Let $f(x)=\left\{\begin{array}{l}|x|,-\infty < x < 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.$ $x=a$…
Let $f(x)=\left\{\begin{array}{l}|x|,-\infty < x < 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.$ $x=a$ is a point where $f(x)$ is continuous but not differentiable and $x=b$ is a point where $f(x)$ is not differentiable $(a \neq b)$. Then, $a+b=$
1
2
-2
0
Solution
$\begin{aligned}|x|=\left\{\begin{array}{cc}-x, & \text { if } x < 0 \\ x, & \text { if } x \geq 0\end{array}\right\} \\ \text { and }|2 x-4|=\left\{\begin{array}{cc}2 x-4, & \text { if } x \geq 2 \\ -(2 x-4), & \text { if } x < 2\end{array}\right.\end{aligned}$
$
f(x)=\left\{\begin{array}{ccc}
-x & , & -\infty < x < 0 \\
x & , & 0 \leq x < 2 \\
2 x-4 & , & 2 \leq x \leq 20
\end{array}\right.
$
At $x=0$
$
\begin{aligned}
& \text { LHL }=\lim _{\text {RHL }}={ }^x \lim _{x \rightarrow 0^{-}} f(x)=0 \\
& f(0)=0
\end{aligned}
$
$f(x)$ is continuous at $x=0$
At $x=2$
$
\begin{aligned}
& \text { LHL }=\lim _{x \rightarrow 2^{-}} f(x)=2 \\
& \text { RHL }=\lim _{x \rightarrow 2^{+}} f(x)=0
\end{aligned}
$
$f(x)$ is not continuous at $x=2$
$
f^{\prime}(x)=\left\{\begin{array}{cc}
-1, & -\infty < x < 0 \\
1, & 0 \leq x < 2 \\
2, & 2 \leq x \leq 20
\end{array}\right.
$
The function is not differentiable at both the points $x=0$ and $x=2$.
Here, $a=0$ and $b=2$
$
\begin{array}{rlrl}
& \therefore & a+b & =0+2 \\
\therefore & a+b & =2
\end{array}
$