JEE Main 2026 Apr 6 Shift 2, Mathematics Q25: Invertible And Composite Functions

JEE Main2026Apr 6, Shift 2Mathematics

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

Your answer

Solution

By definition, $g (f (x)) = {(f (x) - 8)^{1/3} f (x) < 0 (f (x) + 4)^{1/2} f (x) \geq 0$ .

Sign of $f$ . For $x < 0$ : $f (x) = x^{3} + 8$ is negative when $x < - 2$ and non-negative for $- 2 \leq x < 0$ . For $x \geq 0$ : $f (x) = x^{2} - 4$ is negative for $0 \leq x < 2$ and non-negative for $x \geq 2$ .

Build $g \circ f$ piecewise:

For $x < - 2$ : $g (f (x)) = ((x^{3} + 8) - 8)^{1/3} = x$ .

For $- 2 \leq x < 0$ : $g (f (x)) = ((x^{3} + 8) + 4)^{1/2} = (x^{3} + 12)^{1/2}$ .

For $0 \leq x < 2$ : $g (f (x)) = ((x^{2} - 4) - 8)^{1/3} = (x^{2} - 12)^{1/3}$ .

For $x \geq 2$ : $g (f (x)) = ((x^{2} - 4) + 4)^{1/2} = ∣ x ∣ = x$ .

Check breakpoints.

At $x = - 2$ : left limit $= - 2$ , right value $= - 8 + 12 = 2$ → jump (discontinuous).

At $x = 0$ : left limit $= 12$ , right value $= (- 12)^{1/3} = - 312$ → jump (discontinuous).

At $x = 2$ : left limit $= (4 - 12)^{1/3} = - 2$ , right value $= 2$ → jump (discontinuous).

Number of points of discontinuity $= 3$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q25: Invertible And Composite Functions

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