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Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 6 April, Evening Shift — Question 44

Let f(x)={x3+8,x<0x2−4,x≥0f(x) = \begin{cases} x^3+8, & x<0 \\ x^2-4, & x\ge 0 \end{cases} and g(x)={(x−8)1/3,x<0(x+4)1/2,x≥0g(x) = \begin{cases} (x-8)^{1/3}, & x<0 \\ (x+4)^{1/2}, & x\ge 0 \end{cases}. Then the number of points where the function g(f(x))g(f(x)) is discontinuous, is ______.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

g(f(x))={(f(x)−8)1/3,f(x)<0(f(x)+4)1/2,f(x)≥0g(f(x)) = \begin{cases} (f(x)-8)^{1/3} &, f(x) < 0 \\ (f(x)+4)^{1/2} &, f(x) \geq 0 \end{cases}

g(f(x))={(x3)1/3,x<−2(x2−12)1/3,0<x<2(x3+12)1/3,−2≤x≤0(x2)1/2,x≥2g(f(x)) = \begin{cases} (x^3)^{1/3} & , \quad x < -2 \\ (x^2-12)^{1/3} & , \quad 0 < x < 2 \\ (x^3+12)^{1/3} & , \quad -2 \leq x \leq 0 \\ (x^2)^{1/2} & , \quad x \geq 2 \end{cases}

g(f(x))={x,x<−2(x3+12)1/2,−2≤x≤0(x2−12)1/3,0<x<2x,x≥2g(f(x)) = \begin{cases} x & , \quad x < -2 \\ (x^3+12)^{1/2} & , \quad -2 \leq x \leq 0 \\ (x^2-12)^{1/3} & , \quad 0 < x < 2 \\ x & , \quad x \geq 2 \end{cases} Number of points of discontinuity is equal to 3.3 .

Solution figure

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Continuity

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