JEE Main 2026 — Continuity and Differentiability Question with Solution

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 $f(x)$, the only possible point of discontinuity is at $x=0$.
${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}(x^3+8)=8$
${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}(x^2-4)=-4$
Since ${\lim }_{x\to 0^{-}}f(x)\ne {\lim }_{x\to 0^{+}}f(x)$, $f(x)$ is discontinuous at $x=0$.

For $g(x)$, the only possible point of discontinuity is at $x=0$.
${\lim }_{x\to 0^{-}}g(x)={\lim }_{x\to 0^{-}}(x-8{)}^{1/3}=-2$
${\lim }_{x\to 0^{+}}g(x)={\lim }_{x\to 0^{+}}(x+4{)}^{1/2}=2$
Since ${\lim }_{x\to 0^{-}}g(x)\ne {\lim }_{x\to 0^{+}}g(x)$, $g(x)$ is discontinuous at $x=0$.

Next, we find the points where $f(x)$ equals the point of discontinuity of $g(x)$, which is $f(x)=0$.
For $x<0$, $x^3+8=0\Rightarrow x=-2$.
For $x\ge 0$, $x^2-4=0\Rightarrow x=2$.

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

At $x=-2$:
As $x\to -2^{-}$, $f(x)\to 0^{-}$, so ${\lim }_{x\to -2^{-}}g(f(x))={\lim }_{y\to 0^{-}}g(y)=-2$
As $x\to -2^{+}$, $f(x)\to 0^{+}$, so ${\lim }_{x\to -2^{+}}g(f(x))={\lim }_{y\to 0^{+}}g(y)=2$
Since the left-hand limit and right-hand limit are not equal, $g(f(x))$ is discontinuous at $x=-2$.

At $x=0$:
${\lim }_{x\to 0^{-}}g(f(x))=g(8)=(8+4{)}^{1/2}=2\sqrt{3}$
${\lim }_{x\to 0^{+}}g(f(x))=g(-4)=(-4-8{)}^{1/3}=(-12{)}^{1/3}$
Since the limits are not equal, $g(f(x))$ is discontinuous at $x=0$.

At $x=2$:
As $x\to 2^{-}$, $f(x)\to 0^{-}$, so ${\lim }_{x\to 2^{-}}g(f(x))={\lim }_{y\to 0^{-}}g(y)=-2$
As $x\to 2^{+}$, $f(x)\to 0^{+}$, so ${\lim }_{x\to 2^{+}}g(f(x))={\lim }_{y\to 0^{+}}g(y)=2$
Since the limits are not equal, $g(f(x))$ is discontinuous at $x=2$.

Therefore, there are $3$ points of discontinuity.

Answer: $3$