GSEB 2024 Mathematics Question Paper with Answer and Solution

(A) We need to evaluate the integral $I = \int e^x (\tan x + \tan^2 x) \, dx$.
Recall the identity $1 + \tan^2 x = \sec^2 x$,which implies $\tan^2 x = \sec^2 x - 1$.
Substituting this into the integral,we get:
$I = \int e^x (\tan x + \sec^2 x - 1) \, dx$.
Rearranging the terms:
$I = \int e^x ((\tan x - 1) + \sec^2 x) \, dx$.
Let $f(x) = \tan x - 1$.
Then,$f'(x) = \sec^2 x$.
Using the standard integration formula $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we have:
$I = e^x (\tan x - 1) + C$.
Therefore,the correct option is $A$.

(C) Let $f(x) = \sin^7 x \cdot \cos^6 x$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^7(-x) \cdot \cos^6(-x)$
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\sin x)^7 \cdot (\cos x)^6 = -\sin^7 x \cdot \cos^6 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,$\int_{-a}^a f(x) \, dx = 0$ if $f(x)$ is an odd function.
Therefore,$\int_{-1}^1 \sin^7 x \cdot \cos^6 x \, dx = 0$.

(B) To evaluate the integral $I = \int \frac{d x}{x^2+2 x+5}$,we first complete the square in the denominator.
$x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2$.
Now,the integral becomes $I = \int \frac{d x}{(x+1)^2 + 2^2}$.
Using the standard integration formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,where $a = 2$ and the variable is $(x+1)$:
$I = \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C$.
Comparing this with the given options,the correct option is $B$.

(A) To evaluate the integral $I = \int \frac{1}{e^x+1} dx$,we can rewrite the integrand as follows:
$I = \int \frac{1}{e^x(1 + e^{-x})} dx = \int \frac{e^{-x}}{1 + e^{-x}} dx$.
Let $u = 1 + e^{-x}$. Then $du = -e^{-x} dx$,which implies $e^{-x} dx = -du$.
Substituting these into the integral,we get:
$I = \int \frac{-du}{u} = -\log |u| + C = -\log |1 + e^{-x}| + C$.
We can simplify this as:
$I = -\log \left| \frac{e^x + 1}{e^x} \right| + C = -(\log |e^x + 1| - \log |e^x|) + C = -\log |e^x + 1| + x + C$.
Alternatively,using the form $I = \int \frac{1}{e^x+1} dx = \int \frac{e^x+1-e^x}{e^x+1} dx = \int (1 - \frac{e^x}{e^x+1}) dx = x - \log |e^x+1| + C$.
Since $x = \log(e^x)$,we have $I = \log(e^x) - \log |e^x+1| + C = \log \left| \frac{e^x}{e^x+1} \right| + C$.
Thus,the correct option is $A$.

(C) To solve the integral $\int \frac{1}{x + x \log x} \, dx$,we first factor the denominator:
$\int \frac{1}{x(1 + \log x)} \, dx$.
Let $u = 1 + \log x$.
Then,the derivative is $du = \frac{1}{x} \, dx$.
Substituting these into the integral,we get:
$\int \frac{1}{u} \, du = \log |u| + C$.
Substituting back $u = 1 + \log x$,we obtain:
$\log |1 + \log x| + C$.
Thus,the correct option is $C$.

(B) We have the integral $I = \int \frac{\operatorname{cosec}^2 x}{\sec ^2 x} \, dx$.
Using the trigonometric identities $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we get:
$I = \int \frac{\cos^2 x}{\sin^2 x} \, dx = \int \cot^2 x \, dx$.
Using the identity $\cot^2 x = \operatorname{cosec}^2 x - 1$,we have:
$I = \int (\operatorname{cosec}^2 x - 1) \, dx$.
Integrating term by term,we get:
$I = \int \operatorname{cosec}^2 x \, dx - \int 1 \, dx = -\cot x - x + C$.
Thus,the correct option is $B$.

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x) + \sin^{\frac{3}{2}}(\frac{\pi}{2}-x)} \, dx$.
Since $\cos(\frac{\pi}{2}-x) = \sin x$ and $\sin(\frac{\pi}{2}-x) = \cos x$,we get $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \, dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(C) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$.
Since $\cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, 3\pi/2]$,we split the integral:
$A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} -\cos x \, dx$.
Evaluating the first part: $[\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
Evaluating the second part: $-[\sin x]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = -(-2) = 2$.
Total area $A = 1 + 2 = 3$ square units.

(C) The curve is given by $y = x|x|$.
We can define this piecewise as:
$y = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases}$
The area $A$ is given by the integral $\int_{-1}^{1} |y| \, dx$.
Since the function $y = x|x|$ is an odd function,the area bounded by the curve and the $x$-axis from $x=-1$ to $x=1$ is calculated as:
$A = \int_{-1}^{0} | -x^2 | \, dx + \int_{0}^{1} | x^2 | \, dx$
$A = \int_{-1}^{0} x^2 \, dx + \int_{0}^{1} x^2 \, dx$
$A = [\frac{x^3}{3}]_{-1}^{0} + [\frac{x^3}{3}]_{0}^{1}$
$A = (0 - (-\frac{1}{3})) + (\frac{1}{3} - 0)$
$A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ square units.

(B) The curve is given by $y = x|x|$.
Since the interval is $x \in [0, 1]$,we have $x \ge 0$,so $|x| = x$.
Thus,the curve becomes $y = x \cdot x = x^2$ for $x \in [0, 1]$.
The area $A$ bounded by the curve,the $X$-axis,and the lines $x=0$ and $x=1$ is given by the definite integral:
$A = \int_{0}^{1} |y| \, dx = \int_{0}^{1} x^2 \, dx$.
Evaluating the integral:
$A = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$.
Therefore,the area is $\frac{1}{3}$ sq. units.

(C) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
We need to find the area bounded by this curve,the $Y$-axis $(x = 0)$,and the line $y = 3$.
The region is bounded from $y = 0$ to $y = 3$.
The area $A$ is given by the integral $\int_{0}^{3} x \, dy$.
Substituting $x = \frac{y^2}{4}$,we get $A = \int_{0}^{3} \frac{y^2}{4} \, dy$.
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$.
$A = \frac{1}{4} \left( \frac{27}{3} - 0 \right) = \frac{1}{4} \times 9 = \frac{9}{4}$ sq. units.
Therefore,the correct option is $C$.

(B) The area $A$ is given by the integral of the absolute value of the function over the given interval:
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$
In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ except at the boundaries. Specifically,$\cos x \ge 0$ for $x \in [\frac{\pi}{2}, \frac{\pi}{2}]$ (not applicable) and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$.
Thus,$|\cos x| = -\cos x$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$
$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$A = -(\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2}))$
$A = -(-1 - 1) = -(-2) = 2$
Therefore,the area is $2$ sq. units.

(C) Given the differential equation: $\frac{x dy - y dx}{y} = 0$.
Multiplying by $y$ (assuming $y \neq 0$),we get: $x dy - y dx = 0$.
Rearranging the terms: $x dy = y dx$.
Separating the variables: $\frac{dy}{y} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dy}{y} = \int \frac{dx}{x}$.
This gives: $\ln|y| = \ln|x| + \ln|c|$.
Using logarithmic properties: $\ln|y| = \ln|cx|$.
Taking the exponential of both sides: $y = cx$.

(C) The given differential equation is $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$.
Rearranging the terms,we get $\frac{dx}{dy} = -\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$.
Since the equation involves the term $\frac{x}{y}$,it is a homogeneous differential equation of the form $\frac{dx}{dy} = F\left(\frac{x}{y}\right)$.
To solve such an equation,we substitute $x = vy$,where $v$ is a function of $y$.
Therefore,the correct substitution is $x = vy$.

(B) Given the differential equation: $y \log y \, dx - x \, dy = 0$
Rearranging the terms,we get: $y \log y \, dx = x \, dy$
Separating the variables,we have: $\frac{dx}{x} = \frac{dy}{y \log y}$
Integrating both sides: $\int \frac{dx}{x} = \int \frac{dy}{y \log y}$
Let $u = \log y$,then $du = \frac{1}{y} \, dy$.
The integral becomes: $\int \frac{dx}{x} = \int \frac{du}{u}$
$\log |x| = \log |u| + C_1$
$\log |x| = \log |\log y| + C_1$
Taking the exponential of both sides: $|x| = e^{C_1} |\log y|$
Let $e^{C_1} = k$,then $x = k \log y$ or $\log y = \frac{x}{k} = cx$ (where $c = 1/k$).
Therefore,$y = e^{cx}$.

(A) The general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
Since the given differential equation is of the fourth order,the value of $n$ is $4$.
Therefore,the number of arbitrary constants in its general solution is $4$.

(D) The degree of a differential equation is defined only when it is a polynomial equation in terms of its derivatives.
In the given equation,$\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0$,the term $\sin \left(\frac{d y}{d x}\right)$ involves a transcendental function of the derivative $\frac{d y}{d x}$.
Because this term cannot be expressed as a polynomial in terms of the derivatives,the differential equation is not a polynomial equation.
Therefore,the degree of this differential equation is undefined.

(A) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + \frac{2}{x} y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx}$.
$IF = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2$.
Therefore,the integrating factor is $x^2$.

(A) By definition,the general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
However,a particular solution is obtained by assigning specific values to these arbitrary constants,usually to satisfy given initial or boundary conditions.
Therefore,a particular solution contains no arbitrary constants.
Thus,for a fourth-order differential equation,the number of arbitrary constants in its particular solution is $0$.
Hence,the correct option is $A$.

(A) The order of a differential equation is defined as the order of the highest derivative present in the equation.
In the given differential equation $\left(\frac{d^3 y}{d x^3}\right)^4+\left(\frac{d^2 y}{d x^2}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0$,the derivatives present are $\frac{d^3 y}{d x^3}$,$\frac{d^2 y}{d x^2}$,and $\frac{d y}{d x}$.
The highest order derivative is $\frac{d^3 y}{d x^3}$,which is of order $3$.
Therefore,the order of the differential equation is $3$.

(C) Let the vector be $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$.
First,we find the magnitude of the vector $\vec{a}$:
$|\vec{a}| = \sqrt{(1)^2 + (-2)^2 + (3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ of a vector $x\hat{i} + y\hat{j} + z\hat{k}$ are given by $\frac{x}{|\vec{a}|}, \frac{y}{|\vec{a}|}, \frac{z}{|\vec{a}|}$.
Substituting the values,we get:
$l = \frac{1}{\sqrt{14}}$,$m = \frac{-2}{\sqrt{14}}$,$n = \frac{3}{\sqrt{14}}$.
Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.

(D) Let $\vec{r}$ be the resultant vector of $\vec{a}$ and $\vec{b}$.
$\vec{r} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} - \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.
The magnitude of $\vec{r}$ is $|\vec{r}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
The unit vector in the direction of $\vec{r}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{3\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{14}}$.
$A$ vector of magnitude $5$ units parallel to $\vec{r}$ is given by $5\hat{r} = \frac{5}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) = \frac{15}{\sqrt{14}}\hat{i} + \frac{5}{\sqrt{14}}\hat{j} - \frac{10}{\sqrt{14}}\hat{k}$.

(B) Given the equation: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$.
Expanding the left side using the distributive property of the dot product:
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2$.
Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2$,and the dot product is commutative $(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$,we get:
$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$.
Subtracting $|\vec{a}|^2 + |\vec{b}|^2$ from both sides:
$2(\vec{a} \cdot \vec{b}) = 0$.
This implies $\vec{a} \cdot \vec{b} = 0$.
Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$,the dot product of two non-zero vectors is zero if and only if they are perpendicular to each other.
Thus,the correct option is $B$.

(C) We know the properties of unit vectors: $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,and $\hat{k} \times \hat{i} = \hat{j}$.
Also,$\hat{i} \times \hat{k} = -\hat{j}$.
Substituting these values into the expression:
$\hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1$.
$\hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -1$.
$\hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{k} \cdot \hat{k} = 1$.
$\hat{j} \cdot (\hat{j} \times \hat{k}) = \hat{j} \cdot \hat{i} = 0$.
Adding these results: $1 + (-1) + 1 + 0 = 1$.

(D) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$
$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$
$= \sqrt{400 + 25 + 25} = \sqrt{450}$
$= \sqrt{225 \times 2} = 15\sqrt{2}$ square units.
Therefore,the correct option is $D$.

(B) The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Given $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
The dot product is $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$.
The magnitudes are $|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Thus,$\cos \theta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = -\frac{1}{3}$.
Therefore,$\theta = \cos^{-1}\left(-\frac{1}{3}\right)$.
Using the property $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,we get $\theta = \pi - \cos^{-1}\left(\frac{1}{3}\right)$.
Hence,the correct option is $B$.

(A) We know that the magnitude of the cross product of two vectors $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin(\theta)$,where $\theta$ is the angle between the vectors.
Given that $|\bar{a}| = \frac{2}{3}$,$|\bar{b}| = 3$,and $|\bar{a} \times \bar{b}| = 1$.
Substituting these values into the formula:
$1 = (\frac{2}{3}) \times 3 \times \sin(\theta)$
$1 = 2 \sin(\theta)$
$\sin(\theta) = \frac{1}{2}$
Since $\sin(\theta) = \frac{1}{2}$,the angle $\theta$ is $\frac{\pi}{6}$ (or $30^{\circ}$).
Therefore,the correct option is $A$.

(C) The area of a parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is given by the magnitude of their cross product,$|\bar{a} \times \bar{b}|$.
Given $\bar{a} = 0\hat{i} + 1\hat{j} + 2\hat{k}$ and $\bar{b} = 1\hat{i} + 2\hat{j} + 0\hat{k}$.
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$
$= \hat{i}(1 \times 0 - 2 \times 2) - \hat{j}(0 \times 0 - 2 \times 1) + \hat{k}(0 \times 2 - 1 \times 1)$
$= \hat{i}(-4) - \hat{j}(-2) + \hat{k}(-1) = -4\hat{i} + 2\hat{j} - \hat{k}$.
The area is $|\bar{a} \times \bar{b}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$ square units.

(D) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Given $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} + 2\hat{k}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(2) + (2)(3) + (1)(2) = 2 + 6 + 2 = 10$.
Next,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$.
Therefore,the projection is $\frac{10}{\sqrt{17}}$.

(B) The angle $\theta$ between two vectors $\bar{a}$ and $\bar{b}$ is given by the formula $\cos \theta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}| |\bar{b}|}$.
First,calculate the dot product $\bar{a} \cdot \bar{b} = (6)(4) + (2)(-4) + (-8)(2) = 24 - 8 - 16 = 0$.
Since the dot product is $0$,the vectors are perpendicular to each other.
Therefore,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(A) Given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.
First,find the vector $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
$\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.
Next,calculate the magnitude of $\vec{AB}$:
$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
The direction cosines are given by $\frac{x}{|\vec{AB}|}, \frac{y}{|\vec{AB}|}, \frac{z}{|\vec{AB}|}$.
$l = \frac{-2}{6} = -\frac{1}{3}$,$m = \frac{-4}{6} = -\frac{2}{3}$,$n = \frac{4}{6} = \frac{2}{3}$.
Thus,the direction cosines are $(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.

(B) Step $1$: Find the magnitude of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$.
$|\vec{a}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.
Step $2$: Find the unit vector in the direction of $\vec{a}$,denoted by $\hat{a}$.
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{4 \hat{i} + 3 \hat{j} - 2 \hat{k}}{\sqrt{29}}$.
Step $3$: The required vector with magnitude $2 \sqrt{29}$ in the direction of $\vec{a}$ is given by $2 \sqrt{29} \times \hat{a}$.
Vector $= 2 \sqrt{29} \times \left( \frac{4 \hat{i} + 3 \hat{j} - 2 \hat{k}}{\sqrt{29}} \right) = 2(4 \hat{i} + 3 \hat{j} - 2 \hat{k}) = 8 \hat{i} + 6 \hat{j} - 4 \hat{k}$.

(D) The direction ratios of the first line are $a_1 = 2, b_1 = 2, c_1 = 1$.
The direction ratios of the second line are $a_2 = 4, b_2 = 1, c_2 = 8$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values: $\cos \theta = \left| \frac{(2)(4) + (2)(1) + (1)(8)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}} \right|$.
$\cos \theta = \left| \frac{8 + 2 + 8}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}} \right| = \left| \frac{18}{\sqrt{9} \sqrt{81}} \right|$.
$\cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{2}{3}\right)$.

(D) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$ is given by the formula:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$
Here,the point is $(x_1, y_1, z_1) = (5, 2, -4)$ and the direction vector is $\vec{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}$,so $a=3, b=2, c=-8$.
Substituting these values into the formula,we get:
$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z-(-4)}{-8}$
$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$
Thus,the correct option is $D$.

(B) First,we rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line: $\frac{-(x-1)}{3} = \frac{7(y-2)}{2p} = \frac{z-3}{2} \implies \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$.
The direction ratios are $\vec{v_1} = (-3, \frac{2p}{7}, 2)$.
For the second line: $\frac{-7(x-1)}{3p} = \frac{y-5}{1} = \frac{-(z-6)}{5} \implies \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$.
The direction ratios are $\vec{v_2} = (-\frac{3p}{7}, 1, -5)$.
Since the lines are perpendicular,their dot product is zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.
$\frac{11p}{7} = 10$.
$p = \frac{70}{11}$.

(C) The given lines are $\frac{x-1}{-3} = \frac{y-2}{1} = \frac{z-1}{2}$ and $\frac{x-2}{p} = \frac{y-1}{2} = \frac{z-2}{1}$.
The direction ratios of the first line are $\vec{a_1} = (-3, 1, 2)$.
The direction ratios of the second line are $\vec{a_2} = (p, 2, 1)$.
Since the lines are mutually perpendicular,the dot product of their direction ratios must be zero:
$\vec{a_1} \cdot \vec{a_2} = 0$
$(-3)(p) + (1)(2) + (2)(1) = 0$
$-3p + 2 + 2 = 0$
$-3p + 4 = 0$
$3p = 4$
$p = \frac{4}{3}$.

(D) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.
The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.
The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Thus,$\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.
First,rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the third term,$\frac{6-z}{2} = \frac{-(z-6)}{2} = \frac{z-6}{-2}$.
So,the equation becomes $\frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{-2}$.
Comparing this with the standard form,the point $(x_1, y_1, z_1)$ is $(5, -4, 6)$ and the direction ratios $(a, b, c)$ are $(3, 7, -2)$.
The position vector of the point is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$ and the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} - 2\hat{k}$.
The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.
Substituting the values,we get $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$.
Thus,the correct option is $B$.

(A) The given objective function is $Z = 3x + 4y$.
The constraints are $x + y \leq 4$,$x \geq 0$,and $y \geq 0$.
The feasible region is a triangle with vertices at $(0, 0)$,$(4, 0)$,and $(0, 4)$.
We evaluate $Z$ at each vertex:
At $(0, 0)$: $Z = 3(0) + 4(0) = 0$.
At $(4, 0)$: $Z = 3(4) + 4(0) = 12$.
At $(0, 4)$: $Z = 3(0) + 4(4) = 16$.
Comparing these values,the minimum value of $Z$ is $0$.

(A) To find the maximum value of the objective function $Z = 3x + 9y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0, 10)$: $Z = 3(0) + 9(10) = 0 + 90 = 90$
$2$. At $(5, 5)$: $Z = 3(5) + 9(5) = 15 + 45 = 60$
$3$. At $(15, 15)$: $Z = 3(15) + 9(15) = 45 + 135 = 180$
$4$. At $(0, 20)$: $Z = 3(0) + 9(20) = 0 + 180 = 180$
Comparing these values,the maximum value of $Z$ is $180$.

(B) For the minimum value of the objective function $Z = px + qy$ to occur at two corner points $(3,0)$ and $(1,1)$,the value of $Z$ at these points must be equal.
At point $(3,0)$,$Z = p(3) + q(0) = 3p$.
At point $(1,1)$,$Z = p(1) + q(1) = p + q$.
Equating the two values:
$3p = p + q$
$2p = q$
$p = \frac{q}{2}$.
Thus,the condition is $p = \frac{q}{2}$.

(A) To find the maximum value of the objective function $Z = 3x + 2y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(12, 0)$: $Z = 3(12) + 2(0) = 36 + 0 = 36$
$2$. At $(4, 2)$: $Z = 3(4) + 2(2) = 12 + 4 = 16$
$3$. At $(1, 5)$: $Z = 3(1) + 2(5) = 3 + 10 = 13$
$4$. At $(1, 10)$: $Z = 3(1) + 2(10) = 3 + 20 = 23$
Comparing these values,the maximum value is $36$.

(A) To find the minimum value of the objective function $Z = 6x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.

(B) Given the equation: $P(A) + P(B) - P(A \cap B) = P(A)$.
Subtracting $P(A)$ from both sides,we get: $P(B) - P(A \cap B) = 0$.
This implies $P(B) = P(A \cap B)$.
By the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $P(A \cap B) = P(B)$ into the formula,we get $P(A|B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).

(C) Given that $2 P(A) = \frac{5}{13}$,we have $P(A) = \frac{5}{26}$.
Given that $P(B) = \frac{5}{13}$.
We know that $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$\frac{2}{5} = \frac{P(A \cap B)}{5/13}$.
Therefore,$P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.
$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$.
$P(A \cup B) = \frac{5 + 10 - 4}{26} = \frac{11}{26}$.
Thus,the correct option is $C$.