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 ${\int }_0^2\left(f\right(x)-f^{-1}(x\left)\right)dx=2,{\int }_2^4{f}^{-1}\left(x\right)dx=3$, then ${\int }_0^4\left[g\right(x)-max(f\left(x\right),f^{-1}\left(x\right)\left)\right]dx$ is

• A. $4\pi -12$
  
• B. $8\pi -12$
  
• C.
  $4\pi -10$
  
• D. 8

Answer 1

The correct option is C

$4\pi -10$



${\int }_0^2\left(f\right(x)-f^{-1}(x\left)\right)dx=1\Rightarrow A_1=A_2=2$

${\int }_2^4{f}^{-1}\left(x\right)dx=3\Rightarrow A_3=3-2=1$

$\Rightarrow {\int }_0^4\left[g\right(x)-max(f\left(x\right)),f^{-1}(x\left)\right]dx$

$={\int }_0^{4}gxdx-{\int }_0^{4}max\left(f\right(x),f^{-1}(x\left)\right)dx$

$=4\pi -(A_1+PSTR+A_2+A_3$

$=4\pi -(2+5+2+1)=4\pi -10$