Question

The area bounded by the $x-$axis, the curve $y=f\left(x\right)$ and the lines $x=1$ and $x=b$ is equal to $(\sqrt{b^2+1}-\sqrt{2})$ for all $b>1$, then $f\left(x\right)$ is

• A. $\sqrt{x-1}$
• B. $\sqrt{x+1}$
• C. $\sqrt{x^2+1}$
• D. $\frac{x}{\sqrt{x^2+1}}$

Answer 1

The correct option is D $\frac{x}{\sqrt{x^2+1}}$

${\int }_1^{b}f\left(x\right)dx=\sqrt{(b^2+1)}-\sqrt{2}$
Differentiating both sides w.r.t $b$ , we get
$f\left(b\right)=\frac{b}{\sqrt{(b^2+1)}}$
Hence $f\left(x\right)=\frac{x}{\sqrt{(x^2+1)}}$