Question

Area bounded by the line $y=x,$ curve $y=f\left(x\right),(f\left(x\right)>x,x>1)$ and the lines $x=1,x=t$ is $(t+\sqrt{(1+t^2)})-(1+\sqrt{2})$ for all $t>1$. Then $f\left(x\right)=$

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

Answer 1

Answer: (A)

$1+x+\frac{x}{\sqrt{1+x^2}}$

The area bounded by $y=f\left(x\right)$ and $y=x$ between the lines $x=1$ and $x=t$ is
${\int }_1^t(f\left(x\right)-x)dx$
But it is equal to
$(t+\sqrt{1+t^2})-(1+\sqrt{2})$
$\therefore {\int }_1^t(f\left(x\right)-x)dx=(t+\sqrt{1+t^2})-(1+\sqrt{2})$
Differentiating both sides w.r.t $t$ we get
$f\left(t\right)-t=1+\frac{t}{\sqrt{1+t^2}}$
$\Rightarrow f\left(t\right)=1+t+\frac{t}{\sqrt{1+t^2}}$
or $f\left(x\right)=1+x+\frac{x}{\sqrt{1+x^2}}$