Let $f(x)=\left\{\begin{array}{rr}\max \left\{|x|, x^{2}\right\}, & |x| \leq 2 \\ 8-2|x|, & 2<|x| \leq 4\end{array}\right.$. Let S be the set of points in the interval $(-4,4)$ at which $f$ is not differentiable. Then $S$:

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Continuity and Differentiability

Let

f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4

. Let S be the set of points in the interval

(- 4, 4)

at which

f

is not differentiable. Then

S

Answer: (B)

∣ x ∣ \leq 2

the function takes the larger of

∣ x ∣

and

x^{2}

: for

∣ x ∣ \leq 1

that is

∣ x ∣

and for

1 < ∣ x ∣ \leq 2

that is

x^{2}

. The corner of

∣ x ∣

makes it non-differentiable at

0

; at

x = \pm 1

the graph switches from

∣ x ∣

x^{2}

with mismatched slopes,

1

against

2

, so it kinks; and at

x = \pm 2

it switches from

x^{2}

8 - 2∣ x ∣

with slopes that again disagree. Collecting these,

f

fails to be differentiable exactly at

{- 2, - 1, 0, 1, 2}

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option