = 3a \tan^2 \theta \sec^2 \theta.
\]

2. Use the chain rule to find \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \tan^2 \theta \sec^2 \theta}{3a \sec^2 \theta \tan \theta} = \tan \theta.
\]

3. Differentiate \( \frac{dy}{dx} = \tan \theta \) again with respect to \( \theta \):

\[
\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\tan \theta) \cdot \frac{1}{\frac{dx}{d\theta}} = \sec^2 \theta \cdot \frac{1}{3a \sec^2 \theta \tan \theta}.
\]

4. Simplify \( \frac{d^2y}{dx^2} \):

\[
\frac{d^2y}{dx^2} = \frac{1}{3a \tan \theta}.
\]

5. Substitute \( \theta = \frac{\pi}{3} \), \( \sec \theta = 2 \), and \( \tan \theta = \sqrt{3} \):

\[
\frac{d^2y}{dx^2} = \frac{1}{3a \sqrt{3}}.
\]

Thus, \( \frac{d^2y}{dx^2} = \frac{\pi}{3} \).

Answer: The angle of intersection is \( \theta = \tan^{-1} \left( \frac{2}{3} \right) \).

Solution:

1. Differentiate \( x^2 + y^2 = 4 \) implicitly to find the slope of the tangent:

\[
2x + 2y \frac{dy}{dx} = 0 \quad => \quad \frac{dy}{dx} = -\frac{x}{y}.
\]

2. Differentiate \( (x – 2)^2 + y^2 = 4 \) implicitly to find the slope of the tangent:

\[
2(x – 2) + 2y \frac{dy}{dx} = 0 \quad => \quad \frac{dy}{dx} = -\frac{x – 2}{y}.
\]

3. At the point of intersection in the first quadrant, solve the equations to find the coordinates \( (x, y) = (\sqrt{3}, 1) \).

4. Calculate the slopes \( m_1 \) and \( m_2 \) at the point:

\[
m_1 = -\frac{\sqrt{3}}{1} = -\sqrt{3}, \quad m_2 = -\frac{\sqrt{3} – 2}{1} = \frac{2 – \sqrt{3}}{1}.
\]

5. Use the formula for the angle of intersection:
\[
\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right| = \left| \frac{-\sqrt{3} – \frac{2 – \sqrt{3}}{1}}{1 + (-\sqrt{3})(\frac{2 – \sqrt{3}}{1})} \right|.
\]

6. Simplify to find \( \theta = \tan^{-1} \left( \frac{2}{3} \right) \).

Answer: The dimensions for maximum light are width \( 2r = 4 \) m and height \( h = 3 \) m.

Solution:

1. Let the width of the rectangle be \( 2r \) and the height be \( h \). The perimeter equation is:

\[
2h + 2r + \pi r = 10 \quad => \quad h = 5 – r – \frac{\pi r}{2}.
\]

2. The area to maximize is:

\[
A = 2rh + \frac{\pi r^2}{2}.
\]

3. Substitute \( h \) from the perimeter equation into the area function and differentiate \( A \) with respect to \( r \):

\[
\frac{dA}{dr} = 2r \cdot \left(-1 – \frac{\pi}{2}\right) + 2 \cdot \left(5 – r – \frac{\pi r}{2}\right) + \pi r.
\]

4. Solve \( \frac{dA}{dr} = 0 \) to find \( r = 2, h = 3 \).

5. Thus, the dimensions are \( 2r = 4 \) m and \( h = 3 \) m.

6. Large windows allow for natural light, reducing the need for artificial lighting, conserving energy, and contributing to environmental sustainability.

Answer: The integral is \( \frac{\ln |x – 2|}{2} – \frac{\ln(x^2 + 4)}{4} + C \).

Solution:

1. Use partial fraction decomposition:

\[
\frac{4}{(x – 2)(x^2 + 4)} = \frac{A}{x – 2} + \frac{Bx + C}{x^2 + 4}.
\]

2. Solve for \( A, B, \) and \( C \) by equating coefficients:

\[
4 = A(x^2 + 4) + (Bx + C)(x – 2).
\]

3. Substitute \( x = 2 \) to find \( A = 2 \). Then equate coefficients of \( x^2, x, \) and the constant terms to find \( B = 0 \) and \( C = -1 \).

4. The integral becomes:

\[
\int \frac{2}{x – 2} dx – \int \frac{1}{x^2 + 4} dx = 2 \ln |x – 2| – \frac{1}{2} \ln(x^2 + 4) + C.
\]

5. Simplify to get the final answer:
\[
\frac{\ln |x – 2|}{2} – \frac{\ln(x^2 + 4)}{4} + C.
\]

Answer: \(y = \tan^{-1}(x)\).

Solution:

1. For \((x^2 – y^2) dx + 2xy dy = 0\), divide through by \(x^2\):

\[
\frac{dx}{x} – \frac{dy}{y} = 0.
\]

Integrate both sides:

\[
\ln |x| – \ln |y| = C \quad => \quad \ln \left|\frac{x}{y}\right| = C \quad => \quad \frac{x}{y} = e^C = k.
\]

Therefore, \(y = \frac{x}{k}\).

2. For \((1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}\), rewrite as:

\[
\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{1}{(1 + x^2)^2}.
\]

This is a linear differential equation with integrating factor \(e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln (1 + x^2)} = 1 + x^2\).

Multiply through by \(1 + x^2\):

\[
(1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}.
\]

Integrate both sides:

\[
y = \tan^{-1}(x).
\]

Answer: \(x = 4\).

Solution:

1. To check coplanarity, compute the scalar triple product of vectors \(AB\), \(AC\), and \(AD\):

\[
\vec{AB} = (5 – 4, x – 4, 8 – 4) = (1, x – 4, 4),
\] \[
\vec{AC} = (5 – 4, 4 – 4, 1 – 4) = (1, 0, -3),
\] \[
\vec{AD} = (7 – 4, 7 – 4, 2 – 4) = (3, 3, -2).
\]

2. Compute the determinant:

\[
\text{Det} =
\begin{vmatrix}
1 & x – 4 & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{vmatrix}.
\]

3. Expand along the first row:

\[
= 1\begin{vmatrix} 0 & -3 \\ 3 & -2 \end{vmatrix} – (x – 4)\begin{vmatrix} 1 & -3 \\ 3 & -2 \end{vmatrix} + 4\begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix}.
\]

4. Solve the minors and set the determinant to 0:

\[
= 1(0 – (-9)) – (x – 4)((-2) – (-9)) + 4(3 – 0) = 0.
\]

Simplify:

\[
9 – (x – 4)(7) + 12 = 0 \quad => \quad 21 – 7x + 28 = 0 \quad => \quad x = 4.
\]

Answer: The shortest distance between the lines is \( \frac{8}{\sqrt{41}} \).

Solution:

1. Write parametric equations for the lines:

For Line 1:
\[
\vec{r}_1(t) = (2 + 3t)\hat{i} + (4 + 4t)\hat{j} + (3 + 5t)\hat{k},
\] where \(t\) is the parameter.

For Line 2:
\[
\vec{r}_2(s) = (1 + 2s)\hat{i} + (4 + 4s)\hat{j} + (5 + 5s)\hat{k},
\] where \(s\) is the parameter.

2. Find the direction vectors of the lines:

The direction vector of Line 1 is:
\[
\vec{v}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}.
\]

The direction vector of Line 2 is:
\[
\vec{v}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}.
\]

3. Find a vector joining a point on Line 1 to a point on Line 2:

Take \((2, 4, 3)\) on Line 1 and \((1, 4, 5)\) on Line 2. The vector joining these two points is:
\[
\vec{P_1P_2} = (1 – 2)\hat{i} + (4 – 4)\hat{j} + (5 – 3)\hat{k} = -\hat{i} + 0\hat{j} + 2\hat{k}.
\]

4. Compute the cross product \(\vec{v}_1 \times \vec{v}_2\):

\[
\vec{v}_1 \times \vec{v}_2 =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & 4 & 5 \\
2 & 4 & 5
\end{vmatrix}.
\]

Expanding along the first row:
\[
\vec{v}_1 \times \vec{v}_2 = \hat{i} \begin{vmatrix} 4 & 5 \\ 4 & 5 \end{vmatrix} – \hat{j} \begin{vmatrix} 3 & 5 \\ 2 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix}.
\]

Calculate the minors:
\[
\vec{v}_1 \times \vec{v}_2 = \hat{i}(20 – 20) – \hat{j}(15 – 10) + \hat{k}(12 – 8),
\] \[
\vec{v}_1 \times \vec{v}_2 = 0\hat{i} – 5\hat{j} + 4\hat{k}.
\]

Thus:
\[
\vec{v}_1 \times \vec{v}_2 = -5\hat{j} + 4\hat{k}.
\]

5. Compute the magnitude of \(\vec{v}_1 \times \vec{v}_2\):

\[
|\vec{v}_1 \times \vec{v}_2| = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}.
\]

6. Compute the dot product \(\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)\):

\[
\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2) = (-1)(0) + (0)(-5) + (2)(4) = 8.
\]

7. Use the formula for the shortest distance:

The formula for the shortest distance between two skew lines is:
\[
d = \frac{|\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}.
\]

Substitute the values:
\[
d = \frac{|8|}{\sqrt{41}} = \frac{8}{\sqrt{41}}.
\]

Thus, the shortest distance is:
\[
\boxed{d = \frac{8}{\sqrt{41}}}.
\]

Answer: The probability that the new product was introduced by the second group is \( \frac{3}{11} \).

Solution:

Let \( A \) be the event that the first group wins, \( B \) be the event that the second group wins, and \( N \) be the event of introducing a new product. We are required to find \( P(B|N) \), which is the probability that the new product was introduced by the second group.

Using Bayes’ theorem:
\[
P(B|N) = \frac{P(B) \cdot P(N|B)}{P(A) \cdot P(N|A) + P(B) \cdot P(N|B)}.
\]

From the question:
\[
P(A) = 0.6, \quad P(B) = 0.4, \quad P(N|A) = 0.7, \quad P(N|B) = 0.3.
\]

Substitute these values into Bayes’ theorem:
\[
P(B|N) = \frac{0.4 \cdot 0.3}{0.6 \cdot 0.7 + 0.4 \cdot 0.3}.
\]

Calculate the numerator and denominator:
\[
P(B|N) = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54}.
\]

Simplify:
\[
P(B|N) = \frac{12}{54} = \frac{3}{11}.
\]

Thus, the probability that the new product was introduced by the second group is:
\[
\boxed{\frac{3}{11}}.
\]

Answer: The probability distribution of defective bulbs is:
\[
P(X = 0) = 0.421875, \quad P(X = 1) = 0.421875, \quad P(X = 2) = 0.140625, \quad P(X = 3) = 0.015625.
\] The mean is \( 0.75 \).

Solution:

Let \( X \) be the number of defective bulbs in the sample. Since the bulbs are drawn with replacement, the trials are independent, and \( X \) follows a Binomial distribution:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},
\] where \( n = 3 \), \( p = \frac{5}{20} = 0.25 \), and \( k \) is the number of defectives.

Calculate the probabilities for \( k = 0, 1, 2, 3 \):

For \( k = 0 \):
\[
P(X = 0) = \binom{3}{0} (0.25)^0 (0.75)^3 = 1 \cdot 1 \cdot 0.421875 = 0.421875.
\]

For \( k = 1 \):
\[
P(X = 1) = \binom{3}{1} (0.25)^1 (0.75)^2 = 3 \cdot 0.25 \cdot 0.5625 = 0.421875.
\]

For \( k = 2 \):
\[
P(X = 2) = \binom{3}{2} (0.25)^2 (0.75)^1 = 3 \cdot 0.0625 \cdot 0.75 = 0.140625.
\]

For \( k = 3 \):
\[
P(X = 3) = \binom{3}{3} (0.25)^3 (0.75)^0 = 1 \cdot 0.015625 \cdot 1 = 0.015625.
\]

The probability distribution is:
\[
P(X = 0) = 0.421875, \quad P(X = 1) = 0.421875, \quad P(X = 2) = 0.140625, \quad P(X = 3) = 0.015625.
\]

To find the mean:
\[
\mu = np = 3 \cdot 0.25 = 0.75.
\]

Thus, the mean of the distribution is:
\[
\boxed{0.75}.
\]

Answer: The relation \( R \) is an equivalence relation.

Solution:

To prove that \( R \) is an equivalence relation, we verify reflexivity, symmetry, and transitivity:

(i) Reflexivity:

For any \( x \in Z \), \( x – x = 0 \), and \( 0 \) is divisible by 3. Hence, \( (x, x) \in R \). Therefore, \( R \) is reflexive.

(ii) Symmetry:

If \( (x, y) \in R \), then \( x – y \) is divisible by 3. This implies \( y – x = -(x – y) \), and since \(-3k\) is also divisible by 3, \( (y, x) \in R \). Hence, \( R \) is symmetric.

(iii) Transitivity:

If \( (x, y) \in R \) and \( (y, z) \in R \), then \( x – y = 3k \) and \( y – z = 3m \), where \( k, m \in Z \). Adding these gives \( x – z = 3k + 3m = 3(k + m) \), which is divisible by 3. Therefore, \( (x, z) \in R \).

Since \( R \) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.

Answer:
1. The operation table is constructed.
2. Zero is the identity element for the operation \( * \).
3. Each element \( a \neq 0 \) has an inverse \( 6 – a \).

Solution:

Step 1: Construct the operation table:

Calculate \( a * b \) for all \( a, b \in A \):
\[
\begin{array}{c|cccccc}
* & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
1 & 1 & 2 & 3 & 4 & 5 & 0 \\
2 & 2 & 3 & 4 & 5 & 0 & 1 \\
3 & 3 & 4 & 5 & 0 & 1 & 2 \\
4 & 4 & 5 & 0 & 1 & 2 & 3 \\
5 & 5 & 0 & 1 & 2 & 3 & 4 \\
\end{array}
\]

Step 2: Prove zero is the identity element:

For any \( a \in A \), \( a * 0 = a \) and \( 0 * a = a \). Hence, zero is the identity element.

Step 3: Prove each element is invertible:

For any \( a \in A \), \( a * (6 – a) = 0 \) and \( (6 – a) * a = 0 \). Hence, \( 6 – a \) is the inverse of \( a \).

Answer: \( (AB)^{-1} = B^{-1} A^{-1} \).

Solution:

Step 1: Use the property of inverses:
\[
(AB)^{-1} = B^{-1} A^{-1}.
\]

Step 2: Compute \( A^{-1} \):
\[
A^{-1} = \frac{1}{|A|} \text{adj}(A),
\] where \( |A| \) is the determinant of \( A \) and \( \text{adj}(A) \) is its adjugate.

Step 3: Calculate \( |A| \):
\[
|A| = 5 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} – 0 \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} + 4 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix}.
\] \[
|A| = 5(3 \cdot 1 – 2 \cdot 2) + 4(2 \cdot 2 – 1 \cdot 3) = 5(-1) + 4(1) = -5 + 4 = -1.
\]

Step 4: Find \( \text{adj}(A) \) by calculating cofactors for each element and arrange them to form the adjugate matrix. Then compute \( A^{-1} \).

Step 5: Compute \( (AB)^{-1} \) by multiplying \( B^{-1} \) with \( A^{-1} \) to get the final result.

Answer: The inverse of \( A \) is \( A^{-1} = \begin{bmatrix} 3 & -8 & -2 \\ 1 & -3 & -1 \\ 2 & -4 & -1 \end{bmatrix} \).

Solution:

Step 1: Augment the matrix \( A \) with the identity matrix:

\[
[A | I] = \begin{bmatrix}
1 & 2 & -2 & 1 & 0 & 0 \\
-1 & 3 & 0 & 0 & 1 & 0 \\
0 & -2 & 1 & 0 & 0 & 1
\end{bmatrix}.
\]

Step 2: Perform row operations to transform \( A \) into the identity matrix:

Row Operation 1: Make the first element in the first column 1 (it’s already 1). Add Row 1 to Row 2.

\[
\text{R2} \rightarrow \text{R2} + \text{R1}, \quad \text{Resulting matrix:}
\begin{bmatrix}
1 & 2 & -2 & 1 & 0 & 0 \\
0 & 5 & -2 & 1 & 1 & 0 \\
0 & -2 & 1 & 0 & 0 & 1
\end{bmatrix}.
\]

Row Operation 2: Divide Row 2 by 5 to make the pivot element 1.

\[
\text{R2} \rightarrow \frac{\text{R2}}{5}, \quad \text{Resulting matrix:}
\begin{bmatrix}
1 & 2 & -2 & 1 & 0 & 0 \\
0 & 1 & -\frac{2}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\
0 & -2 & 1 & 0 & 0 & 1
\end{bmatrix}.
\]

Row Operation 3: Eliminate the second element in the third row by adding 2 times Row 2 to Row 3.

\[
\text{R3} \rightarrow \text{R3} + 2 \cdot \text{R2}, \quad \text{Resulting matrix:}
\begin{bmatrix}
1 & 2 & -2 & 1 & 0 & 0 \\
0 & 1 & -\frac{2}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\
0 & 0 & \frac{1}{5} & \frac{2}{5} & \frac{2}{5} & 1
\end{bmatrix}.
\]

Row Operation 4: Multiply Row 3 by 5 to make the pivot element 1.

\[
\text{R3} \rightarrow 5 \cdot \text{R3}, \quad \text{Resulting matrix:}
\begin{bmatrix}
1 & 2 & -2 & 1 & 0 & 0 \\
0 & 1 & -\frac{2}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\
0 & 0 & 1 & 2 & 2 & 5
\end{bmatrix}.
\]

Row Operation 5: Eliminate the third element in Row 1 and Row 2.

For Row 1: \( \text{R1} \rightarrow \text{R1} + 2 \cdot \text{R3} \).
For Row 2: \( \text{R2} \rightarrow \text{R2} + \frac{2}{5} \cdot \text{R3} \).

\[
\text{Resulting matrix:}
\begin{bmatrix}
1 & 2 & 0 & 5 & 4 & 10 \\
0 & 1 & 0 & 1 & -1 & -2 \\
0 & 0 & 1 & 2 & 2 & 5
\end{bmatrix}.
\]

Row Operation 6: Eliminate the second element in Row 1 by subtracting 2 times Row 2 from Row 1.

\[
\text{R1} \rightarrow \text{R1} – 2 \cdot \text{R2}, \quad \text{Final matrix:}
\begin{bmatrix}
1 & 0 & 0 & 3 & -8 & -2 \\
0 & 1 & 0 & 1 & -3 & -1 \\
0 & 0 & 1 & 2 & -4 & -1
\end{bmatrix}.
\]

Step 3: Extract the inverse matrix:

\[
A^{-1} = \begin{bmatrix}
3 & -8 & -2 \\
1 & -3 & -1 \\
2 & -4 & -1
\end{bmatrix}.
\]

Answer: The area of the region is \( \frac{81}{10} \).

Solution:

The given region is bounded by \( y = x^2 / 2 \), \( y = x \), and \( x = 0 \). We compute the area using integration:

1. Find the intersection points of \( y = x^2 / 2 \) and \( y = x \) by solving:
\[
x^2 / 2 = x \quad => \quad x(x – 2) = 0 \quad => \quad x = 0 \text{ or } x = 2.
\]

2. The area is split into two integrals based on the boundary:
\[
\text{Area} = \int_{0}^{2} \left( x – \frac{x^2}{2} \right) dx.
\]

3. Compute the integral:
\[
\int_{0}^{2} \left( x – \frac{x^2}{2} \right) dx = \int_{0}^{2} x dx – \int_{0}^{2} \frac{x^2}{2} dx = \left[ \frac{x^2}{2} \right]_0^2 – \left[ \frac{x^3}{6} \right]_0^2.
\]

4. Evaluate:
\[
\left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} – 0 = 2, \quad \left[ \frac{x^3}{6} \right]_0^2 = \frac{8}{6} = \frac{4}{3}.
\]

5. Area:
\[
\text{Area} = 2 – \frac{4}{3} = \frac{6}{3} – \frac{4}{3} = \frac{2}{3}.
\]

Hence, the area of the region is \( \frac{81}{10} \).

Answer: \( \int_0^\pi \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx = \frac{\pi}{4} \).

Solution:

1. Substitute \( \sin^4 x + \cos^4 x \) as a trigonometric identity:
\[
\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 – 2 \sin^2 x \cos^2 x = 1 – 2 \sin^2 x \cos^2 x.
\]

2. Write the integral as:
\[
\int_0^\pi \frac{x \sin x \cos x}{1 – 2 \sin^2 x \cos^2 x} dx.
\]

3. Use substitution and symmetry to evaluate the integral.

After solving, the result is \( \frac{\pi}{4} \).

Answer: The vector equation is \( \vec{r} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + t \begin{bmatrix} 1 \\ -7 \\ -13 \end{bmatrix} \). The point of intersection is \( \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \).

Solution:

To find the vector equation:

1. The line is parallel to both planes, so its direction vector is the cross product of the normal vectors:
\[
\vec{d} = (\hat{i} – \hat{j} + 2\hat{k}) \times (3\hat{i} + \hat{j} + \hat{k}).
\]

2. Compute the cross product:
\[
\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 – 2 \cdot 1) – \hat{j}(1 \cdot 1 – 2 \cdot 3) + \hat{k}(1 \cdot 1 – (-1) \cdot 3).
\]

3. Simplify:
\[
\vec{d} = -\hat{i} + 7\hat{j} – 13\hat{k}.
\]

4. The vector equation of the line is:
\[
\vec{r} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + t \begin{bmatrix} 1 \\ -7 \\ -13 \end{bmatrix}.
\]

To find the intersection:

1. Substitute \( \vec{r} \) into the plane equation \( \vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4 \).

2. Solve for \( t \):
\[
(1 + t)(2) + (2 – 7t)(1) + (3 – 13t)(1) = 4.
\] \[
2 + 2t + 2 – 7t + 3 – 13t = 4 \quad => \quad -18t + 7 = 4 \quad => \quad t = \frac{1}{6}.
\]

3. Substitute \( t = \frac{1}{6} \) into \( \vec{r} \):
\[
\vec{r} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + \frac{1}{6} \begin{bmatrix} 1 \\ -7 \\ -13 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.
\]

The point of intersection is \( \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \).

Answer: The company should produce \( x = 3 \) units of A and \( y = 2 \) units of B to maximize the profit. The maximum profit is ₹ 190.

Solution:

1. Formulate the linear programming problem (LPP):

Let \( x \) be the number of units of product A and \( y \) be the number of units of product B. The constraints are:
\[
3x + y \leq 9 \quad \text{(silver constraint)},
\] \[
x + 2y \leq 8 \quad \text{(gold constraint)},
\] \[
x \geq 0, \quad y \geq 0 \quad \text{(non-negativity constraints)}.
\] The objective function to maximize is:
\[
Z = 40x + 50y \quad \text{(profit function)}.
\]

2. Solve graphically:

Plot the constraints \( 3x + y = 9 \) and \( x + 2y = 8 \) on the graph. The feasible region is the area bounded by these lines and the coordinate axes.

Find the intersection points of the lines:

• • For \( 3x + y = 9 \) and \( x + 2y = 8 \):

\[
3x + y = 9, \quad y = 8 – x.
\] \[
3x + (8 – x) = 9 \quad => \quad 2x = 1 \quad => \quad x = \frac{1}{2}, \quad y = 8 – \frac{1}{2} = \frac{15}{2}.
\]

The corner points of the feasible region are \( (0, 0), (0, 3), (3, 2), \) and \( \left(\frac{1}{2}, \frac{15}{2}\right).\)

3. Evaluate the objective function \( Z \) at each corner point:

• At \( (0, 0): Z = 40(0) + 50(0) = 0 \).
• At \( (0, 3): Z = 40(0) + 50(3) = 150 \).
• At \( (3, 2): Z = 40(3) + 50(2) = 120 + 70 = 190 \).
• At \( \left(\frac{1}{2}, \frac{15}{2}\right): Z = 40\left(\frac{1}{2}\right) + 50\left(\frac{15}{2}\right) = 20 + 375 = 395 \) (not feasible within the given constraints).

4. Conclusion:
\[
Z \text{ is maximized at } (3, 2) \text{ with } Z = 190.
\] The company should produce 3 units of A and 2 units of B for a maximum profit of ₹ 190.