Suppose f is defined from $R\to [-1,1]$  as $f\left(x\right)=\frac{x^2-1}{x^2+1}$  where R is the set of real number. Then the statement which does not hold is

$\pi -2[-1]=\pi +2$$2\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-y^2}}+2\underset{0}{\overset{1}{\int }}\frac{y}{\sqrt{1-y^2}}dy$ $f\left(x\right)=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}$

As f(x) can never be equal to 1. Hence, f(x) is into  $f^{/}\left(x\right)=\frac{4x}{{(x^2+1)}^2}$
$\Rightarrow f^{/}\left(x\right)>0\forall x>0$  and  $f^{/}\left(x\right)<\forall x<0$
x = 0 is point of extremum which is local minima and value of f(0) = –1
 
Area =  $2\underset{0}{\overset{1}{\int }}\sqrt{\frac{2}{1-y}-1}dy=2\underset{0}{\overset{1}{\int }}\sqrt{\frac{1+y}{1-y}}dy$
= ${2{\sin }^{-1}y|}_0^1-{2\sqrt{1-y^2}|}_0^1$
= 
 
Hence, ACD is correct.