Question

Let $f\left(x\right)$ be one-one and onto function, such that $|f\left(x\right)-f^{-1}\left(x\right)|>0$ for $xϵ(0,2)\cup (2,4)$ given $f\left(2\right)=2,f\left(4\right)=0$ and $f\left(0\right)=4$. Let $g\left(x\right)=\sqrt{16-x^2}$. Given $f\left(x\right)<g\left(x\right)\forall xϵ(0,4)$ and graph of $y=f\left(x\right)$ and $y=f^{-1}\left(x\right)$ are symmetrical about the line $x+y=4$.

If $f\left(x\right)-f^{-1}\left(x\right)<0$ for $xϵ(0,2)$ and ${\int }_0^{2}f\left(x\right)dx=5$, then ${\int }_2^4{f}^{-1}\left(x\right)dx$ is

• A. 2
• B. 1
• C. 5
• D. 7

Answer 1

The correct option is B 1


${\int }_0^{2}f\left(x\right)dx=5=$ shaded region PSTR

${\int }_2^4{f}^{-1}\left(x\right)dx=$ area of TRM which is equal to the area of PQR.

$\Rightarrow$ Area of PQR = area of PSTR – area QSTR
$=5-(2\times 2)=5-4=1$