Let the functions \(f: R \rightarrow R\) and \(g: R \rightarrow R\) be defined as : \(f(x)=\left\{\begin{array}{cc}x+2, & x<0 \\ x^{2}, & x \geq 0\end{array}\right.\) and \(g(x)=\left\{\begin{array}{cc}x^{3}, & x<1 \\ 3 x-2, & x \geq 1\end{array}\right.\) Then, the number of points in R where ( \(f \circ g\) ) ( \(x\) ) is NOT differentiable is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Continuity and Differentiability

Let the functions

f : R \to R

and

g : R \to R

be defined as :

f (x) = {x + 2, x^{2}, x < 0 x \geq 0

and

g (x) = {x^{3}, 3 x - 2, x < 1 x \geq 1

Then, the number of points in R where (

f \circ g

) (

x

) is NOT differentiable is equal to :

Answer: (B)

For

x < 0

g (x) = x^{3} < 0

, so

f (g (x)) = x^{3} + 2

; for

0 \leq x < 1

g (x) = x^{3} \geq 0

, so

f (g (x)) = x^{6}

; for

x \geq 1

g (x) = 3 x - 2 \geq 1

, so

f (g (x)) = (3 x - 2)^{2}

(f \circ g) (x) = ⎩ ⎨ ⎧ x^{3} + 2, x^{6}, (3 x - 2)^{2}, x < 0 0 \leq x < 1 x \geq 1

x = 0

the left hand limit is

2

while the value is

0

, so

f \circ g

is discontinuous there and hence not differentiable.
At

x = 1

both one sided values equal

1

, and the derivatives

6 x^{5}

and

6 (3 x - 2)

both equal

6

, so

f \circ g

is differentiable there; on each open piece it is a polynomial.
So

f \circ g

fails to be differentiable at exactly

1

point.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option