GSEB 2022 Mathematics Question Paper with Answer and Solution

(B) relation $R$ on a set $A$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. Reflexive: For every $a \in A$,$(a, a) \in R$. Here,$(1, 1), (2, 2), (3, 3) \in R$,so it is reflexive.
$2$. Symmetric: If $(a, b) \in R$,then $(b, a) \in R$. Since all elements are of the form $(a, a)$,swapping them results in the same pair,so it is symmetric.
$3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. For this relation,this condition is vacuously satisfied as there are no distinct pairs $(a, b)$ and $(b, c)$ that would require a different $(a, c)$.
Since $R$ satisfies all three properties,it is an equivalence relation.

(D) Given $f(x) = (3 - x^3)^{\frac{1}{3}}$.
First,find $(f \circ f)(x) = f(f(x)) = f((3 - x^3)^{\frac{1}{3}})$.
Substitute $f(x)$ into the function: $f(f(x)) = (3 - ((3 - x^3)^{\frac{1}{3}})^3)^{\frac{1}{3}}$.
Simplify: $f(f(x)) = (3 - (3 - x^3))^{\frac{1}{3}} = (3 - 3 + x^3)^{\frac{1}{3}} = (x^3)^{\frac{1}{3}} = x$.
Now,find $f \circ (f \circ f)(x) = f(f(f(x))) = f(x)$.
Therefore,$f \circ (f \circ f)(x) = (3 - x^3)^{\frac{1}{3}}$.

(D) The principal value branch of the inverse cosine function,denoted by $\cos^{-1} x$,is defined as the interval $[0, \pi]$.
Therefore,if $\cos^{-1} x = y$,then the range of $y$ must be $0 \leq y \leq \pi$.
Thus,the correct option is $D$.

(C) We know that $\cos(-\theta) = \cos(\theta)$. Therefore,$\cos(-680^{\circ}) = \cos(680^{\circ})$.
We can express $680^{\circ}$ as $720^{\circ} - 40^{\circ}$,which is $4\pi - 40^{\circ}$.
Since $\cos(2n\pi - \theta) = \cos(\theta)$,we have $\cos(680^{\circ}) = \cos(40^{\circ})$.
The principal value branch of $\cos^{-1}(x)$ is $[0, \pi]$.
Thus,$\cos^{-1}[\cos(40^{\circ})] = 40^{\circ}$.
Converting $40^{\circ}$ to radians: $40 \times \frac{\pi}{180} = \frac{4\pi}{18} = \frac{2\pi}{9}$.
Therefore,the correct option is $C$.

(A) We know the identity for inverse trigonometric functions: $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for all $|x| \geq 1$.
Substituting this into the given expression:
$\cos \left[\sec^{-1} x + \operatorname{cosec}^{-1} x\right] = \cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the value of the expression is $0$.

(B) Let $y = \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.
Then $\cot y = \frac{-1}{\sqrt{3}}$.
We know that the range of the principal value branch of $\cot^{-1} x$ is $(0, \pi)$.
Since $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,we have $\cot\left(\pi - \frac{\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.
Thus,$\cot\left(\frac{2\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.
Since $\frac{2\pi}{3} \in (0, \pi)$,the principal value is $\frac{2\pi}{3}$.

(A) Let $\theta = \tan^{-1} x$.
Then,$\tan \theta = x = \frac{x}{1}$.
We know that in a right-angled triangle,$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.
Here,the opposite side is $x$ and the adjacent side is $1$.
Using the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Now,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{1+x^2}}$.
Therefore,$\cos (\tan^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.
Thus,the correct option is $A$.

(C) We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$ for $x > 0$.
Given expression is $\cot ^{-1}(-\sqrt{3}) - \tan ^{-1}(\sqrt{3})$.
Using the property,$\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3})$.
Since $\cot ^{-1}(\sqrt{3}) = \frac{\pi}{6}$ and $\tan ^{-1}(\sqrt{3}) = \frac{\pi}{3}$,
Substituting these values:
$\cot ^{-1}(-\sqrt{3}) - \tan ^{-1}(\sqrt{3}) = (\pi - \frac{\pi}{6}) - \frac{\pi}{3}$.
$= \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi - 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.
Thus,the correct option is $C$.

(C) We know that the range of the principal value branch of $\sin^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Given expression is $\sin^{-1}\left(\sin \frac{7 \pi}{6}\right)$.
Since $\frac{7 \pi}{6}$ does not lie in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we simplify the expression.
$\sin \frac{7 \pi}{6} = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin \frac{\pi}{6} = \sin\left(-\frac{\pi}{6}\right)$.
Now,$\sin^{-1}\left(\sin \frac{7 \pi}{6}\right) = \sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)$.
Since $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,the value is $-\frac{\pi}{6}$.
Therefore,the correct option is $C$.

(B) Given that $A^2 = A$.
We need to evaluate $(I + A)^3 - 8A$.
Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
Since $I^n = I$ and $I \times A = A$,this simplifies to $I + 3A + 3A^2 + A^3$.
Given $A^2 = A$,we can substitute $A^2$ with $A$:
$A^3 = A^2 \times A = A \times A = A^2 = A$.
Substituting these into the expression:
$(I + A)^3 = I + 3A + 3(A) + A = I + 7A$.
Now,subtract $8A$:
$(I + 7A) - 8A = I - A$.

(B) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Then the transpose $A^{\prime} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Given $A + A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Adding $A$ and $A^{\prime}$:
$A + A^{\prime} = \begin{bmatrix} \sin \alpha + \sin \alpha & -\cos \alpha + \cos \alpha \\ \cos \alpha - \cos \alpha & \sin \alpha + \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix}$.
Equating to $I$:
$\begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $2 \sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.
The principal value of $\alpha$ for $\sin \alpha = \frac{1}{2}$ is $\alpha = \frac{\pi}{6}$.

(C) We are given matrices $A$ and $B$. We need to compute $A - B$.
$A - B = \frac{1}{\pi} \left[ \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} - \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix} \right]$
Subtracting the corresponding elements:
$A - B = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) - (-\cos^{-1}(\pi x)) & \tan^{-1}(\frac{x}{\pi}) - \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) - (-\tan^{-1}(\pi x)) \end{bmatrix}$
Using the trigonometric identities $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ and $\tan^{-1}(\theta) + \cot^{-1}(\theta) = \frac{\pi}{2}$:
$A - B = \frac{1}{\pi} \begin{bmatrix} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$
Thus,the correct option is $C$.

(D) Given the determinant equation: $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
Expanding the determinants on both sides:
$(3 \times 1) - (x \times x) = (3 \times 1) - (2 \times 4)$
$3 - x^2 = 3 - 8$
$3 - x^2 = -5$
Subtracting $3$ from both sides:
$-x^2 = -8$
$x^2 = 8$
Taking the square root on both sides:
$x = \pm \sqrt{8}$
$x = \pm 2\sqrt{2}$
Thus,the correct option is $D$.

(C) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$,and the area is $4$ sq. units.
Substituting the values into the formula:
$4 = \frac{1}{2} |-2(4 - k) + 0(k - 0) + 0(0 - 4)|$
$4 = \frac{1}{2} |-8 + 2k|$
$8 = |-8 + 2k|$
This implies two cases:
Case $1$: $-8 + 2k = 8 \implies 2k = 16 \implies k = 8$
Case $2$: $-8 + 2k = -8 \implies 2k = 0 \implies k = 0$
Thus,the possible values of $k$ are $0$ and $8$.

(C) To find the determinant of matrix $A$,we expand along the first row:
$\operatorname{det}(A) = 1(1 - (-\cos^2 \theta)) - (-\cos \theta)(\cos \theta - (-1)) + (-1)(\cos^2 \theta - 1)$
$\operatorname{det}(A) = 1(1 + \cos^2 \theta) + \cos \theta(\cos \theta + 1) - 1(\cos^2 \theta - 1)$
$\operatorname{det}(A) = 1 + \cos^2 \theta + \cos^2 \theta + \cos \theta - \cos^2 \theta + 1$
$\operatorname{det}(A) = \cos^2 \theta + \cos \theta + 2$
Given $0 < \theta < \frac{\pi}{2}$,we have $0 < \cos \theta < 1$.
Let $f(x) = x^2 + x + 2$ where $x = \cos \theta$.
Since $x \in (0, 1)$,the range of $f(x)$ is $(0^2 + 0 + 2, 1^2 + 1 + 2) = (2, 4)$.
Thus,$\operatorname{det}(A) \in (2, 4)$.

(D) To find the inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we first calculate its determinant $|A| = ad - bc$.
Given $A = \begin{bmatrix} 8 & -2 \\ -4 & 1 \end{bmatrix}$,the determinant is $|A| = (8)(1) - (-2)(-4) = 8 - 8 = 0$.
Since the determinant of the matrix is $0$,the matrix $A$ is a singular matrix.
$A$ singular matrix does not have an inverse.
Therefore,$A^{-1}$ does not exist.

(A) Let $y = \cos^{-1}(\sin x)$.
We know that $\sin x = \cos(\frac{\pi}{2} - x)$.
Substituting this into the expression,we get $y = \cos^{-1}(\cos(\frac{\pi}{2} - x))$.
Since $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$,we have $y = \frac{\pi}{2} - x$.
Now,differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2} - x) = 0 - 1 = -1$.
Thus,the derivative is $-1$.

(D) Given the equation $e^y(x+1)=1$.
Taking the natural logarithm on both sides: $\ln(e^y) + \ln(x+1) = \ln(1)$.
$y + \ln(x+1) = 0$.
Differentiating with respect to $x$: $\frac{d y}{d x} + \frac{1}{x+1} = 0$.
So,$\frac{d y}{d x} = -\frac{1}{x+1} = -(x+1)^{-1}$.
Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$.
Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$.
Therefore,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

(A) To evaluate the integral $I = \int x^2 e^{x^3} dx$,we use the method of substitution.
Let $u = x^3$.
Then,differentiating both sides with respect to $x$,we get $du = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} du$.
Substituting these into the integral,we have:
$I = \int e^u \cdot \frac{1}{3} du$
$I = \frac{1}{3} \int e^u du$
$I = \frac{1}{3} e^u + c$
Substituting back $u = x^3$,we get:
$I = \frac{1}{3} e^{x^3} + c$.
Thus,the correct option is $A$.

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

(B) Let $f(x) = \sin^5 x \cos^4 x$.
We check if the function is even or odd.
$f(-x) = \sin^5(-x) \cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) \, dx = 0$.
Therefore,$\int_{-1}^1 \sin^5 x \cos^4 x \, dx = 0$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$.
We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x dx$.
Recall the property of definite integrals: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is an even function.
For $f_1(x) = x^3$,$f_1(-x) = (-x)^3 = -x^3 = -f_1(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx = 0$.
For $f_2(x) = \cos x$,$f_2(-x) = \cos(-x) = \cos x = f_2(x)$,so it is an even function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx = 2 \int_{0}^{\frac{\pi}{2}} \cos x dx = 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1-0) = 2$.
For $f_3(x) = \tan^5 x$,$f_3(-x) = \tan^5(-x) = -\tan^5 x = -f_3(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x dx = 0$.
Therefore,$I = 0 + 2 + 0 = 2$.

(A) To evaluate $\int \sqrt{x^2+4x+1} \, dx$,first complete the square inside the square root:
$x^2+4x+1 = (x^2+4x+4) - 3 = (x+2)^2 - (\sqrt{3})^2$.
Now the integral becomes $\int \sqrt{(x+2)^2 - (\sqrt{3})^2} \, dx$.
Using the standard formula $\int \sqrt{t^2-a^2} \, dt = \frac{t}{2} \sqrt{t^2-a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2-a^2}| + C$,where $t = x+2$ and $a = \sqrt{3}$:
$= \frac{x+2}{2} \sqrt{(x+2)^2 - 3} - \frac{3}{2} \log |(x+2) + \sqrt{(x+2)^2 - 3}| + C$.
$= \frac{x+2}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \log |x+2 + \sqrt{x^2+4x+1}| + C$.
Comparing this with the given options,the correct option is $A$.

(C) We use the standard formula: $\int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] + C$.
Here,$a = 3$,$b = 4$,and the constant term is $-5$.
Substituting these values into the formula:
$\int e^{3x} \sin(4x-5) dx = \frac{e^{3x}}{3^2 + 4^2} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$.
$= \frac{e^{3x}}{9 + 16} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$.
$= \frac{e^{3x}}{25} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$.
Thus,the correct option is $C$.

(C) We need to evaluate $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.
First,expand the expression inside the square root: $(x-1)(x-2) = x^2 - 3x + 2$.
Complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4}$.
Now the integral becomes $I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx$.
Using the standard formula $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}| + C$,where $t = x - \frac{3}{2}$ and $a = \frac{1}{2}$:
$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + C$.
Simplifying the expression inside the square root back to the original form:
$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + C$.
Thus,the correct option is $C$.

(A) We need to evaluate the definite integral $I = \int_{0}^{\pi} \sin x \, dx$.
The antiderivative of $\sin x$ is $-\cos x$.
Applying the Fundamental Theorem of Calculus:
$I = [-\cos x]_{0}^{\pi}$
$I = -(\cos \pi - \cos 0)$
$I = -(-1 - 1)$
$I = -(-2)$
$I = 2$
Therefore,the correct option is $A$.

(A) To solve the integral $I = \int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx$,we factor out $x^4$ from the expression inside the parenthesis:
$I = \int \frac{(x^4(1+\frac{1}{x^3}))^{\frac{1}{4}}}{x^5} dx$
$I = \int \frac{x(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^5} dx = \int \frac{(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^4} dx$
Let $u = 1 + \frac{1}{x^3}$. Then $du = -\frac{3}{x^4} dx$,which implies $\frac{dx}{x^4} = -\frac{1}{3} du$.
Substituting these into the integral:
$I = \int u^{\frac{1}{4}} (-\frac{1}{3}) du = -\frac{1}{3} \int u^{\frac{1}{4}} du$
$I = -\frac{1}{3} \cdot \frac{u^{\frac{5}{4}}}{\frac{5}{4}} + C = -\frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}} + C$
$I = -\frac{4}{15} (1+\frac{1}{x^3})^{\frac{5}{4}} + C$.
Thus,the correct option is $A$.

(D) Let $I = \int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$.
Factor out $x^3$ from the term inside the cube root:
$I = \int_{\frac{1}{3}}^1 \frac{(x^3(\frac{1}{x^2}-1))^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{x(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3} dx$.
Let $u = \frac{1}{x^2} - 1$. Then $du = -\frac{2}{x^3} dx$,which implies $\frac{dx}{x^3} = -\frac{1}{2} du$.
When $x = \frac{1}{3}$,$u = \frac{1}{(1/3)^2} - 1 = 9 - 1 = 8$.
When $x = 1$,$u = \frac{1}{1^2} - 1 = 0$.
Substituting these into the integral:
$I = \int_8^0 u^{\frac{1}{3}} (-\frac{1}{2}) du = \frac{1}{2} \int_0^8 u^{\frac{1}{3}} du$.
$I = \frac{1}{2} [\frac{u^{\frac{4}{3}}}{4/3}]_0^8 = \frac{1}{2} \cdot \frac{3}{4} [u^{\frac{4}{3}}]_0^8 = \frac{3}{8} (8^{\frac{4}{3}} - 0) = \frac{3}{8} (16) = 6$.
Thus,the correct option is $D$.

(B) Let $I = \frac{1}{2} \int_2^3 \frac{2x}{x^2+1} dx$.
Using the substitution method,let $u = x^2+1$.
Then $du = 2x dx$.
When $x=2$,$u = 2^2+1 = 5$.
When $x=3$,$u = 3^2+1 = 10$.
Substituting these into the integral:
$I = \frac{1}{2} \int_5^{10} \frac{1}{u} du$
$I = \frac{1}{2} [\log |u|]_5^{10}$
$I = \frac{1}{2} (\log 10 - \log 5)$
$I = \frac{1}{2} \log \left(\frac{10}{5}\right)$
$I = \frac{1}{2} \log (2)$.

(C) Let $I = \int \tan ^8 x \sec ^4 x \, dx$.
We can rewrite $\sec ^4 x$ as $\sec ^2 x \cdot \sec ^2 x$.
So,$I = \int \tan ^8 x \cdot \sec ^2 x \cdot \sec ^2 x \, dx$.
Since $\sec ^2 x = 1 + \tan ^2 x$,we have:
$I = \int \tan ^8 x (1 + \tan ^2 x) \sec ^2 x \, dx$.
Let $u = \tan x$,then $du = \sec ^2 x \, dx$.
Substituting these into the integral:
$I = \int u^8 (1 + u^2) \, du = \int (u^8 + u^{10}) \, du$.
Integrating with respect to $u$:
$I = \frac{u^9}{9} + \frac{u^{11}}{11} + c$.
Substituting $u = \tan x$ back:
$I = \frac{\tan ^9 x}{9} + \frac{\tan ^{11} x}{11} + c$.

(A) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
First,rewrite the integrand: $\frac{x^2+1}{(x+1)^2} = \frac{x^2-1+2}{(x+1)^2} = \frac{(x-1)(x+1)+2}{(x+1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}$.
Let $f(x) = \frac{x-1}{x+1}$.
Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$.
Thus,the integral becomes $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c = \left( \frac{x-1}{x+1} \right) e^x + c$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 36$.
Dividing both sides by $36$, we get:
$\frac{9x^2}{36} + \frac{4y^2}{36} = 1$
$\frac{x^2}{4} + \frac{y^2}{9} = 1$
This is of the form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a^2 = 9$ and $b^2 = 4$.
So, $a = 3$ and $b = 2$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values, we get:
$A = \pi \times 3 \times 2 = 6\pi$ sq. units.

(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.
The latus rectum is the line $x = a = 3$.
The area bounded by the parabola and its latus rectum is given by $A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{3} \sqrt{12x} \, dx$.
$A = 2 \times \sqrt{12} \int_{0}^{3} x^{1/2} \, dx = 2 \times 2\sqrt{3} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3}$.
$A = 4\sqrt{3} \times \frac{2}{3} \times (3)^{3/2} = \frac{8\sqrt{3}}{3} \times 3\sqrt{3} = 8 \times 3 = 24$ sq. units.

(B) To find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$,we use the method of separation of variables.
Step $1$: Separate the variables $x$ and $y$:
$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$
Step $2$: Integrate both sides:
$\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$
Step $3$: Apply the standard integration formula $\int \frac{du}{1+u^2} = \tan^{-1} u + c$:
$\tan^{-1} y = \tan^{-1} x + c$
Thus,the general solution is $\tan^{-1} y = \tan^{-1} x + c$.

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

(C) Given the differential equation: $1+(\frac{dy}{dx})^2=\sqrt{\frac{d^2y}{dx^2}}$
To find the order and degree,we first eliminate the radical by squaring both sides:
$(1+(\frac{dy}{dx})^2)^2 = \frac{d^2y}{dx^2}$
The highest order derivative present in the equation is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $1$,so the degree is $1$.
Therefore,the order and degree are $2$ and $1$ respectively.

(D) Given the differential equation: $\left(\frac{d^2 y}{d x^2}\right)^{1/3} = \left(\frac{d^3 y}{d x^3}\right)^{1/2}$.
To eliminate the fractional exponents,raise both sides to the power of $6$ (the least common multiple of $2$ and $3$):
$\left(\left(\frac{d^2 y}{d x^2}\right)^{1/3}\right)^6 = \left(\left(\frac{d^3 y}{d x^3}\right)^{1/2}\right)^6$.
This simplifies to: $\left(\frac{d^2 y}{d x^2}\right)^2 = \left(\frac{d^3 y}{d x^3}\right)^3$.
The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The power to which the highest order derivative is raised is $3$,so the degree is $3$.
Therefore,the order and degree are $3$ and $3$ respectively.

(A) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = kx$.
Dividing both sides by $(1-x^2)$,we get:
$\frac{dy}{dx} + \frac{x}{1-x^2} y = \frac{kx}{1-x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{1-x^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx}$.
$IF = e^{\int \frac{x}{1-x^2} dx}$.
Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$IF = e^{-\frac{1}{2} \int \frac{1}{u} du} = e^{-\frac{1}{2} \ln|u|} = e^{\ln|u|^{-1/2}} = |u|^{-1/2} = \frac{1}{\sqrt{1-x^2}}$.
Thus,the correct option is $A$.

(C) We know the properties of the cross product of unit vectors:
$\hat{j} \times \hat{k} = \hat{i}$
$\hat{i} \times \hat{k} = -\hat{j}$
$\hat{i} \times \hat{j} = \hat{k}$
Substituting these into the expression:
$\hat{i} \cdot (\hat{i}) + \hat{j} \cdot (-\hat{j}) + \hat{k} \cdot (\hat{k})$
$= (\hat{i} \cdot \hat{i}) - (\hat{j} \cdot \hat{j}) + (\hat{k} \cdot \hat{k})$
Since the dot product of a unit vector with itself is $1$:
$= 1 - 1 + 1 = 1$
Thus,the correct option is $C$.

(D) 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 $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$.
Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Now,substitute these values into the formula: $\cos \theta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{3}\right)$.
Comparing this with the given options,the correct answer is not explicitly listed as $\cos^{-1}(1/3)$,so the correct choice is $D$.

(C) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$
$= \hat{i}(1(1) - 4(-1)) - \hat{j}(3(1) - 4(1)) + \hat{k}(3(-1) - 1(1))$
$= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)$
$= 5\hat{i} + \hat{j} - 4\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2}$
$= \sqrt{25 + 1 + 16} = \sqrt{42}$ sq. units.
Thus,the correct option is $C$.

(C) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the relationship between their dot product and cross product is given by the identity:
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
Given values are $|\vec{a}|=10$,$|\vec{b}|=2$,and $\vec{a} \cdot \vec{b}=12$.
Substituting these values into the identity:
$|\vec{a} \times \vec{b}|^2 + (12)^2 = (10)^2 (2)^2$
$|\vec{a} \times \vec{b}|^2 + 144 = 100 \times 4$
$|\vec{a} \times \vec{b}|^2 + 144 = 400$
$|\vec{a} \times \vec{b}|^2 = 400 - 144$
$|\vec{a} \times \vec{b}|^2 = 256$
Taking the square root on both sides:
$|\vec{a} \times \vec{b}| = \sqrt{256} = 16$
Therefore,the correct option is $C$.

(B) Given that $\vec{a}$ is a unit vector,so $|\vec{a}| = 1$.
The given equation is $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$.
Using the dot product property $(A-B) \cdot (A+B) = |A|^2 - |B|^2$,we get:
$|\vec{x}|^2 - |\vec{a}|^2 = 8$.
Since $|\vec{a}| = 1$,we have $|\vec{a}|^2 = 1$.
Substituting this into the equation:
$|\vec{x}|^2 - 1 = 8$.
$|\vec{x}|^2 = 9$.
Taking the square root on both sides,$|\vec{x}| = 3$ (since magnitude is always non-negative).
Therefore,the correct option is $B$.

(A) Let the given vector be $\vec{a} = 5 \hat{i} - \hat{j} + 2 \hat{k}$.
First,find the magnitude of vector $\vec{a}$:
$|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}$.
The unit vector in the direction of $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{5 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{30}}$.
$A$ vector of magnitude $8$ units in the direction of $\vec{a}$ is $8 \hat{a} = 8 \times \left( \frac{5 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{30}} \right) = \frac{40}{\sqrt{30}} \hat{i} - \frac{8}{\sqrt{30}} \hat{j} + \frac{16}{\sqrt{30}} \hat{k}$.
Thus,the correct option is $A$.

(D) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Given the point is $(2, 3, 4)$,so $x_1 = 2, y_1 = 3, z_1 = 4$.
The line is parallel to the $Y$-axis. The direction ratios of the $Y$-axis are $(0, 1, 0)$.
Therefore,the direction ratios of the line are $(a, b, c) = (0, 1, 0)$.
Substituting these values into the equation,we get $\frac{x-2}{0} = \frac{y-3}{1} = \frac{z-4}{0}$.
Thus,the correct option is $D$.

(D) The given lines are $\frac{x+3}{2}=\frac{y}{-3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$.
The direction ratios of the first line are $\vec{b_1} = (2, -3, -6)$ and the direction ratios of the second line are $\vec{b_2} = (10, -2, 11)$.
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 $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_1^2}}$.
Calculating the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(10) + (-3)(-2) + (-6)(11) = 20 + 6 - 66 = -40$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$|\vec{b_2}| = \sqrt{10^2 + (-2)^2 + 11^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.
Thus,$\cos \theta = \frac{|-40|}{7 \times 15} = \frac{40}{105} = \frac{8}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{8}{21}\right)$.

(C) According to the Fundamental Theorem of Linear Programming,if the feasible region for a linear programming problem is bounded,then the objective function $Z = ax + by$ must attain both a maximum and a minimum value at the corner points (vertices) of the feasible region. Therefore,the correct option is $C$.

(B) To find the maximum value of the objective function $Z = 4x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0,0): Z = 4(0) + 3(0) = 0$
$2$. At $(25,5): Z = 4(25) + 3(5) = 100 + 15 = 115$
$3$. At $(16,16): Z = 4(16) + 3(16) = 64 + 48 = 112$
$4$. At $(5,24): Z = 4(5) + 3(24) = 20 + 72 = 92$
Comparing these values,the maximum value is $115$,which occurs at the point $(25,5)$.

(A) For the maximum of $Z = qx + py$ to occur at two corner points $(3,4)$ and $(0,5)$,the value of $Z$ must be equal at these two points.
At $(3,4)$,$Z = q(3) + p(4) = 3q + 4p$.
At $(0,5)$,$Z = q(0) + p(5) = 5p$.
Equating the two values: $3q + 4p = 5p$.
Subtracting $4p$ from both sides,we get $3q = p$,or $p = 3q$.
Thus,the correct option is $A$.

(D) To find the maximum value of the objective function $Z = 4x + y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0,0)$: $Z = 4(0) + 0 = 0$
$2$. At $(30,0)$: $Z = 4(30) + 0 = 120$
$3$. At $(20,30)$: $Z = 4(20) + 30 = 80 + 30 = 110$
$4$. At $(0,50)$: $Z = 4(0) + 50 = 50$
Comparing these values,the maximum value of $Z$ is $120$ at the point $(30,0)$.