Maths MCQ

(a): The area bounded by the curve $y=-x^2, x$-axis, $x=1$ and $x=4$ is shown below.
Area can be calculated as
$
\begin{aligned}
& A=\left|\int_1^4-x^2 d x\right| \\
& =\left|\left[\frac{-x^3}{3}\right]_1^4\right| \quad  \\
& =\left|\frac{-64}{3}+\frac{1}{3}\right|=21 \text { sq. units }
\end{aligned}
$

(a) : The given lines are $2 y+x=8, x$-axis, $x=2$ and $x=4$.
$
\begin{aligned}
& \text { Required area }=\int_2^4 y d x \\
& =\int_2^4\left(\frac{8-x}{2}\right) d x=\frac{1}{2}\left[8 x-\frac{x^2}{2}\right]_2^4=\frac{1}{2}[32-8-16+2]
\end{aligned}
$
$
=5 \text { sq. units }
$

(b) : We have, $x^2+y^2=9 \ldots$ (i), a circle with centre $(0,0)$ and radius 3 .
and $y^2=8 x \ldots$ (ii) a parabola with vertex $(0,0)$.
Points of intersection of given curves are $(1,2 \sqrt{2})$ and $(1,-2 \sqrt{2})$.
$\begin{array}{r}\text { Required area }=2\left[\int_0^1 \sqrt{8 x} d x+\int_1^3 \sqrt{9-x^2} d x\right] \\ =2\left(\sqrt{8} \times \frac{2}{3}\right)+2\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_1^3 \\ =\left[\frac{2 \sqrt{2}}{3}+\frac{9 \pi}{2}-9 \sin ^{-1}\left(\frac{1}{3}\right)\right] \text { sq. units }\end{array}$

(a) : Required area is shaded in the figure.
Points of intersection of curve $y=2 x-x^2$ and line $y=x$ are $(0,0)$ and $(1,1)$.
$\therefore$ Required area $=\int_0^1\left\{\left(2 x-x^2\right)-x\right\} d x$
$=\int_0^1 2 x d x-\int_0^1 x^2 d x-\int_0^1 x d x$
$=2\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1-\left[\frac{x^2}{2}\right]_0^1=1-\frac{1}{3}-\frac{1}{2}=\frac{1}{6}$ square units

1. (d) : We have $x^2+y^2=2 \ldots$ (i) a circle with centre $(0,0)$ and radius $\sqrt{2}$.
and $x^2+4 y^2=4 \ldots$ (ii) an ellipse with centre $(0,0)$ and vertex $(2,0)$ and $(-2,0)$.
Point of intersection of (i) and (ii) is $\left(\frac{2}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}\right)$



$\begin{aligned} & \text { Required area }=4\left[\frac{1}{2} \int_0^{\frac{2}{\sqrt{3}}} \sqrt{4-x^2} d x+\int_{\frac{2}{\sqrt{3}}}^{\sqrt{2}} \sqrt{2-x^2} d x\right] \\ & =4\left[\left(\frac{1}{4} x \sqrt{4-x^2}+\sin ^{-1} \frac{x}{2}\right)_0^{\frac{2}{\sqrt{3}}}\right. \\ & \left.+\left(\frac{x}{2} \sqrt{2-x^2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right)_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}\right] \\ & =4\left[\frac{1}{2 \sqrt{3}} \sqrt{4-\frac{4}{3}}+\sin ^{-1} \frac{1}{\sqrt{3}}+\frac{\pi}{2}-\frac{1}{\sqrt{3}} \sqrt{2-\frac{4}{3}}-\sin ^{-1} \frac{\sqrt{2}}{\sqrt{3}}\right] \\ & =4\left[\sin ^{-1} \frac{1}{\sqrt{3}}+\cos ^{-1} \sqrt{\frac{2}{3}}\right] \\ & =4\left[2 \tan ^{-1} \frac{1}{\sqrt{2}}\right]=4 \tan ^{-1} 2 \sqrt{2} \text { sq. units }\end{aligned}$