Question

Consider the function $f\left(x\right)$ and $g\left(x\right)$ on $R,$ defined as $f\left(x\right)=2x-x^2$ and $g\left(x\right)=x^n$ where $n\in N.$ If the area between $y=f\left(x\right)$ and $y=g\left(x\right)$ in the first quadrant is $\frac{1}{2}$ sq. unit, then $n$ is a divisor of

• A. $12$
• B. $15$
• C. $20$
• D. $30$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $15$
C $20$
D $30$

Given $f\left(x\right)=2x-x^2$ and $g\left(x\right)=x^n$
To get intersection points, solving $f\left(x\right)$ and $g\left(x\right)$
we have $2x-x^2=x^n$
$\Rightarrow x=0$ and $x=1$


Area $=\underset{0}{\overset{1}{\int }}\left(f\right(x)-g(x\left)\right)dx$
$=\underset{0}{\overset{1}{\int }}(2x-x^2-x^n)dx$
$={[x^2-\frac{x^3}{3}-\frac{x^{n+1}}{n+1}]}_0^1$
$=[1-\frac{1}{3}-\frac{1}{n+1}]=\frac{2}{3}-\frac{1}{n+1}$

Now, $\frac{2}{3}-\frac{1}{n+1}=\frac{1}{2}$
$\Rightarrow \frac{1}{6}=\frac{1}{n+1}$
$\Rightarrow n+1=6\Rightarrow n=5$
Hence, $n$ is divisor of $15,20,30$