JEE Main 2019 — Continuity and Differentiability Question with Solution

$f\left(x\right)=15-\left|x-10\right|$

$\Rightarrow g\left(x\right)=f\left(f\left(x\right)\right)$

$\Rightarrow g\left(x\right)=15-\left|f\left(x\right)-10\right|$

$=15-\left|15-\left|x-10\right|-10\right|$

$=15-\left|5-\left|x-10\right|\right|$

$=15-\left|\left|x-10\right|-5\right|$

We know that, $\left|x\right|=\left\{\begin{array}{cc} x,& x\ge 0\\ -x,& x<0\end{array}\begin{array}{}\end{array}\right.$

Case $1:$

When $x<10$

$g\left(x\right)=15-\left|10-x-5\right|$

$\Rightarrow g\left(x\right)=15-\left|5-x\right|$

Now if $x<5$, then $g\left(x\right)=15-5+x=x+10$

And if $x\ge 5$, then $g\left(x\right)=15+5-x=20-x$

Case $2:$ $x\ge 10$

$g\left(x\right)=15-\left|x-10-5\right|$

$\Rightarrow g\left(x\right)=15-\left|x-15\right|$

Now if $10\le x<15$

$\Rightarrow g\left(x\right)=15-15+x$

So, $g\left(x\right)=x$

Now for $x\ge 15$, then $g\left(x\right)=15-x+15=30-x$

 
$g\left(x\right)=\left\{\begin{array}{ll}x+10,& x<5\\ 20-x,& 5\le x<10\\ x,& 10<x\le 15\\ 30-x,& x\ge 15\end{array}\right.$

Now, it is clear from the graph of $g\left(x\right)$ has sharp corners at$x=5, 10, 15$ and hence this function is not differentiable at $x=5, 10, 15.$