Let \(f(x)=\left\{\begin{array}{ll}x^{3}+8 & x<0 \\ x^{2}-4 & x \geq 0\end{array}\right.\) and \(\mathrm{g}(\mathrm{x})= \begin{cases}(\mathrm{x}-8)^{1 / 3} ; & \mathrm{x}<0 \\ (\mathrm{x}+4)^{1 / 2} ; & \mathrm{x} \geq 0\end{cases}\) Then the number of points, where the function gof is discontinuous, is ______.

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Continuity and Differentiability

Let

f (x) = {x^{3} + 8 x^{2} - 4 x < 0 x \geq 0

and

g (x) = {(x - 8)^{1/3}; (x + 4)^{1/2}; x < 0 x \geq 0

Then the number of points, where the function gof is discontinuous, is ______.

Answer: 3

Where

f

changes sign the branch of

g

switches, so examine

g (f (x))

on the pieces set by

f (x) = 0

x = - 2

(from

x^{3} + 8

) and

x = 2

(from

x^{2} - 4

). This gives

g (f (x)) = x

for

x < - 2

x^{3} + 12

for

- 2 \leq x < 0

(x^{2} - 12)^{1/3}

for

0 \leq x < 2

, and

x

for

x \geq 2

. At

x = - 2

the left value is

- 2

and the right value is

4 = 2

; at

x = 0

the left value is

12

and the right value is

(- 12)^{1/3}

; at

x = 2

the left value is

(- 8)^{1/3} = - 2

and the right value is

2

. Each junction gives unequal one-sided limits, so

g \circ f

is discontinuous at exactly

3

points.

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer