GSEB 2013 Mathematics Question Paper with Answer and Solution

(C) To find the area bounded by the curve $y = x^2 - x - 6$ and the $X$-axis,we first find the points where the curve intersects the $X$-axis by setting $y = 0$.
$x^2 - x - 6 = 0$
$(x - 3)(x + 2) = 0$
So,the intersection points are $x = -2$ and $x = 3$.
The area $A$ is given by the integral of the absolute value of the function from $x = -2$ to $x = 3$:
$A = \int_{-2}^{3} |x^2 - x - 6| \, dx$
Since the parabola opens upward and lies below the $X$-axis between its roots,the value of $y$ is negative in this interval.
$A = - \int_{-2}^{3} (x^2 - x - 6) \, dx$
$A = - [\frac{x^3}{3} - \frac{x^2}{2} - 6x]_{-2}^{3}$
Evaluating the definite integral:
At $x = 3$: $(\frac{27}{3} - \frac{9}{2} - 18) = 9 - 4.5 - 18 = -13.5$
At $x = -2$: $(\frac{-8}{3} - \frac{4}{2} + 12) = -2.666 - 2 + 12 = 7.333$
$A = - (-13.5 - 7.333) = - (-20.833) = 20.833$
Converting to fraction: $20.833 = \frac{125}{6}$ sq. units.

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

(C) The area $A$ is given by the definite integral of the function $y = |x-5|$ from $x=5$ to $x=6$.
Since $x \ge 5$ in the interval $[5, 6]$,we have $|x-5| = x-5$.
Therefore,$A = \int_{5}^{6} (x-5) \, dx$.
Let $u = x-5$,then $du = dx$. When $x=5, u=0$ and when $x=6, u=1$.
$A = \int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 0.5$ sq. units.
Thus,the correct option is $C$.

(D) The region is bounded by $y=x^2$ and $y=4$.
First,find the points of intersection by setting $x^2 = 4$,which gives $x = \pm 2$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -2$ to $x = 2$:
$A = \int_{-2}^{2} (4 - x^2) dx$
Since the function is symmetric about the $y$-axis,$A = 2 \int_{0}^{2} (4 - x^2) dx$.
$A = 2 [4x - \frac{x^3}{3}]_{0}^{2}$
$A = 2 [(4(2) - \frac{2^3}{3}) - (0)]$
$A = 2 [8 - \frac{8}{3}] = 2 [\frac{24-8}{3}] = 2 [\frac{16}{3}] = \frac{32}{3}$ sq. units.

(C) Given the equation of the curve $x+2y=8$,we can express $y$ in terms of $x$ as:
$2y = 8 - x$
$y = \frac{8-x}{2} = 4 - \frac{x}{2}$
To find the area bounded by the curve,the $X$-axis,and the lines $x=1$ and $x=5$,we use the definite integral:
$\text{Area} = \int_{1}^{5} y \, dx$
$\text{Area} = \int_{1}^{5} (4 - \frac{x}{2}) \, dx$
Integrating term by term:
$\text{Area} = [4x - \frac{x^2}{4}]_{1}^{5}$
Evaluating at the limits:
$\text{Area} = (4(5) - \frac{5^2}{4}) - (4(1) - \frac{1^2}{4})$
$\text{Area} = (20 - \frac{25}{4}) - (4 - \frac{1}{4})$
$\text{Area} = 20 - 6.25 - 4 + 0.25$
$\text{Area} = 16 - 6 = 10$
Thus,the area is $10$ sq. units.
The correct option is $C$.

(D) The given curve is $y = |x - 5|$.
Since the interval is $x \in [0, 2]$,we have $x - 5 < 0$,so $|x - 5| = -(x - 5) = 5 - x$.
The area $A$ is given by the integral:
$A = \int_{0}^{2} (5 - x) \, dx$
$A = [5x - \frac{x^2}{2}]_{0}^{2}$
$A = (5(2) - \frac{2^2}{2}) - (0 - 0)$
$A = 10 - 2 = 8$ sq. units.
Thus,the correct option is $D$.

(D) The given equation is $y = 2 \sqrt{1 - x^2}$.
Squaring both sides,we get $y^2 = 4(1 - x^2)$,which implies $x^2 + \frac{y^2}{4} = 1$.
This is the equation of an ellipse with semi-major axis $a = 2$ (along the $Y$-axis) and semi-minor axis $b = 1$ (along the $X$-axis).
Since the curve is $y = 2 \sqrt{1 - x^2}$,it represents the upper half of the ellipse.
The area bounded by the curve and the $X$-axis is the area of the upper semi-ellipse.
The area of the full ellipse is given by $A = \pi ab = \pi \times 1 \times 2 = 2 \pi$.
Therefore,the area of the upper semi-ellipse is $\frac{1}{2} \times 2 \pi = \pi$ sq. units.

(C) To find the area bounded by the curve $y = 2x - x^2$ and the $X$-axis,we first find the points of intersection with the $X$-axis by setting $y = 0$.
$2x - x^2 = 0$
$x(2 - x) = 0$
So,$x = 0$ and $x = 2$.
The area $A$ is given by the integral of the curve from $x = 0$ to $x = 2$:
$A = \int_{0}^{2} (2x - x^2) \, dx$
$A = [x^2 - \frac{x^3}{3}]_{0}^{2}$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = 4 - \frac{8}{3}$
$A = \frac{12 - 8}{3} = \frac{4}{3}$ sq. units.