GSEB 2017 Mathematics Question Paper with Answer and Solution

(A) We are given the equation: $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$.
We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,which implies $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$.
Substituting this into the given equation:
$4 \cos^{-1} x + (\frac{\pi}{2} - \cos^{-1} x) = \frac{\pi}{2}$.
Simplifying the expression:
$3 \cos^{-1} x + \frac{\pi}{2} = \frac{\pi}{2}$.
Subtracting $\frac{\pi}{2}$ from both sides:
$3 \cos^{-1} x = 0$.
Dividing by $3$:
$\cos^{-1} x = 0$.
Taking the cosine of both sides:
$x = \cos(0) = 1$.

(A) Let $x = \cos^{-1} \frac{1}{3} + \cos^{-1} \frac{1}{5}$ and $y = \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{1}{5}$.
We know that $\cos^{-1} z = \frac{\pi}{2} - \sin^{-1} z$.
Thus,$x = (\frac{\pi}{2} - \sin^{-1} \frac{1}{3}) + (\frac{\pi}{2} - \sin^{-1} \frac{1}{5}) = \pi - (\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{1}{5}) = \pi - y$.
Therefore,$\cos(x) = \cos(\pi - y) = -\cos(y)$.
This implies $\cos(x) + \cos(y) = -\cos(y) + \cos(y) = 0$.

(C) We know that $\cos \theta = \sin \left(\frac{\pi}{2} - \theta\right)$ and $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$.
Substituting these into the expression:
$\sin ^{-1}\left(\sin \left(\frac{\pi}{2} - \frac{\pi}{13}\right)\right) + \cos ^{-1}\left(\cos \left(\frac{\pi}{2} - \frac{\pi}{13}\right)\right)$
Since $\frac{\pi}{2} - \frac{\pi}{13} = \frac{11\pi}{26}$,which lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$ and $[0, \pi]$ for $\cos^{-1}$,we have:
$= \left(\frac{\pi}{2} - \frac{\pi}{13}\right) + \left(\frac{\pi}{2} - \frac{\pi}{13}\right)$
$= \pi - \frac{2\pi}{13}$
$= \frac{13\pi - 2\pi}{13} = \frac{11\pi}{13}$.

(C) Given the equation $AB = 4I$,where $I$ is the identity matrix.
To find $A^{-1}$,we multiply both sides of the equation by $A^{-1}$ from the left:
$A^{-1}(AB) = A^{-1}(4I)$
$(A^{-1}A)B = 4A^{-1}$
Since $A^{-1}A = I$,we have:
$IB = 4A^{-1}$
$B = 4A^{-1}$
Dividing both sides by $4$,we get:
$A^{-1} = \frac{1}{4}B$
Therefore,the correct option is $C$.

(C) matrix $A$ is symmetric if $A = A^T$,which implies $A_{ij} = A_{ji}$ for all $i, j$.
For the given matrix $A = \begin{bmatrix} 5 & 2x+3 \\ x-2 & x+1 \end{bmatrix}$,the condition for symmetry is $A_{12} = A_{21}$.
Equating the elements,we get $2x + 3 = x - 2$.
Subtracting $x$ from both sides,we get $x + 3 = -2$.
Subtracting $3$ from both sides,we get $x = -5$.
Therefore,the correct value of $x$ is $-5$.

(D) Given the matrix equation: $\begin{bmatrix} a_1+a_2 & 4 \\ 3 & a_3+a_4 \end{bmatrix} - \begin{bmatrix} 3a_2 & 3a_1 \\ 3a_4 & 3a_3 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \\ -2a_4 & 1 \end{bmatrix}$.
Performing the subtraction,we get: $\begin{bmatrix} a_1-2a_2 & 4-3a_1 \\ 3-3a_4 & a_3-2a_4 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \\ -2a_4 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$1) \; a_1 - 2a_2 = -6$
$2) \; 4 - 3a_1 = -a_1 \implies 4 = 2a_1 \implies a_1 = 2$
$3) \; 3 - 3a_4 = -2a_4 \implies a_4 = 3$
$4) \; a_3 - 2a_4 = 1 \implies a_3 - 2(3) = 1 \implies a_3 = 7$
From equation $(1)$,substitute $a_1 = 2$: $2 - 2a_2 = -6 \implies -2a_2 = -8 \implies a_2 = 4$.
Now,calculate the sum $\sum_{i=1}^4 a_i = a_1 + a_2 + a_3 + a_4 = 2 + 4 + 7 + 3 = 16$.

(C) Let $\Delta = \left|\begin{array}{ccc}1 & a & b \\ 1 & a+b & b \\ 1 & a & a+b\end{array}\right|$.
Applying the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = \left|\begin{array}{ccc}1 & a & b \\ 0 & b & 0 \\ 0 & 0 & a\end{array}\right|$.
Expanding along the first column:
$\Delta = 1 \times (b \times a - 0 \times 0) = ab$.
Thus,the correct option is $C$.

(B) Let the given determinant be $\Delta = \left|\begin{array}{ccc}x & 3 & 5 \\ 2 & 6 & 10 \\ 7 & 21 & 35\end{array}\right|$.
Observe the columns of the determinant:
Column $2$ is $C_2 = [3, 6, 21]^T$ and Column $3$ is $C_3 = [5, 10, 35]^T$.
Notice that $C_3 = \frac{5}{3} C_2$.
Since two columns are proportional,the value of the determinant is $0$ for any value of $x$.
Alternatively,observe that row $2$ is $R_2 = [2, 6, 10]$ and row $3$ is $R_3 = [7, 21, 35]$.
$R_3 = 3.5 \times R_2$,which confirms the determinant is identically zero regardless of $x$.
Thus,the equation holds for all real numbers $x \in R$.

(B) Let $y = \frac{2^x + 3^x}{4^x} = \frac{2^x}{4^x} + \frac{3^x}{4^x} = \left(\frac{2}{4}\right)^x + \left(\frac{3}{4}\right)^x = \left(\frac{1}{2}\right)^x + \left(\frac{3}{4}\right)^x$.
Now,differentiate with respect to $x$ using the formula $\frac{d}{dx}(a^x) = a^x \log a$:
$\frac{dy}{dx} = \frac{d}{dx}\left(\left(\frac{1}{2}\right)^x\right) + \frac{d}{dx}\left(\left(\frac{3}{4}\right)^x\right)$
$\frac{dy}{dx} = \left(\frac{1}{2}\right)^x \log \frac{1}{2} + \left(\frac{3}{4}\right)^x \log \frac{3}{4}$.
Thus,the correct option is $B$.

(A) To find the derivative of $y = 3^{1-2 x}$,we use the chain rule and the formula $\frac{d}{d x}(a^u) = a^u \cdot \log a \cdot \frac{d u}{d x}$.
Here,$a = 3$ and $u = 1 - 2x$.
First,find the derivative of the exponent: $\frac{d}{d x}(1 - 2x) = -2$.
Now,apply the formula:
$\frac{d}{d x}(3^{1-2 x}) = 3^{1-2 x} \cdot \log 3 \cdot \frac{d}{d x}(1 - 2x)$
$= 3^{1-2 x} \cdot \log 3 \cdot (-2)$
$= -2 \cdot 3^{1-2 x} \log 3$.
Thus,the correct option is $A$.

(C) To determine the nature of the function $f(x) = \tan^{-1} x - x$,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(\tan^{-1} x - x) = \frac{1}{1+x^2} - 1$
Simplifying the expression:
$f'(x) = \frac{1 - (1+x^2)}{1+x^2} = \frac{-x^2}{1+x^2}$
Since $x^2 \ge 0$ and $1+x^2 > 0$ for all $x \in R$,the derivative $f'(x) = \frac{-x^2}{1+x^2} \le 0$ for all $x \in R$.
Since $f'(x) \le 0$ for all $x \in R$,the function $f(x)$ is a decreasing function on $R$.

(A) Given the curve equation is $y^2 = 18x$.
Differentiating both sides with respect to $t$,we get:
$2y \frac{dy}{dt} = 18 \frac{dx}{dt}$
$y \frac{dy}{dt} = 9 \frac{dx}{dt}$
According to the problem,the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate,i.e.,$\frac{dy}{dt} = 2 \frac{dx}{dt}$.
Substituting this into the differentiated equation:
$y(2 \frac{dx}{dt}) = 9 \frac{dx}{dt}$
Since $\frac{dx}{dt} \neq 0$,we can divide by $\frac{dx}{dt}$:
$2y = 9 \implies y = \frac{9}{2}$.
Now,substitute $y = \frac{9}{2}$ into the original curve equation $y^2 = 18x$:
$\left(\frac{9}{2}\right)^2 = 18x$
$\frac{81}{4} = 18x$
$x = \frac{81}{4 \times 18} = \frac{81}{72} = \frac{9}{8}$.
Thus,the required point is $\left(\frac{9}{8}, \frac{9}{2}\right)$.

(B) The velocity $v$ of a particle is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.
Given $s = t^3 - 6t^2 + 9t$.
Differentiating with respect to $t$:
$v = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9$.
To find the velocity at $t = 2 \ s$,substitute $t = 2$ into the velocity equation:
$v(2) = 3(2)^2 - 12(2) + 9$
$v(2) = 3(4) - 24 + 9$
$v(2) = 12 - 24 + 9 = -3 \ m/s$.
Therefore,the velocity at $t = 2 \ s$ is $-3 \ m/s$.

(B) The area $A$ of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum $x = a$ is given by the integral:
$A = 2 \int_{0}^{a} \sqrt{4ax} \, dx$
$A = 2 \times 2\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$
$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$
$A = 4\sqrt{a} \times \frac{2}{3} \times a^{3/2} = \frac{8}{3} a^2$
Given that the area is $24$ sq. units,we have:
$\frac{8}{3} a^2 = 24$
$a^2 = 24 \times \frac{3}{8} = 9$
$a = \pm 3$
Thus,the correct option is $B$.

(D) The area $A$ is given by the integral of the absolute value of the function $y = x^2 - x - 6$ from $x = -1$ to $x = 1$.
First,we check if the curve crosses the $x$-axis in the interval $[-1, 1]$.
Setting $y = 0$,we get $x^2 - x - 6 = 0$,which factors as $(x - 3)(x + 2) = 0$.
The roots are $x = 3$ and $x = -2$.
Since neither $3$ nor $-2$ lies in the interval $[-1, 1]$,the function $y = x^2 - x - 6$ does not change sign in this interval.
For $x \in [-1, 1]$,$x^2 - x - 6$ is negative (e.g.,at $x = 0$,$y = -6$).
Thus,the area is $A = \int_{-1}^{1} |x^2 - x - 6| \, dx = \int_{-1}^{1} -(x^2 - x - 6) \, dx$.
$A = - \left[ \frac{x^3}{3} - \frac{x^2}{2} - 6x \right]_{-1}^{1}$.
$A = - \left[ (\frac{1}{3} - \frac{1}{2} - 6) - (-\frac{1}{3} - \frac{1}{2} + 6) \right]$.
$A = - \left[ \frac{1}{3} - \frac{1}{2} - 6 + \frac{1}{3} + \frac{1}{2} - 6 \right]$.
$A = - \left[ \frac{2}{3} - 12 \right] = - \left[ \frac{2 - 36}{3} \right] = - \left[ -\frac{34}{3} \right] = \frac{34}{3}$ sq. units.

(D) The area $A$ is given by the integral of the absolute value of the function over the interval $[1, 3]$.
$A = \int_{1}^{3} |\sin(\pi x)| \, dx$.
Since $\sin(\pi x)$ changes sign at $x = 2$,we split the integral:
$A = \int_{1}^{2} |\sin(\pi x)| \, dx + \int_{2}^{3} |\sin(\pi x)| \, dx$.
In the interval $[1, 2]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$.
In the interval $[2, 3]$,$\sin(\pi x) \geq 0$,so $|\sin(\pi x)| = \sin(\pi x)$.
$A = \int_{1}^{2} -\sin(\pi x) \, dx + \int_{2}^{3} \sin(\pi x) \, dx$.
Evaluating the integrals:
$A = \left[ \frac{\cos(\pi x)}{\pi} \right]_{1}^{2} + \left[ -\frac{\cos(\pi x)}{\pi} \right]_{2}^{3}$.
$A = \frac{1}{\pi} (\cos(2\pi) - \cos(\pi)) - \frac{1}{\pi} (\cos(3\pi) - \cos(2\pi))$.
$A = \frac{1}{\pi} (1 - (-1)) - \frac{1}{\pi} (-1 - 1) = \frac{2}{\pi} + \frac{2}{\pi} = \frac{4}{\pi}$.
Thus,the correct option is $D$.

(B) The area $A$ is given by the integral of the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$
In the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2}]$.
Thus,$|\cos x| = -\cos x$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$
$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$A = -[\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})]$
$A = -[-1 - 1]$
$A = -[-2] = 2$ sq. units.
Therefore,the correct option is $B$.

(A) 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 formula $A = \int_{a}^{b} |f(x)| \, dx$.
Given the curve $y = -2\sqrt{x}$,the lines $x = 0$ and $x = 1$,and the $x$-axis $(y = 0)$,the area is:
$A = \int_{0}^{1} |-2\sqrt{x}| \, dx$
$A = \int_{0}^{1} 2\sqrt{x} \, dx$
$A = 2 \int_{0}^{1} x^{1/2} \, dx$
$A = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$
$A = 2 \times \frac{2}{3} [x^{3/2}]_{0}^{1}$
$A = \frac{4}{3} (1^{3/2} - 0^{3/2})$
$A = \frac{4}{3}$ sq. units.
Thus,the correct option is $A$.

(C) 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 $A = \int_{a}^{b} y \, dx$.
Given the curve $y = x^2 + 2$,the limits $x = 1$ and $x = 2$,the area is:
$A = \int_{1}^{2} (x^2 + 2) \, dx$
Integrating the function:
$A = [\frac{x^3}{3} + 2x]_{1}^{2}$
Substituting the upper and lower limits:
$A = (\frac{2^3}{3} + 2(2)) - (\frac{1^3}{3} + 2(1))$
$A = (\frac{8}{3} + 4) - (\frac{1}{3} + 2)$
$A = (\frac{8 + 12}{3}) - (\frac{1 + 6}{3})$
$A = \frac{20}{3} - \frac{7}{3} = \frac{13}{3}$
Thus,the area is $\frac{13}{3}$ sq. units.

(B) The given equation of the ellipse is $25x^2 + 16y^2 = 400$.
Dividing both sides by $400$,we get:
$\frac{25x^2}{400} + \frac{16y^2}{400} = 1$
$\frac{x^2}{16} + \frac{y^2}{25} = 1$
Comparing this with the standard equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$.
Thus,$a = 5$ and $b = 4$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,$A = \pi \times 5 \times 4 = 20\pi$ sq. units.