Question

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

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

Answer 1

Answer: (A)

It is given that, $f(x)>x$, for all $x>1.$ So, area bounded by $y=f(x),y=x$ and the lines $x=1,x=t$ is given by
$\underset{1}{\overset{t}{\int }}\{f(x)-x\}dx$
But this area is given equal to $\left(t+\sqrt{1+t^2}-\sqrt{2}-1\right)$ sq. unit. Therefore,
$\underset{1}{\overset{t}{\int }}\{f(x)-x\}dx=t+\sqrt{1+t^2}-\sqrt{2}-1$, for all $t>1$
On differentiating both sides w.r.t. $t$, we get
$f(t)-t=1+\frac{t}{\sqrt{1+t^2}}$ for all $t>1$
$\Rightarrow f(t)=t+1+\frac{t}{\sqrt{1+t^2}}$ for all $t>1$
Hence, $f(x)=x+1+\frac{x}{\sqrt{1+x^2}}$ for all $x>1$