Let $f: \mathbb{R} \rightarrow \mathbb{R}$ 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 $g: \mathbb{R} \rightarrow \mathbb{R}$ be given by $g(x) = f(x+2) - f(x-2)$. If $n$ and $m$ denote the number of points in $\mathbb{R}$ where $g$ is not continuous and not differentiable, respectively, then $n + m$ is equal to ____

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

be a function defined as

f (x) = {3 (1 - \frac{∣ x ∣}{2}) 0 if ∣ x ∣ \leq 2 if ∣ x ∣ > 2

Let

g : R \to 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 ____

Answer: 4

The graph of

f

is a triangle of height

3

over

[- 2, 2]

and zero outside, so

f (x + 2)

is that triangle over

[- 4, 0]

and

f (x - 2)

is it over

[0, 4]

. The two supports meet only at

x = 0

, where both vanish, so

g (x) = ⎩ ⎨ ⎧ \frac{3 ( x + 4 )}{2} - \frac{3 x}{2} \frac{3 ( x - 4 )}{2} 0, - 4 \leq x \leq - 2, - 2 \leq x \leq 2, 2 \leq x \leq 4, ∣ x ∣ > 4

The pieces agree at every junction, so

g

is continuous on the whole line and

n = 0

. Its slope jumps at

x = - 4

x = - 2

x = 2

and

x = 4

, whereas at

x = 0

the slope is

- \frac{3}{2}

on both sides, so

m = 4

. Hence

n + m = 4

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer