JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 18)
If $$[t]$$ denotes the greatest integer $$\leq t$$, then the number of points, at which the function $$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$$ is not differentiable in the open interval $$(-20,20)$$, is __________.
Answer
79
Explanation
$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$
$$ =4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x]-240 $$
$f(x)$ is non differentiable at $x=-\frac{3}{2}$
and $f(x)$ is discontinuous at $\{-19,-18, \ldots ., 18,19\}$
as well as $\left\{-\frac{39}{2},-\frac{37}{2}, \ldots,-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{39}{2}\right\}$,
at same point they are also non differentiable
$$ \begin{aligned} \therefore & \text { Total number of points of non differentiability } \\ &=39+40 \\ &=79 \end{aligned} $$
$$ =4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x]-240 $$
$f(x)$ is non differentiable at $x=-\frac{3}{2}$
and $f(x)$ is discontinuous at $\{-19,-18, \ldots ., 18,19\}$
as well as $\left\{-\frac{39}{2},-\frac{37}{2}, \ldots,-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{39}{2}\right\}$,
at same point they are also non differentiable
$$ \begin{aligned} \therefore & \text { Total number of points of non differentiability } \\ &=39+40 \\ &=79 \end{aligned} $$
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