1. Tardigrade
  2. Question
  3. Mathematics
  4. Let f: R arrow R be a function defined as f(x)= begincases3(1-(|x|/2)) text if |x| ≤ 2 0 text if |x| > 2 endcases Let g: R arrow R be given by g(x)=f(x+2)-f(x-2). If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n+m is equal to .

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Q. Let $f: R \rightarrow R$ be a function defined
as $f(x)=\begin{cases}3\left(1-\frac{|x|}{2}\right) & \text { if }|x| \leq 2 \\ 0 & \text { if }|x| > 2\end{cases}$
Let $g: R \rightarrow R$ be given by $g(x)=f(x+2)-f(x-2)$.
If $n$ and $m$ denote the number of points in $R$ where $g$ is not continuous and not differentiable, respectively, then $n+m$ is equal to _______.

JEE MainJEE Main 2021Binomial Theorem

Solution:

$f(x-2) \begin{cases}\frac{3 x}{2} & -4 \leq x \leq-2 \\ -\frac{3 x}{2} & -2 < x \leq 0 \\ 0 & x \in(-\infty,-4) \cup(0,+\infty)\end{cases}$
$f(x-2) \begin{cases}\frac{3 x}{2} & 0 \leq x \leq 2 \\ -\frac{3 x}{2}+6 & 2 \leq x \leq 4 \\ 0 & x \in(-\infty, 0) \cup(4,+\infty)\end{cases}$
$g(x)=f(x+2)-f(x-2)\begin{cases}3 x 2+6 & -4 \leq x \leq-2 \\ -\frac{3 x}{2} & -2 < x < 2 \\ \frac{3 x}{2}-6 & 2 \leq x \leq 4 \\ 0 & x \in(-\infty,-4) \cup(4, \infty)\end{cases}$
image
$n=0$
$m=4 \Rightarrow(n+m=4)$