JEE Main 2026 — Mathematics Calculus

We are given the function:

$$f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}$$f(x)={ex−1,x2−5x+6,​x<0x≥0​

We need to analyze the continuity and differentiability of $g(x) = f(|x|) + |f(x)|$g(x)=f(∣x∣)+∣f(x)∣.

For $x < 0$x<0, $|x| = -x > 0$∣x∣=−x>0. Thus, $f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6$f(∣x∣)=(−x)2−5(−x)+6=x2+5x+6.

Also, for $x < 0$x<0, $f(x) = e^{x-1} > 0$f(x)=ex−1>0, so $|f(x)| = e^{x-1}$∣f(x)∣=ex−1.

Therefore, for $x < 0$x<0, $g(x) = x^2 + 5x + 6 + e^{x-1}$g(x)=x2+5x+6+ex−1.

For $x \geq 0$x≥0, $|x| = x$∣x∣=x. Thus, $f(|x|) = f(x) = x^2 - 5x + 6$f(∣x∣)=f(x)=x2−5x+6.

Therefore, for $x \geq 0$x≥0, $g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|$g(x)=x2−5x+6+∣x2−5x+6∣.

Let us check the continuity of $g(x)$g(x) at $x = 0$x=0:

$$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x^2 + 5x + 6 + e^{x-1}) = 6 + \dfrac{1}{e}$$x→0−lim​g(x)=x→0−lim​(x2+5x+6+ex−1)=6+e1​

$$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x^2 - 5x + 6 + |x^2 - 5x + 6|) = 6 + 6 = 12$$x→0+lim​g(x)=x→0+lim​(x2−5x+6+∣x2−5x+6∣)=6+6=12

$$g(0) = 12$$g(0)=12

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. For all other $x$x, $g(x)$g(x) is a sum of continuous functions and is therefore continuous.

Thus, the number of points of discontinuity is $\alpha = 1$α=1.

Now, let us check the differentiability of $g(x)$g(x).

Since $g(x)$g(x) is discontinuous at $x = 0$x=0, it is not differentiable at $x = 0$x=0.

For $x < 0$x<0, $g(x) = x^2 + 5x + 6 + e^{x-1}$g(x)=x2+5x+6+ex−1, which is differentiable everywhere in its domain.

For $x > 0$x>0, we can rewrite $g(x)$g(x) by analyzing the sign of $x^2 - 5x + 6 = (x-2)(x-3)$x2−5x+6=(x−2)(x−3):

$$g(x) = \begin{cases} 2(x^2 - 5x + 6), & x \in (0, 2] \cup [3, \infty) \\ 0, & x \in (2, 3) \end{cases}$$g(x)={2(x2−5x+6),0,​x∈(0,2]∪[3,∞)x∈(2,3)​

Differentiating $g(x)$g(x) for $x > 0, x \neq 2, 3$x>0,x=2,3:

$$g'(x) = \begin{cases} 4x - 10, & x \in (0, 2) \cup (3, \infty) \\ 0, & x \in (2, 3) \end{cases}$$g′(x)={4x−10,0,​x∈(0,2)∪(3,∞)x∈(2,3)​

Checking differentiability at $x = 2$x=2:

$$g'(2^-) = 4(2) - 10 = -2$$g′(2−)=4(2)−10=−2

$$g'(2^+) = 0$$g′(2+)=0

Since $g'(2^-) \neq g'(2^+)$g′(2−)=g′(2+), $g(x)$g(x) is not differentiable at $x = 2$x=2.

Checking differentiability at $x = 3$x=3:

$$g'(3^-) = 0$$g′(3−)=0

$$g'(3^+) = 4(3) - 10 = 2$$g′(3+)=4(3)−10=2

Since $g'(3^-) \neq g'(3^+)$g′(3−)=g′(3+), $g(x)$g(x) is not differentiable at $x = 3$x=3.

Thus, $g(x)$g(x) is not differentiable at exactly three points: $x = 0, 2, 3$x=0,2,3.

So, the number of points of non-differentiability is $\beta = 3$β=3.

Finally, $\alpha + \beta = 1 + 3 = 4$α+β=1+3=4.

Answer: $4$4