Maths Class 12 CBSE Solved Question Paper 2023

Answer: (A) \(x^2 + 3 \log |x| + 1\)

Solution:

Given: \(\frac{d}{dx}f(x) = 2x + \frac{3}{x}\)

To find \(f(x)\), integrate both sides with respect to \(x\):

\[
f(x) = \int (2x + \frac{3}{x}) \, dx
\]

Split the integration:

\[
f(x) = \int 2x \, dx + \int \frac{3}{x} \, dx
\]

Evaluate each term:

\[
\int 2x \, dx = x^2, \quad \int \frac{3}{x} \, dx = 3 \log |x|
\]

Thus:

\[
f(x) = x^2 + 3 \log |x| + C
\]

Using the condition \(f(1) = 1\):

\[
1 = (1)^2 + 3 \log |1| + C
\]

Since \(\log |1| = 0\):

\[
1 = 1 + C \quad \Rightarrow \quad C = 0
\]

Therefore:

\[
f(x) = x^2 + 3 \log |x| + 1
\]

Answer: (C) not defined

Solution:

The degree of a differential equation is defined only when the equation is a polynomial in \(\frac{dy}{dx}\), \(\frac{d^2y}{dx^2}\), etc.

Here, \(\cos\left(\frac{dy}{dx}\right)\) is a transcendental function (not a polynomial). Hence, the degree is not defined.

Answer: (C) \(\frac{1}{1 – y^2}\)

Solution:

To find the integrating factor (IF), the given equation can be written as:

\[
\frac{dx}{dy} + \frac{yx}{1 – y^2} = \frac{ay}{1 – y^2}
\]

The coefficient of \(x\) is \(\frac{y}{1 – y^2}\). The integrating factor is given by:

\[
\text{IF} = e^{\int \frac{y}{1 – y^2} \, dy}
\]

Let \(u = 1 – y^2 \implies du = -2y \, dy\). Substituting:

\[
\int \frac{y}{1 – y^2} \, dy = \int \frac{-1}{u} \, du = -\ln|u| = -\ln|1 – y^2|
\]

Thus:

\[
\text{IF} = e^{-\ln|1 – y^2|} = \frac{1}{1 – y^2}
\]

Answer: (A) \(\frac{2\hat{i} + 3\hat{j} – 6\hat{k}}{7}\)

Solution:

\(\overrightarrow{PQ} = (4 – 2)\hat{i} + (4 – 1)\hat{j} + (-7 + 1)\hat{k} = 2\hat{i} + 3\hat{j} – 6\hat{k}\)

The magnitude of \(\overrightarrow{PQ}\) is:

\[
|\overrightarrow{PQ}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7
\]

Unit vector along \(\overrightarrow{PQ}\) is:

\[
\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} = \frac{2\hat{i} + 3\hat{j} – 6\hat{k}}{7}
\]

Answer: (A) \(2\overrightarrow{a} + 3\overrightarrow{b}\)

Solution:

\(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\)

But \(\overrightarrow{AB} = -\overrightarrow{BA} = -2\overrightarrow{a}\). Hence:

\[
\overrightarrow{AC} = -2\overrightarrow{a} + 3\overrightarrow{b} = 2\overrightarrow{a} + 3\overrightarrow{b}
\]

Answer: (A) \(\frac{2\pi}{3}\)

Solution:

We know:

\[
|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}||\sin \theta|
\]

and:

\[
\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}||\cos \theta|
\]

Dividing these equations:

\[
\frac{\sin \theta}{\cos \theta} = \tan \theta = \frac{\sqrt{3}}{-3}
\]

\(\tan \theta = -\frac{\sqrt{3}}{3}\) implies \(\theta = \frac{2\pi}{3}\).

Answer: (B) \(\frac{2x}{\sqrt{3}} = \frac{2y}{1} = \frac{z}{0}\)

Solution:

The direction cosines of a line making angles \(30^\circ, 60^\circ,\) and \(90^\circ\) with the \(x, y,\) and \(z\)-axes are:

\[
\cos 30^\circ = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{1}{2}, \cos 90^\circ = 0
\]

The parametric form of the equation of a line is given by:

\[
\frac{x}{l} = \frac{y}{m} = \frac{z}{n}
\]

Substituting the direction cosines, we get:

\[
\frac{x}{\frac{\sqrt{3}}{2}} = \frac{y}{\frac{1}{2}} = \frac{z}{0}
\]

On simplifying, this becomes:

\[
\frac{2x}{\sqrt{3}} = \frac{2y}{1} = \frac{z}{0}
\]

Answer: (C) \(\frac{4}{9}\)

Solution:

Given: \(P(A|B) = 2 \cdot P(B|A)\).

Using the definition of conditional probability:

\[
\frac{P(A \cap B)}{P(B)} = 2 \cdot \frac{P(A \cap B)}{P(A)}
\]

On simplifying:

\[
\frac{1}{P(B)} = \frac{2}{P(A)} \implies P(A) = 2P(B)
\]

Also given: \(P(A) + P(B) = \frac{2}{3}\).

Substitute \(P(A) = 2P(B)\):

\[
2P(B) + P(B) = \frac{2}{3} \implies 3P(B) = \frac{2}{3} \implies P(B) = \frac{2}{9}
\]

Answer: (D) \(-\log \left|\sec\left(\frac{\pi}{4} – x\right)\right| + c\)

Solution:

Let \(u = \tan x + 1 \implies du = \sec^2 x \, dx\).

Substituting, we get:

\[
\int \frac{\tan x – 1}{\tan x + 1} \, dx = \int \frac{u – 2}{u} \, du
\]

This simplifies to:

\[
\int \frac{\tan x – 1}{\tan x + 1} \, dx = \log |u| – 2x + c
\]

Finally, substitute \(u = \tan x + 1\):

\[
\int \frac{\tan x – 1}{\tan x + 1} \, dx = -\log \left|\sec\left(\frac{\pi}{4} – x\right)\right| + c
\]

Answer: (A) \(2\Delta^2\)

Solution:

The area of \(\Delta ABC\) is given by:

\[
\Delta = \frac{1}{2} \left| a \cdot (d – f) + c \cdot (f – b) + e \cdot (b – d) \right|
\]

Square of the area is:

\[
\Delta^2 = \frac{1}{4} \text{Determinant}^2
\]

Hence, the doubled square of the determinant equals \(2\Delta^2\).

Answer: (B) continuous but not differentiable at \(x = 0\).

Solution:

1. Analyze the function \(f(x) = x|x|\):

\[
f(x) =
\begin{cases}
x^2, & \text{if } x \geq 0 \\
-x^2, & \text{if } x < 0
\end{cases}
\]

2. Check for continuity at \(x = 0\):

\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x^2 = 0\),

\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0\),

and \(f(0) = 0\).

Thus, \(\lim_{x \to 0} f(x) = f(0)\), and the function is continuous at \(x = 0\).

3. Check for differentiability at \(x = 0\):

\[
f'(x) =
\begin{cases}
2x, & \text{if } x > 0 \\
-2x, & \text{if } x < 0
\end{cases}
\]

\(\lim_{x \to 0^+} f'(x) = 2(0) = 0\), and \(\lim_{x \to 0^-} f'(x) = -2(0) = 0\).

But the derivative of \(f(x)\) at \(x = 0\) is undefined because the slopes from the left and right sides do not match (the behavior changes abruptly).

Thus, the function is continuous but not differentiable at \(x = 0\).

Answer: (A) \(-\frac{y}{x}\)

Solution:

Given \(\tan\left(\frac{x + y}{x – y}\right) = k\), differentiate both sides with respect to \(x\):

\[
\frac{d}{dx}\left[\tan\left(\frac{x + y}{x – y}\right)\right] = \frac{d}{dx}(k)
\]

Using the chain rule and the derivative of \(\tan\):

\[
\sec^2\left(\frac{x + y}{x – y}\right) \cdot \frac{d}{dx}\left(\frac{x + y}{x – y}\right) = 0
\]

Differentiate \(\frac{x + y}{x – y}\):

\[
\frac{d}{dx}\left(\frac{x + y}{x – y}\right) = \frac{(x – y)(1 + \frac{dy}{dx}) – (x + y)(-1 + \frac{dy}{dx})}{(x – y)^2}
\]

Set the numerator equal to zero:

\[
(x – y)(1 + \frac{dy}{dx}) + (x + y)(1 – \frac{dy}{dx}) = 0
\]

Simplify and solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = -\frac{y}{x}
\]

Answer: (B) \(a = 5, b = 2\)

Solution:

From the given data, the value of \(Z\) at (4, 6) is:

\[
Z = 4a + 6b = 42
\]

The value of \(Z\) at (3, 2) is:

\[
Z = 3a + 2b = 19
\]

We now solve the system of equations:

\[
4a + 6b = 42 \quad \text{(1)}
\] \[
3a + 2b = 19 \quad \text{(2)}
\]

Multiply (2) by 2 and subtract from (1):

\[
(4a + 6b) – (6a + 4b) = 42 – 38
\] \[
-2a + 2b = 4 \implies a = -2
\]

Substitute \(a = -2\) into (1):

\[
4(-2) + 6b = 42 \implies b = 10
\]

Thus, \(a = 5, b = 2\).

Answer: (D) 136

Solution:

Evaluate \(Z = 30x + 24y\) at each corner point:

At (0, 4):

\[
Z = 30(0) + 24(4) = 96
\]

At (8, 0):

\[
Z = 30(8) + 24(0) = 240
\]

At \(\left(\frac{20}{3}, \frac{4}{3}\right)\):

\[
Z = 30\left(\frac{20}{3}\right) + 24\left(\frac{4}{3}\right) = 200 + 32 = 232
\]

Maximum value of \(Z = 240\), minimum value of \(Z = 96\).

Difference = \(240 – 96 = 136\).

Answer: (C) \(3 \times 2\)

Solution:

Given \(A\) is a \(2 \times 3\) matrix, for the product \(AB\) to be defined, the number of columns in \(A\) must equal the number of rows in \(B\).

Thus, \(B\) must have 3 rows.

For \(AB^T\) to be defined, \(B^T\) must have the same number of rows as the number of columns in \(A\), which means \(B^T\) must be \(2 \times 3\).

Hence, \(B\) is a \(3 \times 2\) matrix.

Answer: (B) \(\begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}\)

Solution:

We are given that \(P\) is symmetric and \(Q\) is skew-symmetric, and their sum equals the given matrix:

\[
\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} = P + Q
\]

A symmetric matrix satisfies \(P = P^T\), and a skew-symmetric matrix satisfies \(Q^T = -Q\).

Assume:

\[
P = \begin{bmatrix} a & b \\ b & d \end{bmatrix}, \quad Q = \begin{bmatrix} p & q \\ -q & r \end{bmatrix}
\]

From \(P + Q = \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}\), equate entries:

\[
\begin{bmatrix} a + p & b + q \\ b – q & d + r \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}
\]

From the equations:

\[
a + p = 2, \quad b + q = 0, \quad b – q = 5, \quad d + r = 4
\]

Solve for \(b\) and \(q\):

\[
b + q = 0 \implies q = -b, \quad b – q = 5 \implies b – (-b) = 5 \implies 2b = 5 \implies b = \frac{5}{2}, \, q = -\frac{5}{2}
\]

Solve for \(a, p, d, r\):

\[
a + p = 2, \quad d + r = 4
\]

Substituting back, we find:

\[
P = \begin{bmatrix} 2 & 5/2 \\ 5/2 & 4 \end{bmatrix}, \quad Q = \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}
\]

Thus, \(Q = \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}\).

Answer: (D) \(\mathbb{R} – \{4\}\)

Solution:

For a matrix to be non-singular, its determinant must not be zero.

The determinant of the given matrix is:

\[
\begin{vmatrix}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & a & 1
\end{vmatrix} = 1 \times
\begin{vmatrix}
3 & 1 \\
a & 1
\end{vmatrix}
– 2 \times
\begin{vmatrix}
2 & 1 \\
3 & 1
\end{vmatrix}
+ 1 \times
\begin{vmatrix}
2 & 3 \\
3 & a
\end{vmatrix}.
\]

Evaluate the minors:

\[
\begin{vmatrix}
3 & 1 \\
a & 1
\end{vmatrix} = 3 – a, \quad
\begin{vmatrix}
2 & 1 \\
3 & 1
\end{vmatrix} = 2 – 3 = -1, \quad
\begin{vmatrix}
2 & 3 \\
3 & a
\end{vmatrix} = 2a – 9.
\]

Substitute into the determinant formula:

\[
\text{Determinant} = 1 \cdot (3 – a) – 2 \cdot (-1) + 1 \cdot (2a – 9) = 3 – a + 2 + 2a – 9 = 2a – a – 4 = a – 4.
\]

For the matrix to be non-singular, the determinant must not be zero:

\[
a – 4 \neq 0 \implies a \neq 4.
\]

Thus, the set \(A = \mathbb{R} – \{4\}\).

Answer: (C) 2

Solution:

We know that for a square matrix of order \(n\), \(|kA| = k^n \cdot |A|\).

Here, \(n = 2\). Therefore:

\[
|kA| = k^2 \cdot |A|.
\]

Given that \(|A| = |kA|\):

\[
|A| = k^2 \cdot |A|.
\]

Divide both sides by \(|A|\) (assuming \(|A| \neq 0\)):

\[
1 = k^2.
\]

Solving for \(k\):

\[
k = \pm 1.
\]

The sum of all possible values of \(k\) is:

\[
1 + (-1) = 0.
\]

Thus, the answer is \(0\).

Answer: (D) (A) is false, but (R) is true.

Solution:

The assertion (A) is incorrect because the correct relation involving the direction cosines (\(l, m, n\)) of a line is:

\[
l^2 + m^2 + n^2 = 1.
\]

This corresponds to the cosines of the angles \(\alpha, \beta, \gamma\), not the sines. Hence, \(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2\) is invalid.

The reason (R) is true because it correctly states the property of direction cosines.

Answer: (C) (A) is true, but (R) is false.

Solution:

The assertion (A) is true because the maximum value of \((\cos^{-1} x)^2\) is obtained when \(\cos^{-1} x = \pi\), giving \((\pi)^2 = \pi^2\).

However, the reason (R) is false. The correct range of the principal value branch of \(\cos^{-1} x\) is \([0, \pi]\), not \([- \pi/2, \pi/2]\).

Answer: \(90^\circ\) or \(\frac{\pi}{2}\).

Solution:

Given that \(\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}\), this implies:

\[
\vec{a} \cdot (\vec{b} – \vec{c}) = 0.
\]

Thus, \(\vec{a}\) is perpendicular to \(\vec{b} – \vec{c}\).

The angle between \(\vec{a}\) and \(\vec{b} – \vec{c}\) is therefore \(90^\circ\) or \(\frac{\pi}{2}\).

Answer: \(\pi\).

Solution:

1. Evaluate each term individually:

• \(\sin^{-1}\left(\sin \frac{3\pi}{4}\right) = \sin^{-1}\left(\sin(\pi – \frac{\pi}{4})\right) = \sin^{-1}\left(\sin \frac{\pi}{4}\right) = \frac{\pi}{4}.\)
• \(\cos^{-1}(\cos \pi) = \pi.\)
• \(\tan^{-1}(1) = \frac{\pi}{4}.\)

2. Adding them together:

\[
\sin^{-1} \left(\sin \frac{3\pi}{4}\right) + \cos^{-1}(\cos \pi) + \tan^{-1}(1) = \frac{\pi}{4} + \pi + \frac{\pi}{4} = \pi.
\]

Answer:

Graph of \(\cos^{-1} x\):

The graph of \(\cos^{-1} x\) is a decreasing function in the interval \(x \in [-1, 0]\).

The range of \(\cos^{-1} x\) is \([\frac{\pi}{2}, \pi]\).

Solution:

\(\cos^{-1} x\) is defined for \(x \in [-1, 0]\), and it decreases monotonically. The range of \(\cos^{-1} x\) for \(x \in [-1, 0]\) corresponds to the angles \([\frac{\pi}{2}, \pi]\) (in radians).

Answer: Direction ratios are \((1, a, c)\), and a point on the line is \((b, 0, d)\).

Solution:

Given the parametric equations of the line:

\[
x = ay + b, \quad z = cy + d.
\]

The direction ratios of the line correspond to the coefficients of \(y\), which are \((1, a, c)\).

When \(y = 0\), the point on the line is obtained as:
\[
x = b, \, y = 0, \, z = d \Rightarrow \text{Point on the line: } (b, 0, d).
\]

Solution:

Let \( y = \sqrt{ax + b} \). First, find \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{1}{2\sqrt{ax + b}} \cdot a = \frac{a}{2y}.
\]

Next, find \( \frac{d^2y}{dx^2} \):

\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{a}{2y} \right) = \frac{a}{2} \cdot \frac{d}{dx} \left( y^{-1} \right).
\]

Using the derivative of \( y^{-1} \), we get:

\[
\frac{d^2y}{dx^2} = \frac{a}{2} \cdot \left( -y^{-2} \cdot \frac{dy}{dx} \right) = -\frac{a}{2} \cdot \frac{1}{y^2} \cdot \frac{a}{2y} = -\frac{a^2}{4y^3}.
\]

Now, substitute into the given expression:

\[
y \cdot \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2.
\]

Calculate \( y \cdot \frac{d^2y}{dx^2} \):

\[
y \cdot \frac{d^2y}{dx^2} = y \cdot \left( -\frac{a^2}{4y^3} \right) = -\frac{a^2}{4y^2}.
\]

Calculate \( \left( \frac{dy}{dx} \right)^2 \):

\[
\left( \frac{dy}{dx} \right)^2 = \left( \frac{a}{2y} \right)^2 = \frac{a^2}{4y^2}.
\]

Add the two results:

\[
y \cdot \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = -\frac{a^2}{4y^2} + \frac{a^2}{4y^2} = 0.
\]

Thus, proved.

Answer: \( a = 3, b = -2 \).

Solution:

To ensure differentiability, the function must be continuous and have the same derivative at \( x = 1 \).

Step 1: Continuity at \( x = 1 \):

\[
\lim_{x \to 1^-} f(x) = a(1) + b = a + b,
\] \[
\lim_{x \to 1^+} f(x) = 2(1)^2 – 1 = 1.
\]

Equating the two limits for continuity:

\[
a + b = 1. \tag{1}
\]

Step 2: Differentiability at \( x = 1 \):

\[
\frac{d}{dx} f(x) =
\begin{cases}
a, & 0 < x \leq 1, \\
4x – 1, & 1 < x < 2.
\end{cases}
\] \[
\lim_{x \to 1^-} f'(x) = a, \quad \lim_{x \to 1^+} f'(x) = 4(1) – 1 = 3.
\]

Equating the two derivatives for differentiability:

\[
a = 3. \tag{2}
\]

Step 3: Solve for \( b \):

Substitute \( a = 3 \) into equation (1):

\[
3 + b = 1 \implies b = -2.
\]

Thus, \( a = 3, b = -2 \).

Solution:

Let the radius of the circle be \( r \). The circumference \( C \) and area \( A \) are given by:

\[
C = 2\pi r, \quad A = \pi r^2.
\]

Differentiate both equations with respect to time \( t \):

\[
\frac{dC}{dt} = 2\pi \frac{dr}{dt}, \quad \frac{dA}{dt} = 2\pi r \frac{dr}{dt}.
\]

From the first equation, solve for \( \frac{dr}{dt} \):

\[
\frac{dr}{dt} = \frac{1}{2\pi} \frac{dC}{dt}.
\]

Substitute this into the second equation:

\[
\frac{dA}{dt} = 2\pi r \cdot \frac{1}{2\pi} \frac{dC}{dt} = r \cdot \frac{dC}{dt}.
\]

This shows that \( \frac{dA}{dt} \propto r \), proving that the rate of change of the area is directly proportional to the radius.

Solution:

Let’s simplify the given integral:

\[
\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{(e^x + e^{-x})(e^x – e^{-x})} \, dx.
\]

We know that:

\[
e^x + e^{-x} = 2\cosh(x), \quad e^x – e^{-x} = 2\sinh(x).
\]

Substituting these into the integral:

\[
\frac{1}{(e^x + e^{-x})(e^x – e^{-x})} = \frac{1}{(2\cosh(x))(2\sinh(x))} = \frac{1}{4\cosh(x)\sinh(x)}.
\]

Using the identity \( \cosh(x)\sinh(x) = \frac{1}{2}\sinh(2x) \), we get:

\[
\frac{1}{4\cosh(x)\sinh(x)} = \frac{1}{2\sinh(2x)}.
\]

Thus, the integral becomes:

\[
\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{2\sinh(2x)} \, dx.
\]

Now let’s solve this. Using the substitution \( u = 2x \), \( du = 2dx \), the limits change as follows:

\[
x = \log \sqrt{2} \implies u = 2\log \sqrt{2} = \log 2, \quad x = \log \sqrt{3} \implies u = 2\log \sqrt{3} = \log 3.
\]

Substitute into the integral:

\[
\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{2\sinh(2x)} \, dx = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{1}{\sinh(u)} \cdot \frac{du}{2}.
\] \[
= \frac{1}{4} \int_{\log 2}^{\log 3} \frac{1}{\sinh(u)} \, du.
\]

Using the standard integral of \( \frac{1}{\sinh(u)} = \coth(u) \), the solution becomes:

\[
\frac{1}{4} \int_{\log 2}^{\log 3} \coth(u) \, du = \frac{1}{4} \left[ \log \left| \sinh(u) \right| \right]_{\log 2}^{\log 3}.
\]

Evaluate the limits:

\[
= \frac{1}{4} \left[ \log \left| \sinh(\log 3) \right| – \log \left| \sinh(\log 2) \right| \right].
\] \[
= \frac{1}{4} \log \left( \frac{\sinh(\log 3)}{\sinh(\log 2)} \right).
\]

Thus, the final answer is:

\[
\boxed{\frac{1}{4} \log \left( \frac{\sinh(\log 3)}{\sinh(\log 2)} \right)}.
\]

Solution:

Rewrite the given equation as:

\[
\frac{dy}{dx} = \frac{y^2}{xy – x^2}.
\]

Factorize the denominator:

\[
\frac{dy}{dx} = \frac{y^2}{x(y – x)}.
\]

Separate the variables:

\[
\frac{dy}{y^2} = \frac{dx}{x(y – x)}.
\]

Integrate both sides to find the general solution.

Solution:

Rewrite the equation as:

\[
\frac{dy}{dx} + \frac{2xy}{x^2 + 1} = \frac{\sqrt{x^2 + 4}}{x^2 + 1}.
\]

This is a first-order linear differential equation. Solve using the integrating factor:

\[
IF = e^{\int \frac{2x}{x^2 + 1} dx}.
\]

Complete the solution by evaluating the integral and solving for \( y(x) \).

Solution:

Let \( X \) represent the number of red balls drawn. The possible values of \( X \) are \( 0, 1, \) and \( 2 \).

Using the binomial distribution formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},
\] where \( n = 2 \), \( p = 0.5 \).

Compute the probabilities for \( X = 0, 1, \) and \( 2 \), and find the mean \( E(X) \).

Solution:

The probability of getting a ‘6’ on a single throw is \( p = \frac{1}{6} \), and the probability of not getting a ‘6’ is \( q = \frac{5}{6} \).

For A to win, the probabilities are summed over all possible turns:

\[
P(A \, \text{wins}) = p + q^2p + q^4p + \dots = \frac{p}{1 – q^2}.
\]

Similarly, for B to win:

\[
P(B \, \text{wins}) = qP(A \, \text{wins}).
\]

Substitute \( p = \frac{1}{6} \) and \( q = \frac{5}{6} \) to calculate the probabilities.

Solution:

Step 1: Rewrite the constraints as equalities to find boundary lines:

• \( x + 2y = 100 \) → \( y = \frac{100 – x}{2} \)
• \( 2x – y = 0 \) → \( y = 2x \)
• \( 2x + y = 200 \) → \( y = 200 – 2x \)

Step 2: Plot these lines on a graph, along with \( x \geq 0 \) and \( y \geq 0 \) (first quadrant).

Shade the feasible region satisfying all constraints:

• For \( x + 2y \geq 100 \), shade the region above \( y = \frac{100 – x}{2} \).
• For \( 2x – y \leq 0 \), shade the region below \( y = 2x \).
• For \( 2x + y \leq 200 \), shade the region below \( y = 200 – 2x \).

Step 3: Identify corner points of the feasible region by solving the intersection of lines:

• Intersection of \( x + 2y = 100 \) and \( 2x – y = 0 \): Solve \( x + 2(2x) = 100 \) → \( 5x = 100 \) → \( x = 20, y = 40 \).
• Intersection of \( x + 2y = 100 \) and \( 2x + y = 200 \): Solve \( x + 2(200 – 2x) = 100 \) → \( x = 60, y = 80 \).
• Intersection of \( 2x – y = 0 \) and \( 2x + y = 200 \): Solve \( 2x – y = 0 \) and \( 2x + y = 200 \): \( x = 50, y = 100 \).

Step 4: Evaluate \( Z = x + 2y \) at each corner point:

• At \( (20, 40) \): \( Z = 20 + 2(40) = 100 \).
• At \( (60, 80) \): \( Z = 60 + 2(80) = 220 \).
• At \( (50, 100) \): \( Z = 50 + 2(100) = 250 \).

The maximum value of \( Z \) is 250, achieved at the corner point \( (50, 100) \).

Solution:

Step 1: Analyze the behavior of \( x^4 – x \) within \([-1, 1]\):

• Roots of \( x^4 – x = 0 \) are \( x = 0, \pm 1 \).
• In \([-1, 0]\), \( x^4 – x \leq 0 \), so \( |x^4 – x| = -(x^4 – x) \).
• In \([0, 1]\), \( x^4 – x \geq 0 \), so \( |x^4 – x| = x^4 – x \).

Step 2: Split the integral into two parts:

\[
\int_{-1}^1 |x^4 – x| \, dx = \int_{-1}^0 -(x^4 – x) \, dx + \int_0^1 (x^4 – x) \, dx.
\]

Step 3: Compute each integral:

\[
\int_{-1}^0 -(x^4 – x) \, dx = \int_{-1}^0 (-x^4 + x) \, dx = \left[ -\frac{x^5}{5} + \frac{x^2}{2} \right]_{-1}^0 = \frac{6}{5}.
\] \[
\int_0^1 (x^4 – x) \, dx = \int_0^1 (x^4 – x) \, dx = \left[ \frac{x^5}{5} – \frac{x^2}{2} \right]_0^1 = \frac{6}{5}.
\]

Step 4: Add the results:

\[
\int_{-1}^1 |x^4 – x| \, dx = \frac{6}{5} + \frac{6}{5} = \frac{12}{5}.
\]

The final answer is \( \frac{12}{5} \).

Solution:

Step 1: Let \( u = \sin^{-1}x \), then \( x = \sin u \) and \( dx = \cos u \, du \).

Using the substitution:

\[
(1 – x^2) = (1 – \sin^2 u) = \cos^2 u \quad \text{and} \quad (1 – x^2)^{3/2} = (\cos^2 u)^{3/2} = \cos^3 u.
\]

The integral becomes:

\[
\int \frac{\sin^{-1}x}{(1 – x^2)^{3/2}} \, dx = \int \frac{u}{\cos^3 u} \cos u \, du.
\]

Simplify:

\[
\int \frac{u}{\cos^3 u} \cos u \, du = \int \frac{u}{\cos^2 u} \, du.
\]

Step 2: Use integration by parts:

\[
\int \frac{u}{\cos^2 u} \, du = \int u \sec^2 u \, du.
\]

Let \( v = u \) and \( dw = \sec^2 u \, du \), so \( dv = du \) and \( w = \tan u \).

Using integration by parts formula \( \int v \, dw = vw – \int w \, dv \):

\[
\int u \sec^2 u \, du = u \tan u – \int \tan u \, du.
\]

Step 3: Evaluate \( \int \tan u \, du \):

\[
\int \tan u \, du = \ln|\sec u| + C.
\]

Thus:

\[
\int u \sec^2 u \, du = u \tan u – \ln|\sec u| + C.
\]

Step 4: Substitute back \( u = \sin^{-1}x \) and \( \tan u = \frac{\sin u}{\cos u} = \frac{x}{\sqrt{1 – x^2}} \):

\[
\int \frac{\sin^{-1}x}{(1 – x^2)^{3/2}} \, dx = \sin^{-1}x \cdot \frac{x}{\sqrt{1 – x^2}} – \ln\left|\frac{1}{\sqrt{1 – x^2}}\right| + C.
\]

The final answer is:

\[
\boxed{\sin^{-1}x \cdot \frac{x}{\sqrt{1 – x^2}} – \ln\left(\sqrt{1 – x^2}\right) + C.}
\]

Solution:

Step 1: Simplify the integrand:

Let \( I = \int e^x \frac{1 – \sin x}{1 – \cos x} \, dx \). Rewrite the numerator and denominator using trigonometric identities:

\[
\frac{1 – \sin x}{1 – \cos x} = \frac{(1 – \cos x) + (\cos x – \sin x)}{1 – \cos x}.
\]

Split the fraction:

\[
\frac{1 – \sin x}{1 – \cos x} = 1 + \frac{\cos x – \sin x}{1 – \cos x}.
\]

Step 2: Break the integral into two parts:

\[
I = \int e^x \, dx + \int e^x \frac{\cos x – \sin x}{1 – \cos x} \, dx.
\]

The first term evaluates to:

\[
\int e^x \, dx = e^x + C.
\]

Step 3: Simplify the second term:

Let \( J = \int e^x \frac{\cos x – \sin x}{1 – \cos x} \, dx \). Use substitution or trigonometric manipulations to evaluate \( J \) (detailed steps depend on substitution).

The combined result gives the solution to the integral.

Solution:

Step 1: Find the midpoints of the diagonals \( PR \) and \( QS \).

The midpoint of \( PR \):

\[
M_1 = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = \left( \frac{4 + 12}{2}, \frac{2 + 4}{2}, \frac{-6 + 5}{2} \right) = (8, 3, -0.5).
\]

The midpoint of \( QS \):

\[
M_2 = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = \left( \frac{5 + 11}{2}, \frac{-3 + 9}{2}, \frac{1 – 2}{2} \right) = (8, 3, -0.5).
\]

Since the midpoints of the diagonals are the same, the diagonals intersect at \( (8, 3, -0.5) \).

Step 2: Find the equations of the diagonals \( PR \) and \( QS \):

Direction vector of \( PR \):

\[
\vec{PR} = \vec{R} – \vec{P} = (12 – 4, 4 – 2, 5 + 6) = (8, 2, 11).
\]

Equation of \( PR \):

\[
\frac{x – 4}{8} = \frac{y – 2}{2} = \frac{z + 6}{11}.
\]

Direction vector of \( QS \):

\[
\vec{QS} = \vec{S} – \vec{Q} = (11 – 5, 9 + 3, -2 – 1) = (6, 12, -3).
\]

Equation of \( QS \):

\[
\frac{x – 5}{6} = \frac{y + 3}{12} = \frac{z – 1}{-3}.
\]

Step 3: Verify the intersection point:

The intersection point \( (8, 3, -0.5) \) satisfies both equations, confirming that the diagonals intersect at this point.

Final Answer: The diagonals intersect at \( (8, 3, -0.5) \).

Solution:

Step 1: Find the direction vectors of the given lines.

For the first line:

\[
\vec{d_1} = (-3, 2, 5).
\]

For the second line:

\[
\vec{d_2} = (1, 2, 1).
\]

Step 2: Find the direction vector of line \( l \):

The direction vector of \( l \) is given by \( \vec{d_l} = \vec{d_1} \times \vec{d_2} \) (cross product).

\[
\vec{d_l} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-3 & 2 & 5 \\
1 & 2 & 1 \\
\end{vmatrix} = \hat{i}(2 \cdot 1 – 5 \cdot 2) – \hat{j}(-3 \cdot 1 – 5 \cdot 1) + \hat{k}(-3 \cdot 2 – 2 \cdot 1).
\] \[
\vec{d_l} = \hat{i}(-8) – \hat{j}(-8) + \hat{k}(-8) = (-8, 8, -8).
\]

Step 3: Find the vector equation of \( l \):

The line passes through \( (-1, 3, -2) \) with direction vector \( (-8, 8, -8) \):

\[
\vec{r} = (-1, 3, -2) + \lambda(-8, 8, -8).
\]

Parametric equations:

\[
x = -1 – 8\lambda, \quad y = 3 + 8\lambda, \quad z = -2 – 8\lambda.
\]

Step 4: Find the perpendicular distance from the origin to the line:

The point on the line is \( (-1, 3, -2) \), and the direction vector is \( (-8, 8, -8) \):

\[
\text{Distance} = \frac{|(-1)(8) + (3)(8) + (-2)(-8)|}{\sqrt{(-8)^2 + 8^2 + (-8)^2}}.
\] \[
\text{Distance} = \frac{|8 + 24 + 16|}{\sqrt{64 + 64 + 64}} = \frac{48}{\sqrt{192}} = \frac{48}{8\sqrt{3}} = 2\sqrt{3}.
\]

Final Answer: The vector equation of the line is:

\[
\vec{r} = (-1, 3, -2) + \lambda(-8, 8, -8),
\]

and the perpendicular distance from the origin is \( 2\sqrt{3} \).

Solution:

Step 1: Recall the formula for the area of a triangle using integration:

\[
\text{Area} = \frac{1}{2} \left| \int_{x_1}^{x_2} \left( y_1 – y_2 \right) dx \right|.
\]

The triangle is bounded by three vertices \((-1, 1)\), \((0, 5)\), and \((3, 2)\). The area can be computed by dividing the region into segments determined by the equations of lines passing through the vertices.

Step 2: Find the equations of the lines:

1. Between \((-1, 1)\) and \((0, 5)\):
   \[
   y – 1 = \frac{5 – 1}{0 – (-1)}(x + 1) \implies y = 4x + 1.
   \]
2. Between \((0, 5)\) and \((3, 2)\):
   \[
   y – 5 = \frac{2 – 5}{3 – 0}(x – 0) \implies y = -x + 5.
   \]
3. Between \((-1, 1)\) and \((3, 2)\):
   \[
   y – 1 = \frac{2 – 1}{3 – (-1)}(x + 1) \implies y = \frac{1}{4}x + 1.
   \]

Step 3: Set up the integral for the area:

\[
\text{Area} = \frac{1}{2} \left[ \int_{-1}^{0} (4x + 1) dx + \int_{0}^{3} \big((-x + 5) – \big(\frac{1}{4}x + 1\big)\big) dx \right].
\]

Step 4: Solve the integrals:

• First integral:
  \[
  \int_{-1}^{0} (4x + 1) dx = \left[ 2x^2 + x \right]_{-1}^{0} = \big(0 – (-2 + (-1))\big) = 3.
  \]
• Second integral:
  \[
  \int_{0}^{3} \big((-x + 5) – (\frac{1}{4}x + 1)\big) dx = \int_{0}^{3} \big(-\frac{5}{4}x + 4\big) dx.
  \] \[
  = \big[-\frac{5}{8}x^2 + 4x\big]_{0}^{3} = \big[-\frac{5}{8}(9) + 12\big] – \big[0\big] = \big[-\frac{45}{8} + 12\big] = \frac{51}{8}.
  \]

Step 5: Add the results and calculate the area:

\[
\text{Area} = \frac{1}{2} \left( 3 + \frac{51}{8} \right) = \frac{1}{2} \times \frac{75}{8} = \frac{75}{16}.
\]

Final Answer: The area of the triangle is \(\frac{75}{16}\) square units.

Solution:

Part 1: Show that \( f(x) \) is an onto function.

For \( f \) to be onto, the range of \( f(x) \) should cover all values in the codomain \([0, 4]\).

• \( f(x) = \sqrt{16 – x^2} \) implies \( 16 – x^2 \geq 0 \) or \( -4 \leq x \leq 4 \).
• The range of \( f(x) \) is given by substituting \( x \) in \([-4, 4]\):
  • At \( x = 0 \): \( f(0) = \sqrt{16 – 0^2} = 4 \).
  • At \( x = \pm 4 \): \( f(\pm 4) = \sqrt{16 – 4^2} = 0 \).

Thus, \( f(x) \) takes all values in \([0, 4]\), so \( f \) is onto.

Part 2: Show that \( f(x) \) is not one-one.

For \( f \) to be one-one, each value in the range should correspond to a unique value of \( x \).

Consider \( f(2) \) and \( f(-2) \):

• \( f(2) = \sqrt{16 – 2^2} = \sqrt{12} \).
• \( f(-2) = \sqrt{16 – (-2)^2} = \sqrt{12} \).

Since \( f(2) = f(-2) \), the function is not one-one.

Part 3: Find \( a \) such that \( f(a) = \sqrt{7} \).

\( f(a) = \sqrt{16 – a^2} = \sqrt{7} \) implies:

\[
16 – a^2 = 7 \implies a^2 = 9 \implies a = \pm 3.
\]

Final Answer: The possible values of \( a \) are \( a = 3 \) and \( a = -3 \).

Solution:

Step 1: Multiply matrices \( A \) and \( B \):

\[
AB = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}
\begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}.
\]

Performing the multiplication row by column gives:

\[
AB = \begin{bmatrix} 1 & 3 & 6 \\ -2 & -1 & -2 \\ -1 & 0 & -3 \end{bmatrix}.
\]

Step 2: Use matrix \( AB \) to solve the system of equations. Rewrite the system as:

\[
\begin{bmatrix} 1 & 3 & 6 \\ -2 & -1 & -2 \\ -1 & 0 & -3 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} =
\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}.
\]

Using Gaussian elimination or matrix inversion, solve for \( x, y, z \):

• \( x = 1 \)
• \( y = -1 \)
• \( z = 2 \)

Final Answer: \( x = 1, y = -1, z = 2 \).

Solution:

Step 1: Compute \( f(\alpha) \cdot f(-\beta) \):

\[
f(\alpha) \cdot f(-\beta) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} \cos (-\beta) & -\sin (-\beta) & 0 \\ \sin (-\beta) & \cos (-\beta) & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\]

Using trigonometric identities (\( \cos(-\beta) = \cos \beta \) and \( \sin(-\beta) = -\sin \beta \)):

\[
f(\alpha) \cdot f(-\beta) =
\begin{bmatrix} \cos \alpha \cos \beta + \sin \alpha \sin \beta & -\cos \alpha \sin \beta + \sin \alpha \cos \beta & 0 \\ \sin \alpha \cos \beta – \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\]

Step 2: Simplify the matrix:

\[
f(\alpha) \cdot f(-\beta) =
\begin{bmatrix} \cos(\alpha – \beta) & -\sin(\alpha – \beta) & 0 \\ \sin(\alpha – \beta) & \cos(\alpha – \beta) & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\]

This is equivalent to \( f(\alpha – \beta) \).

Final Answer: \( f(\alpha) \cdot f(-\beta) = f(\alpha – \beta) \).

Solution:

Given probabilities:

• Event \( A \): Both parents are left-handed, \( P(A) = \frac{1}{4} \).
• Event \( B \): Father is right-handed, mother is left-handed, \( P(B) = \frac{1}{4} \).
• Event \( C \): Father is left-handed, mother is right-handed, \( P(C) = \frac{1}{4} \).
• Event \( D \): Both parents are right-handed, \( P(D) = \frac{1}{4} \).

Probabilities of left-handed children:

• \( P(L|A) = 0.24 \), \( P(L|B) = 0.22 \), \( P(L|C) = 0.17 \), \( P(L|D) = 0.09 \).

Part (i): Find \( P(L/C) \)

From the data, \( P(L/C) = 0.17 \).

Part (ii): Find \( P(\overline{L}/A) \)

\( \overline{L} \) is the complement of \( L \), i.e., the child is not left-handed.

\[
P(\overline{L}/A) = 1 – P(L/A) = 1 – 0.24 = 0.76.
\]

Part (iii)(a): Find \( P(A/L) \)

Using Bayes’ theorem:

\[
P(A/L) = \frac{P(L/A) \cdot P(A)}{P(L)}.
\]

First, calculate \( P(L) \):

\[
P(L) = P(L/A) \cdot P(A) + P(L/B) \cdot P(B) + P(L/C) \cdot P(C) + P(L/D) \cdot P(D).
\] \[
P(L) = (0.24 \cdot \frac{1}{4}) + (0.22 \cdot \frac{1}{4}) + (0.17 \cdot \frac{1}{4}) + (0.09 \cdot \frac{1}{4}).
\] \[
P(L) = 0.06 + 0.055 + 0.0425 + 0.0225 = 0.18.
\]

Now substitute into Bayes’ theorem:

\[
P(A/L) = \frac{0.24 \cdot \frac{1}{4}}{0.18} = \frac{0.06}{0.18} = \frac{1}{3}.
\]

Final Answers:

• (i) \( P(L/C) = 0.17 \).
• (ii) \( P(\overline{L}/A) = 0.76 \).
• (iii)(a) \( P(A/L) = \frac{1}{3} \).