JEE Main 2026 — Mathematics Calculus

The possible points of discontinuity for the composite function $g(f(x))$g(f(x)) are the points where $f(x)$f(x) is discontinuous and the points where $f(x)$f(x) is equal to a point of discontinuity of $g(x)$g(x).

First, we find the points of discontinuity of $f(x)$f(x) and $g(x)$g(x).

For $f(x)$f(x), the only possible point of discontinuity is at $x = 0$x=0.

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^3 + 8) = 8$$x→0−lim​f(x)=x→0−lim​(x3+8)=8

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 4) = -4$$x→0+lim​f(x)=x→0+lim​(x2−4)=−4

Since $\displaystyle \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$x→0−lim​f(x)=x→0+lim​f(x), $f(x)$f(x) is discontinuous at $x = 0$x=0.

For $g(x)$g(x), the only possible point of discontinuity is at $x = 0$x=0.

$$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x - 8)^{1/3} = -2$$x→0−lim​g(x)=x→0−lim​(x−8)1/3=−2

$$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x + 4)^{1/2} = 2$$x→0+lim​g(x)=x→0+lim​(x+4)1/2=2

Since $\displaystyle \lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$x→0−lim​g(x)=x→0+lim​g(x), $g(x)$g(x) is discontinuous at $x = 0$x=0.

Next, we find the points where $f(x)$f(x) equals the point of discontinuity of $g(x)$g(x), which is $f(x) = 0$f(x)=0.

For $x < 0$x<0, $x^3 + 8 = 0 \Rightarrow x = -2$x3+8=0⇒x=−2.

For $x \geq 0$x≥0, $x^2 - 4 = 0 \Rightarrow x = 2$x2−4=0⇒x=2.

Thus, the possible points of discontinuity for $g(f(x))$g(f(x)) are $x = -2$x=−2, $x = 0$x=0, and $x = 2$x=2. We check the continuity at each of these points.

At $x = -2$x=−2:

As $x \to -2^-$x→−2−, $f(x) \to 0^-$f(x)→0−, so $\displaystyle \lim_{x \to -2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$x→−2−lim​g(f(x))=y→0−lim​g(y)=−2

As $x \to -2^+$x→−2+, $f(x) \to 0^+$f(x)→0+, so $\displaystyle \lim_{x \to -2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$x→−2+lim​g(f(x))=y→0+lim​g(y)=2

Since the left-hand limit and right-hand limit are not equal, $g(f(x))$g(f(x)) is discontinuous at $x = -2$x=−2.

At $x = 0$x=0:

$$\lim_{x \to 0^-} g(f(x)) = g(8) = (8 + 4)^{1/2} = 2\sqrt{3}$$x→0−lim​g(f(x))=g(8)=(8+4)1/2=23​

$$\lim_{x \to 0^+} g(f(x)) = g(-4) = (-4 - 8)^{1/3} = (-12)^{1/3}$$x→0+lim​g(f(x))=g(−4)=(−4−8)1/3=(−12)1/3

Since the limits are not equal, $g(f(x))$g(f(x)) is discontinuous at $x = 0$x=0.

At $x = 2$x=2:

As $x \to 2^-$x→2−, $f(x) \to 0^-$f(x)→0−, so $\displaystyle \lim_{x \to 2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$x→2−lim​g(f(x))=y→0−lim​g(y)=−2

As $x \to 2^+$x→2+, $f(x) \to 0^+$f(x)→0+, so $\displaystyle \lim_{x \to 2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$x→2+lim​g(f(x))=y→0+lim​g(y)=2

Since the limits are not equal, $g(f(x))$g(f(x)) is discontinuous at $x = 2$x=2.

Therefore, there are $3$3 points of discontinuity.

Answer: $3$3