JEE Main 2019 — Mathematics Calculus

$$f(x)=15-|x-10|$$f(x)=15−∣x−10∣

$$\Rightarrow g(x)=f(f(x))$$⇒g(x)=f(f(x))

$$\Rightarrow g(x)=15-|f(x)-10|$$⇒g(x)=15−∣f(x)−10∣

$$=15-|15-|x-10|-10|$$=15−∣15−∣x−10∣−10∣

$$=15-|5-|x-10||$$=15−∣5−∣x−10∣∣

$$=15-||x-10|-5|$$=15−∣∣x−10∣−5∣

We know that, |x|={\begin{matrix}x, & x\geq 0 \\ -x, & x<0\end{matrix}\begin{matrix}\end{matrix}

Case $1:$1:

When $x<10$x<10

$$g(x)=15-|10-x-5|$$g(x)=15−∣10−x−5∣

$$\Rightarrow g(x)=15-|5-x|$$⇒g(x)=15−∣5−x∣

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

And if $x\geq 5$x≥5, then $g(x)=15+5-x=20-x$g(x)=15+5−x=20−x

Case $2:$2: $x\geq 10$x≥10

$$g(x)=15-|x-10-5|$$g(x)=15−∣x−10−5∣

$$\Rightarrow g(x)=15-|x-15|$$⇒g(x)=15−∣x−15∣

Now if $10\leq x<15$10≤x<15

$$\Rightarrow g(x)=15-15+x$$⇒g(x)=15−15+x

So, $g(x)=x$g(x)=x

Now for $x\geq 15$x≥15, then $g(x)=15-x+15=30-x$g(x)=15−x+15=30−x

g(x)={\begin{matrix}x+10, & x<5 \\ 20-x, & 5\leq x<10 \\ x, & 10<x\leq 15 \\ 30-x, & x\geq 15\end{matrix}

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