Question

Let $f:R\to R$ be a differentiable function satisfying the condition $f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}\forall$ real values of $x$ & $y$ and $f^{\prime }\left(2\right)=2$
If $h\left(x\right)=|f\left(\left|x\right|\right)-5|\forall x\in R$ then the function $h\left(x\right)$ is non differentiable at number of points

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}\to 1$

Put $x=y=0$

$f\left(0\right)=\frac{2+2f\left(0\right)}{3}$

$\Rightarrow f\left(0\right)=2$

Differentiate $1$ partially w.r.t. x

$f\left(\frac{x+y}{3}\right)\left(\frac{1}{3}\right)=\frac{f^{\prime }\left(x\right)}{3}$

$\Rightarrow f^{\prime }\left(\frac{x+y}{3}\right)=f^{\prime }\left(x\right)$

Put $x=0$

$f^{\prime }\left(\frac{y}{3}\right)=f^{\prime }\left(0\right)=k$

$\Rightarrow f\left(\frac{y}{3}\right)=ky+C$

$f\left(y\right)=3ky+C(k\ne 0)$

We know that $f\left(0\right)=2$

$\Rightarrow 2=0+C\Rightarrow c=2$

$\Rightarrow f\left(y\right)=3ky+2$

$\Rightarrow f\left(x\right)=3kx+2$

$f^{\prime }\left(x\right)=3k$

Given $f^{\prime }\left(2\right)=2$

$\Rightarrow 2=3k\Rightarrow k=\frac{2}{3}$

$\Rightarrow f\left(x\right)=2x+2,f\left(\left|x\right|\right)=2\left|x\right|+2$

$h\left(x\right)=|f\left(\left|x\right|\right)-5|$

$h^{\prime }\left(x\right)=\frac{|f\left(\left|x\right|\right)-5|}{f\left(\left|x\right|\right)-5}\times f^{\prime }\left(x\right)\times \frac{\left|x\right|}{x}$

$h\left(x\right)$ will be non differentiable when $h^{\prime }\left(x\right)$ is not defined

$h^{\prime }\left(x\right)$ is not defined when

$x(f\left(\left|x\right|\right)-5)=0$

$\Rightarrow x=0,f\left(\left|x\right|\right)=5$

$2\left|x\right|=3$

$\left|x\right|=\frac{3}{2}\Rightarrow x=\pm \frac{3}{2}$

At $3$ points $(0,\pm \frac{3}{2})h\left(x\right)$ is not derivable.