Question

If $A_1$ = Area enclosed by the curve $y=-x^2$ and the straight line $x+y+2=0$
$A_2$ = Area bounded by the curve $y=\sqrt{x},x=2y+3$ in the first quadrant and $x$-axis.
Then which of the following is $CORRECT?$

• A. $A_1+A_2=9$
• B. $A_1-A_2=2$
• C. $A_1\times A_2=9$
• D. $\frac{A_2}{A_1}=2$

Answer 1

Answer: (D)

We have, $y=-x^2\dots \left(i\right)$,
and $x+y+2=\mathrm{0...}\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $-x-2=-x^2$
$\Rightarrow x^2-x-2=0$
$\Rightarrow \left(x-2\right)\left(x+1\right)=0$
$\Rightarrow x=2,-1$

Thus the area of the shaded region
$A_1=\left|\underset{1}{\overset{2}{\int }}\left(-x-2+x^2\right)dx\right|$
$=\left|\underset{-1}{\overset{2}{\int }}\left(x^2-x-2\right)dx\right|$
$=\left|{\left[\frac{x^3}{3}-\frac{x^2}{2}-2x\right]}_{-1}^2\right|$
$=\left|\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2\right|$
$=\left|\frac{16-12-24+2+3-12}{6}\right|$
$=\left|-\frac{27}{6}\right|=\frac{9}{2}$ sq. units
Also, we have, $y=\sqrt{x}\dots \left(i\right)$
and $x=2y+3\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $y=\sqrt{2y+3}$
Squaring on both sides, we get $y^2-2y-3=0$
$\Rightarrow \left(y+1\right)\left(y-3\right)=0$
$\Rightarrow y=-1,3$

Thus the required area of the shaded region
$A_2=\underset{0}{\overset{3}{\int }}\left(2y+3-y^2\right)dy$
$={\left[y^2+3y-\frac{y^3}{3}\right]}_0^3$
$=\left[9+9-9-0\right]=9$ sq. units