Question

Let $f^{\prime }\left(x^2\right)=\frac{1}{x}$ for $x>0,$ $f\left(1\right)=1$ and $g^{\prime }({\sin }^{2}x-1)={\cos }^{2}x+p$ for all $x\in R,$ $g(-1)=0.$ If $h\left(x\right)=\{\begin{array}{cc}f\left(x\right),& x>0\\ g\left(x\right),& -1\le x\le 0\end{array}$ is a continuous function, then the absolute value of $2p$ is

Answer 1

$f^{\prime }\left(x^2\right)=\frac{1}{x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\sqrt{x}}$ for $x>0$
$\Rightarrow f\left(x\right)=2\sqrt{x}+C$
$f\left(1\right)=1\Rightarrow C=-1$
$\therefore f\left(x\right)=2\sqrt{x}-1$ for $x>0$

and $g^{\prime }({\sin }^{2}x-1)={\cos }^{2}x+p\forall x\in R$
$\Rightarrow g^{\prime }(-{\cos }^{2}x)={\cos }^{2}x+p$
$\Rightarrow g^{\prime }\left(x\right)=p-x,\forall x\in [-1,0]$
$\Rightarrow g\left(x\right)=px-\frac{x^2}{2}+K$
$g(-1)=0$
$\Rightarrow 0=-p-\frac{1}{2}+K\Rightarrow K=\frac{1}{2}+p$
$\therefore g\left(x\right)=px-\frac{x^2}{2}+\frac{1}{2}+p$

$\therefore h\left(x\right)=\{\begin{array}{cc}2\sqrt{x}-1,& x>0\\ px-\frac{x^2}{2}+\frac{1}{2}+p,& -1\le x\le 0\end{array}$

At $x=0,$
$L.H.L.=R.H.L.=f\left(0\right)$
$\Rightarrow -1=\frac{1}{2}+p$
$\Rightarrow p=\frac{-3}{2}$
Hence, $2p=-3$
Absolute value of $2p$ is $3.$