Question

The smaller area enclosed by $y=f\left(x\right)$, where $f\left(x\right)$ is a polynomial of least degree satisfying $\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x^3}]}^{1/x}=e$ and the circle $x^2+y^2=2$ above the x-axis is:

• A. $\frac{\pi }{2}+\frac{3}{5}$
• B. $\frac{\pi }{2}-\frac{3}{5}$
• C. $\frac{\pi }{2}-\frac{6}{5}$
• D. none of these

Answer 1

Answer: (B)

$\frac{\pi }{2}-\frac{3}{5}$

Since $\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x^3}]}^{1/x}$ exists,

So $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=0$

$\therefore f\left(x\right)=a_4{x}^4+a_5{x}^5+...+a_n{x}^n,a_n\ne 0,n\ge 4$

Since, $f\left(x\right)$ is of least degree $\Rightarrow f\left(x\right)=a_4{x}^4$

Again, $\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x^3}]}^{1/x}=e\Rightarrow a_4=1\Rightarrow f\left(x\right)=x^4$

The graph of $y=x^4$ and $x^2+y^2=2$ are shown in the fugure

$\therefore$ The required area

$=2{\int }_0^1(\sqrt{2-x^2}-x^4)dx=\frac{\pi }{2}-\frac{3}{5}$