JEE Mains · Maths · STD 12 - 5. continuity and differentiation

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

A 3
B 6
C 9
D 12

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Answer & Solution

Correct Answer

(A) 3

Step-by-step Solution

Detailed explanation

The possible points of discontinuity for the composite function \(g(f(x))\) are the points where \(f(x)\) is discontinuous and the points where \(f(x)\) is equal to a point of discontinuity of \(g(x)\). First, we find the points of discontinuity of \(f(x)\) and \(g(x)\). For…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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

Step-by-step Solution

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