GSEB 2015 Mathematics Question Paper with Answer and Solution

(A) Given the matrix equation:
$2\begin{bmatrix} 5 & x \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
First,multiply the first matrix by the scalar $2$:
$\begin{bmatrix} 10 & 2x \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
Now,add the two matrices on the left side:
$\begin{bmatrix} 10+0 & 2x+1 \\ 6+1 & 8+y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
$\begin{bmatrix} 10 & 2x+1 \\ 7 & 8+y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
By comparing the corresponding elements:
$2x + 1 = 5 \implies 2x = 4 \implies x = 2$
$8 + y = 0 \implies y = -8$
Therefore,$x = 2$ and $y = -8$.

(A) Given the equations:
$A - B = \begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix}$ (Equation $1$)
$A + B = \begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(A - B) + (A + B) = \begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$
$2A = \begin{bmatrix} 2+6 & 5+3 \\ 9+(-1) & 0+0 \end{bmatrix}$
$2A = \begin{bmatrix} 8 & 8 \\ 8 & 0 \end{bmatrix}$
Dividing by $2$:
$A = \frac{1}{2} \begin{bmatrix} 8 & 8 \\ 8 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 0 \end{bmatrix}$
Thus,the correct option is $A$.

(C) First,multiply the first two matrices:
$\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(1)+x(0)+1(0) & 1(3)+x(5)+1(3) & 1(2)+x(1)+1(2) \end{bmatrix} = \begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix}$
Now,multiply this result by the third matrix:
$\begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} = 1(1) + (5x+6)(1) + (x+4)(x) = 0$
$1 + 5x + 6 + x^2 + 4x = 0$
$x^2 + 9x + 7 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-9 \pm \sqrt{81 - 4(1)(7)}}{2} = \frac{-9 \pm \sqrt{81 - 28}}{2} = \frac{-9 \pm \sqrt{53}}{2}$
We need to find $2x + 9$:
$2x + 9 = 2 \left( \frac{-9 \pm \sqrt{53}}{2} \right) + 9 = -9 \pm \sqrt{53} + 9 = \pm \sqrt{53}$
Thus,the correct option is $C$.

(D) The question implies a standard matrix equality or identity related to complex numbers or algebraic forms. Given the options,the expression $x = a^2+b^2$ and $y = 2ab$ is a standard identity form often associated with the modulus and product properties in matrix representations of complex numbers. Therefore,the correct option is $D$.

(D) Given the matrix equality:
$\left[\begin{array}{cc}x-1 & 2y \\ x+y & 3\end{array}\right]=\left[\begin{array}{cc}3x-7 & y^2-3 \\ 6 & y\end{array}\right]$
By comparing the corresponding elements,we get:
$1) \ x-1 = 3x-7 \implies 2x = 6 \implies x = 3$
$2) \ 3 = y \implies y = 3$
$3) \ 2y = y^2-3 \implies 2(3) = (3)^2-3 \implies 6 = 9-3 \implies 6 = 6$ (Satisfied)
$4) \ x+y = 6 \implies 3+3 = 6$ (Satisfied)
Thus,the solution is $(x, y) = (3, 3)$.

(A) For a $2 \times 2$ matrix $A = [a_{ij}]$,the elements are given by $a_{ij} = \frac{i + 2j^2}{3}$.
For $i=1, j=1$: $a_{11} = \frac{1 + 2(1)^2}{3} = \frac{1 + 2}{3} = \frac{3}{3} = 1$.
For $i=1, j=2$: $a_{12} = \frac{1 + 2(2)^2}{3} = \frac{1 + 8}{3} = \frac{9}{3} = 3$.
For $i=2, j=1$: $a_{21} = \frac{2 + 2(1)^2}{3} = \frac{2 + 2}{3} = \frac{4}{3}$.
For $i=2, j=2$: $a_{22} = \frac{2 + 2(2)^2}{3} = \frac{2 + 8}{3} = \frac{10}{3}$.
Thus,the matrix $A = \left[\begin{array}{cc} 1 & 3 \\ \frac{4}{3} & \frac{10}{3} \end{array}\right]$.

(B) For any skew-symmetric matrix $A$ of order $n$,we have $A^T = -A$.
Taking the determinant on both sides,we get $|A^T| = |-A|$.
Since $|A^T| = |A|$ and $|-A| = (-1)^n |A|$,we have $|A| = (-1)^n |A|$.
For a $3 \times 3$ matrix,$n = 3$,so $|A| = (-1)^3 |A| = -|A|$.
This implies $2|A| = 0$,which means $|A| = 0$.

(B) The marginal revenue is defined as the derivative of the total revenue function $R(x)$ with respect to $x$,denoted as $MR = \frac{dR}{dx}$.
Given $R(x) = 10x^2 + 20x + 1500$.
Differentiating with respect to $x$:
$\frac{dR}{dx} = \frac{d}{dx}(10x^2 + 20x + 1500) = 20x + 20$.
To find the marginal revenue when $x = 1500$:
$MR = 20(1500) + 20 = 30000 + 20 = 30020$.
Thus,the correct option is $B$.

(B) Let the radius of the sphere be $r$.
The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$.
The surface area $S$ of the sphere is given by $S = 4 \pi r^2$.
From the surface area formula,we have $r^2 = \frac{S}{4 \pi}$,which implies $r = \sqrt{\frac{S}{4 \pi}} = \frac{1}{2} \sqrt{\frac{S}{\pi}}$.
We need to find $\frac{dV}{dS}$.
Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$.
$\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \pi r^3) = 4 \pi r^2$.
$\frac{dS}{dr} = \frac{d}{dr} (4 \pi r^2) = 8 \pi r$.
Therefore,$\frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Substituting the value of $r$,we get $\frac{dV}{dS} = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{S}{\pi}}) = \frac{1}{4} \sqrt{\frac{S}{\pi}}$.
Thus,the correct option is $B$.

(D) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} |f(x)| \, dx$.
Here,$f(x) = 5 \sin x$,$a = 0$,and $b = \frac{\pi}{2}$.
Since $\sin x \geq 0$ for $0 \leq x \leq \frac{\pi}{2}$,we have:
$A = \int_{0}^{\frac{\pi}{2}} 5 \sin x \, dx$
$A = 5 [-\cos x]_{0}^{\frac{\pi}{2}}$
$A = 5 [-\cos(\frac{\pi}{2}) - (-\cos(0))]$
$A = 5 [0 + 1]$
$A = 5 \times 1 = 5$ sq. units.
Thus,the correct option is $D$.

(B) To find the area of the region bounded by the curve $y=x$ and the lines $x=1$ and $x=10$,we use the definite integral:
Area $= \int_{a}^{b} y \, dx$
Here,$a = 1$,$b = 10$,and $y = x$.
Area $= \int_{1}^{10} x \, dx$
$= \left[ \frac{x^2}{2} \right]_{1}^{10}$
$= \frac{10^2}{2} - \frac{1^2}{2}$
$= \frac{100}{2} - \frac{1}{2}$
$= \frac{99}{2} \text{ sq. units.}$