GSEB 2016 Mathematics Question Paper with Answer and Solution

(A) The given system of linear equations is:
$ax - y = -2$
$x + ay = -3$
For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Here,$a_1 = a$,$b_1 = -1$,$a_2 = 1$,and $b_2 = a$.
Substituting these values into the condition:
$\frac{a}{1} \neq \frac{-1}{a}$
$a^2 \neq -1$
Since $a^2$ is always non-negative for any real number $a$,$a^2$ can never be equal to $-1$.
Therefore,the condition $a^2 \neq -1$ is satisfied for all real numbers $a$.
Thus,the set of values of $a$ is $R$.

(C) The slope $m$ of the line passing through points $(x_1, y_1) = (-7, 8)$ and $(x_2, y_2) = (5, 2)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 8}{5 - (-7)} = \frac{-6}{12} = -\frac{1}{2}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(-7, 8)$:
$y - 8 = -\frac{1}{2}(x - (-7))$
$2(y - 8) = -(x + 7)$
$2y - 16 = -x - 7$
$x + 2y - 9 = 0$.
Therefore,the correct option is $C$.

(B) Given the function $f(x) = 3 \sin x - 4 \sin^3 x$.
We know the trigonometric identity $\sin(3x) = 3 \sin x - 4 \sin^3 x$.
Therefore,$f(x) = \sin(3x)$.
Now,substitute $x = \frac{\pi}{3}$ into the function:
$f\left(\frac{\pi}{3}\right) = \sin\left(3 \times \frac{\pi}{3}\right) = \sin(\pi)$.
Since $\sin(\pi) = 0$,the value is $0$.
Thus,the correct option is $B$.

(A) To determine which statement is true,we first calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = 4I$.
Since $A^2 = 4I$,option $A$ is correct.
We can also check the determinant: $|A| = 0(0 - 0) - 0(0 - 0) - 2(0 - 4) = -2(-4) = 8$.
Since $|A| \neq 0$,$A^{-1}$ exists.
$A$ is not a diagonal matrix because it has non-zero elements off the main diagonal.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2A$.
Next,calculate $A^3 = A^2 \times A = (2A) \times A = 2A^2 = 2(2A) = 2^2 A$.
Similarly,$A^4 = A^3 \times A = (2^2 A) \times A = 2^2 A^2 = 2^2(2A) = 2^3 A$.
By induction,we can generalize that $A^n = 2^{n-1} A$ for any positive integer $n \geq 1$.
Therefore,for $n = 100$,$A^{100} = 2^{100-1} A = 2^{99} A$.

(A) The characteristic equation of a matrix $A$ is given by $|A - \lambda I| = 0$.
For $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{bmatrix}$,the characteristic equation is:
$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & 1 \\ 2 & 1-\lambda & 3 \\ 1 & 1 & -\lambda \end{vmatrix} = 0$.
Expanding the determinant:
$(1-\lambda)[(1-\lambda)(-\lambda) - 3] - 2[2(-\lambda) - 3] + 1[2 - (1-\lambda)] = 0$.
$(1-\lambda)(\lambda^2 - \lambda - 3) - 2(-2\lambda - 3) + (1 + \lambda) = 0$.
$(\lambda^2 - \lambda - 3 - \lambda^3 + \lambda^2 + 3\lambda) + 4\lambda + 6 + 1 + \lambda = 0$.
$-\lambda^3 + 2\lambda^2 + 7\lambda + 4 = 0$.
Multiplying by $-1$,we get $\lambda^3 - 2\lambda^2 - 7\lambda - 4 = 0$.
By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation:
$A^3 - 2A^2 - 7A - 4I_3 = 0$.
Comparing this with the given equation $A^3 - 2A^2 + kA - 4I_3 = 0$,we find $k = -7$.

(C) Let $x = \frac{1}{m}$ and $y = \frac{1}{n}$.
Then the given equations become:
$5x + 2y = 9$ --- $(1)$
$3x + 4y = 11$ --- $(2)$
Multiply equation $(1)$ by $2$ to eliminate $y$:
$10x + 4y = 18$ --- $(3)$
Subtract equation $(2)$ from equation $(3)$:
$(10x - 3x) + (4y - 4y) = 18 - 11$
$7x = 7 \implies x = 1$
Substitute $x = 1$ into equation $(1)$:
$5(1) + 2y = 9$
$5 + 2y = 9 \implies 2y = 4 \implies y = 2$
Since $x = \frac{1}{m} = 1$,we get $m = 1$.
Since $y = \frac{1}{n} = 2$,we get $n = \frac{1}{2}$.
Thus,the values are $m = 1$ and $n = \frac{1}{2}$.

(D) We know that for any invertible matrix $A$,the product $A \cdot A^{-1} = I$,where $I$ is the identity matrix of the same order.
Given $A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$.
Therefore,$A \cdot A^{-1} = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Performing matrix multiplication:
$\begin{bmatrix} (2x)(1) + (0)(-1) & (2x)(0) + (0)(2) \\ (x)(1) + (x)(-1) & (x)(0) + (x)(2) \end{bmatrix} = \begin{bmatrix} 2x & 0 \\ 0 & 2x \end{bmatrix}$.
Equating this to the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we get:
$2x = 1$.
Solving for $x$,we get $x = \frac{1}{2}$.
Thus,the correct option is $D$.

(C) Let the given determinants be $D_1, D_2$ and $D_3$ respectively.
$D_1 = \left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right| = x(3+54) - 4(2+45) + 6(12-15) = 57x - 188 - 18 = 57x - 206$.
$D_2 = \left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right| = 5(-36-15) - 6(-54-10) + 1(18-8) = 5(-51) - 6(-64) + 10 = -255 + 384 + 10 = 139$.
$D_3 = \left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2x & -8 & -11 \\ 5 & 6 & 1\end{array}\right| = 2(-8+66) - 3(1-2x+55) - 9(6-12x+40) = 2(58) - 3(56-2x) - 9(46-12x) = 116 - 168 + 6x - 414 + 108x = 114x - 466$.
Given $D_1 + D_2 = D_3$,so $(57x - 206) + 139 = 114x - 466$.
$57x - 67 = 114x - 466$.
$466 - 67 = 114x - 57x$.
$399 = 57x$.
$x = \frac{399}{57} = 7$.
Thus,the correct option is $C$.

(D) First,evaluate the trigonometric values:
$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$.
Substituting these into the determinant:
$D = \left| \begin{array}{ccc} 2(\frac{\sqrt{3}}{2}) & 1 & 0 \\ 1 & 2(\frac{\sqrt{3}}{2}) & 1 \\ 0 & 1 & 2(\frac{\sqrt{3}}{2}) \end{array} \right| = \left| \begin{array}{ccc} \sqrt{3} & 1 & 0 \\ 1 & \sqrt{3} & 1 \\ 0 & 1 & \sqrt{3} \end{array} \right|$.
Expanding along the first row:
$D = \sqrt{3} ((\sqrt{3})(\sqrt{3}) - (1)(1)) - 1 ((1)(\sqrt{3}) - (1)(0)) + 0
= \sqrt{3} (3 - 1) - 1 (\sqrt{3})
= \sqrt{3} (2) - \sqrt{3}
= 2\sqrt{3} - \sqrt{3}
= \sqrt{3}$.
Since the calculated value is $\sqrt{3}$,and the provided options were incorrect,the correct value is $\sqrt{3}$.

(B) Given the matrix $A = \begin{bmatrix} 5x & 10 \\ 8 & 7 \end{bmatrix}$.
The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $|A| = ad - bc$.
Applying this to matrix $A$:
$|A| = (5x)(7) - (10)(8)$
$|A| = 35x - 80$
We are given that $|A| = 25$. Therefore:
$35x - 80 = 25$
$35x = 25 + 80$
$35x = 105$
$x = \frac{105}{35}$
$x = 3$
Thus,the correct value is $3$.

(A) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = \alpha^2 - (2 \times 2) = \alpha^2 - 4$.
We are given that $|A^3| = 125$.
Using the property of determinants $|A^n| = |A|^n$,we have $|A|^3 = 125$.
Taking the cube root on both sides,we get $|A| = \sqrt[3]{125} = 5$.
Now,substitute the value of $|A|$ back into the equation:
$\alpha^2 - 4 = 5$.
$\alpha^2 = 9$.
$\alpha = \pm 3$.
Therefore,the correct option is $A$.

(A) Given $y = 5 \sin x + 6 \cos x$.
First,find the derivative $y_1 = \frac{dy}{dx} = 5 \cos x - 6 \sin x$.
Now,calculate $y^2 + (y_1)^2$:
$y^2 = (5 \sin x + 6 \cos x)^2 = 25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x$.
$(y_1)^2 = (5 \cos x - 6 \sin x)^2 = 25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x$.
Adding these two expressions:
$y^2 + (y_1)^2 = (25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x) + (25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x)$.
$y^2 + (y_1)^2 = 25(\sin^2 x + \cos^2 x) + 36(\cos^2 x + \sin^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,we get:
$y^2 + (y_1)^2 = 25(1) + 36(1) = 61$.

(A) Let $y = x^{2 x}$.
Taking the natural logarithm on both sides,we get $\log y = \log(x^{2 x})$.
Using the property $\log(a^b) = b \log a$,we have $\log y = 2 x \log x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{d y}{d x} = 2 \cdot \frac{d}{d x}(x \log x)$.
$\frac{1}{y} \frac{d y}{d x} = 2 \left( x \cdot \frac{1}{x} + \log x \cdot 1 \right)$.
$\frac{1}{y} \frac{d y}{d x} = 2(1 + \log x)$.
Therefore,$\frac{d y}{d x} = y \cdot 2(1 + \log x)$.
Substituting $y = x^{2 x}$,we get $\frac{d y}{d x} = 2 x^{2 x}(1 + \log x)$.
Thus,the correct option is $A$.

(C) Given $y = (\cos^{-1} x)^2$.
Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$.
So,$\sqrt{1-x^2} y_1 = -2 \cos^{-1} x$.
Squaring both sides,we get $(1-x^2) y_1^2 = 4 (\cos^{-1} x)^2 = 4y$.
Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2y_1 y_2 + y_1^2 (-2x) = 4y_1$.
Dividing by $2y_1$ (assuming $y_1 \neq 0$),we get $(1-x^2) y_2 - x y_1 = 2$.
Thus,$(1-x^2) y_2 - x y_1 - 2 = 0$.
Comparing this with $(1-x^2) y_2 + p x y_1 + q = 0$,we get $p = -1$ and $q = -2$.
Therefore,$p+q = -1 + (-2) = -3$.

(C) Let $y = \tan^{-1} \left( \frac{x}{1+6x^2} \right)$.
We can rewrite the argument of $\tan^{-1}$ as $\frac{3x - 2x}{1 + (3x)(2x)}$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$,we get:
$y = \tan^{-1}(3x) - \tan^{-1}(2x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1}(3x)) - \frac{d}{dx} (\tan^{-1}(2x))$.
Using the chain rule $\frac{d}{dx} (\tan^{-1}(u)) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = \frac{1}{1+(3x)^2} \cdot 3 - \frac{1}{1+(2x)^2} \cdot 2$.
$\frac{dy}{dx} = \frac{3}{1+9x^2} - \frac{2}{1+4x^2}$.
Thus,the correct option is $C$.

(A) Let $y = e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.
We know that $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.
So,the first term becomes $e^{\pi/2}$,which is a constant. The derivative of a constant is $0$.
Now,differentiate the remaining terms: $\frac{d}{dx} \left[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin ^{-1} \frac{x}{a} \right]$.
Using the product rule for the first part: $\frac{d}{dx} \left( \frac{x}{2} \sqrt{a^2-x^2} \right) = \frac{1}{2} \sqrt{a^2-x^2} + \frac{x}{2} \cdot \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x) = \frac{1}{2} \sqrt{a^2-x^2} - \frac{x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-x^2-x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-2x^2}{2\sqrt{a^2-x^2}}$.
For the second part: $\frac{d}{dx} \left( \frac{a^2}{2} \sin ^{-1} \frac{x}{a} \right) = \frac{a^2}{2} \cdot \frac{1}{\sqrt{1-(x/a)^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{1}{\sqrt{(a^2-x^2)/a^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{a}{\sqrt{a^2-x^2}} \cdot \frac{1}{a} = \frac{a^2}{2\sqrt{a^2-x^2}}$.
Adding these results: $\frac{a^2-2x^2}{2\sqrt{a^2-x^2}} + \frac{a^2}{2\sqrt{a^2-x^2}} = \frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}} = \sqrt{a^2-x^2}$.

(A) To find the derivative of $f(x) = \sin(x^2) + \cos(x^2)$,we use the chain rule.
Let $u = x^2$,then $\frac{du}{dx} = 2x$.
The derivative is given by $\frac{d}{dx}(\sin(u) + \cos(u)) = \frac{d}{du}(\sin(u) + \cos(u)) \cdot \frac{du}{dx}$.
$= (\cos(u) - \sin(u)) \cdot 2x$.
Substituting $u = x^2$ back into the expression,we get:
$= 2x(\cos(x^2) - \sin(x^2))$.
Thus,the correct option is $A$.

(A) Let $y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$.
Expanding the expression using $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$y = (\sqrt{x})^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2$
$y = x + 2 + \frac{1}{x}$
Now,differentiate with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(x) + \frac{d}{d x}(2) + \frac{d}{d x}(x^{-1})$
$\frac{d y}{d x} = 1 + 0 - x^{-2}$
$\frac{d y}{d x} = 1 - \frac{1}{x^2}$
Thus,the correct option is $A$.

(D) To determine the interval where the function $f(x) = x^x$ is decreasing,we find its derivative.
Let $y = x^x$.
Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(x)$.
Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.
Thus,$f'(x) = \frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)$.
For the function to be decreasing,we require $f'(x) < 0$.
Since $x^x > 0$ for all $x \in R^{+}$,the condition $f'(x) < 0$ implies $\ln(x) + 1 < 0$.
This gives $\ln(x) < -1$,which means $x < e^{-1}$ or $x < 1/e$.
Given $x \in R^{+}$,the interval is $(0, 1/e)$.

(A) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x$ in that interval.
Given $f(x) = x^2 + ax + 1$,the derivative is $f'(x) = 2x + a$.
For $f(x)$ to be increasing on $[1, 2]$,we require $f'(x) \ge 0$ for all $x \in [1, 2]$.
This implies $2x + a \ge 0$ for all $x \in [1, 2]$.
Since $2x + a$ is an increasing linear function,its minimum value on the interval $[1, 2]$ occurs at the smallest value of $x$,which is $x = 1$.
Therefore,we need $f'(1) \ge 0$.
Substituting $x = 1$ into the derivative: $2(1) + a \ge 0$.
$2 + a \ge 0$,which gives $a \ge -2$.
Thus,the function is increasing on $[1, 2]$ for $a \ge -2$.

(B) To determine the nature of the function $f(x) = \log_{10} \cos x$,we find its derivative with respect to $x$.
$f'(x) = \frac{d}{dx} (\log_{10} \cos x) = \frac{1}{\cos x \cdot \ln 10} \cdot (-\sin x) = -\frac{\tan x}{\ln 10}$.
In the interval $\left(0, \frac{\pi}{2}\right)$,$\tan x > 0$ and $\ln 10 > 0$.
Therefore,$f'(x) = -\frac{\tan x}{\ln 10} < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since the derivative is negative in the given interval,the function $f(x)$ is a decreasing function.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula: $V = \frac{4}{3} \pi r^3$.
To find the rate of change of the volume with respect to its radius,we differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) = 4 \pi r^2$.
Now,we evaluate this derivative at $r = 7 \ cm$:
$\frac{dV}{dr} \Big|_{r=7} = 4 \pi (7)^2 = 4 \pi (49) = 196 \pi$.
The unit of the rate of change of volume with respect to radius is $\frac{cm^3}{cm} = cm^2$.
Therefore,the rate of change is $196 \pi \ cm^2$.

(D) The area $A$ is given by the integral of the absolute value of the function:
$A = \int_{0}^{3\pi} |\cos x| \, dx$
Since the function $\cos x$ changes sign at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ and $x = \frac{5\pi}{2}$,we split the integral:
$A = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{5\pi/2} \cos x \, dx - \int_{5\pi/2}^{3\pi} \cos x \, dx$
Evaluating each part:
$|\sin x|_{0}^{\pi/2} = |1 - 0| = 1$
$|\sin x|_{\pi/2}^{3\pi/2} = |-1 - 1| = |-2| = 2$
$|\sin x|_{3\pi/2}^{5\pi/2} = |1 - (-1)| = |2| = 2$
$|\sin x|_{5\pi/2}^{3\pi} = |0 - 1| = |-1| = 1$
Summing these values: $A = 1 + 2 + 2 + 1 = 6$ sq. units.
Thus,the correct option is $D$.

(A) The given curve is $5y = 5 - x$,which can be written as $y = 1 - \frac{x}{5}$.
To find the area bounded by the curve,the $X$-axis,and the lines $x = 1$ and $x = 4$,we use the definite integral:
$\text{Area} = \int_{1}^{4} y \, dx$
$\text{Area} = \int_{1}^{4} (1 - \frac{x}{5}) \, dx$
Integrating with respect to $x$:
$\text{Area} = [x - \frac{x^2}{10}]_{1}^{4}$
Substitute the limits:
$\text{Area} = (4 - \frac{16}{10}) - (1 - \frac{1}{10})$
$\text{Area} = (4 - 1.6) - (1 - 0.1)$
$\text{Area} = 2.4 - 0.9 = 1.5$
Thus,the area is $1.5$ sq. units,which is $\frac{3}{2}$ sq. units.

(D) The given curve is $y = |x - 3|$.
For the interval $x \in [0, 2]$,the expression $(x - 3)$ is always negative,so $|x - 3| = -(x - 3) = 3 - x$.
The area $A$ is given by the integral $\int_{0}^{2} (3 - x) \, dx$.
Evaluating the integral: $A = [3x - \frac{x^2}{2}]_{0}^{2}$.
Substituting the limits: $A = (3(2) - \frac{2^2}{2}) - (3(0) - \frac{0^2}{2})$.
$A = (6 - 2) - 0 = 4$ sq. units.
Therefore,the correct option is $D$.

(B) The equation of the circle is $x^2 + y^2 = 16$, which can be written as $x^2 + y^2 = 4^2$. This is a circle centered at the origin $(0, 0)$ with a radius $r = 4$.
In the first quadrant, the equation of the circle is $y = \sqrt{16 - x^2}$.
The region is bounded by $x = 0$ and $x = 4$ in the first quadrant.
The area $A$ is given by the integral: $A = \int_{0}^{4} y \, dx = \int_{0}^{4} \sqrt{16 - x^2} \, dx$.
Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$, we get:
$A = [\frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}(\frac{x}{4})]_{0}^{4}$.
$A = [\frac{4}{2} \sqrt{16 - 16} + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(0)]$.
$A = [0 + 8(\frac{\pi}{2})] - [0 + 0] = 4\pi$.
Thus, the area is $4\pi$ sq. units.

(B) To find the area bounded by the curve $y = x^2 - x$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y = 0$.
$x^2 - x = 0 \implies x(x - 1) = 0$.
So,the intersection points are $x = 0$ and $x = 1$.
The curve $y = x^2 - x$ lies below the $X$-axis in the interval $(0, 1)$ because for any $x \in (0, 1)$,$x^2 < x$,so $y < 0$.
The area $A$ is given by the integral:
$A = \left| \int_{0}^{1} (x^2 - x) \, dx \right|$
$A = \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} \right|$
$A = \left| \left( \frac{1}{3} - \frac{1}{2} \right) - (0 - 0) \right|$
$A = \left| -\frac{1}{6} \right| = \frac{1}{6}$ sq. units.

(B) Given the curve equation $2y = -x + 8$,we can express $y$ as $y = \frac{-x + 8}{2} = -\frac{1}{2}x + 4$.
To find the area bounded by the curve,the $X$-axis,and the lines $x = 3$ and $x = 5$,we use the definite integral:
$\text{Area} = \int_{3}^{5} y \, dx = \int_{3}^{5} (-\frac{1}{2}x + 4) \, dx$.
Integrating the function:
$\int (-\frac{1}{2}x + 4) \, dx = [-\frac{1}{2} \cdot \frac{x^2}{2} + 4x] = [-\frac{x^2}{4} + 4x]$.
Now,applying the limits from $3$ to $5$:
$\text{Area} = [-\frac{5^2}{4} + 4(5)] - [-\frac{3^2}{4} + 4(3)]$.
$\text{Area} = [-\frac{25}{4} + 20] - [-\frac{9}{4} + 12]$.
$\text{Area} = [-\frac{25}{4} + \frac{80}{4}] - [-\frac{9}{4} + \frac{48}{4}]$.
$\text{Area} = \frac{55}{4} - \frac{39}{4} = \frac{16}{4} = 4$.
Thus,the area is $4$ sq. units.
Therefore,the correct option is $B$.

(A) The area $A$ bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} |f(x)| \, dx$.
Given the curve $y = 2x^2$,the $X$-axis,and the line $x = 1$,the region is bounded from $x = 0$ to $x = 1$.
Thus,the area $A = \int_{0}^{1} 2x^2 \, dx$.
Evaluating the integral: $A = 2 \left[ \frac{x^3}{3} \right]_{0}^{1}$.
$A = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$ sq. units.
Therefore,the correct option is $A$.