Question

The graphs of $f\left(x\right)=-x^2+2$ and $g\left(x\right)=x^3-x^2-kx+2,k>0$ are shown below. The graphs intersect and create two closed regions $A$ and $B.$


Which of the following is (are) CORRECT?

• A. $\text{Area}\left(A\right)=\text{Area}\left(B\right)\forall k>0$
• B. $\text{Area}\left(A\right)=\frac{1}{2}\text{Area}\left(B\right)\forall k>0$
• C. $\text{Area}\left(A\right)=4\text{sq. unit when}k=4$
• D. $\text{Area}\left(B\right)=1\text{sq. unit when}k=2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\text{Area}\left(A\right)=\text{Area}\left(B\right)\forall k>0$
C $\text{Area}\left(A\right)=4\text{sq. unit when}k=4$
D $\text{Area}\left(B\right)=1\text{sq. unit when}k=2$

$f\left(x\right)=-x^2+2$ and $g\left(x\right)=x^3-x^2-kx+2$
Intersection point of $f\left(x\right)$ and $g\left(x\right):$
$-x^2+2=x^3-x^2-kx+2$
$\Rightarrow x^3-kx=0$
$\Rightarrow x=0,\pm \sqrt{k}$

From the graph, we can conclude that
$g\left(x\right)\ge f\left(x\right)\text{for}-\sqrt{k}\le x\le 0f\left(x\right)\ge g\left(x\right)\text{for}0\le x\le \sqrt{k}$
So,
$A=\underset{-\sqrt{k}}{\overset{0}{\int }}\left[g\right(x)-f(x\left)\right]dxB=\underset{0}{\overset{\sqrt{k}}{\int }}\left[f\right(x)-g(x\left)\right]dx$

Since, $g\left(x\right)-f\left(x\right)=x^3-kx$ is an odd function, so
$\underset{-\sqrt{k}}{\overset{\sqrt{k}}{\int }}\left[g\right(x)-f(x\left)\right]dx=0\Rightarrow \underset{-\sqrt{k}}{\overset{0}{\int }}\left[g\right(x)-f(x\left)\right]dx+{\int }_0^{\sqrt{k}}\left[g\right(x)-f(x\left)\right]dx=0\Rightarrow A-\underset{\sqrt{k}}{\overset{0}{\int }}\left[g\right(x)-f(x\left)\right]dx=0\Rightarrow A-B=0\therefore A=B$

Now,
$B=\underset{0}{\overset{\sqrt{k}}{\int }}(kx-x^3)dx\Rightarrow B={[\frac{k{x}^2}{2}-\frac{x^4}{4}]}_0^{\sqrt{k}}\Rightarrow B=\frac{k^2}{4}$

Therefore,
When $k=4$
$A=B=4\text{sq. unit}$
When $k=2$
$A=B=1\text{sq. unit}$