Continuity & DifferentiabilityhardPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Equilateral Triangle on Half-Lines: α²/R = 9 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

Let

f(x)=\begin{cases}e^{x-1}, & x<0\\ x^2-5x+6, & x\ge0\end{cases}

and

g(x)=f(|x|)+|f(x)|

. If the number of points where

g

is not continuous and is not differentiable are

\alpha

and

\beta

respectively, then

\alpha+\beta

is equal to

Solution

Answer: 4 (± 0.01)

Step 1:

f(|x|)

: for

x\ge0

f(|x|)=x^2-5x+6

; for

x<0

f(|x|)=x^2+5x+6

. Both give

6

x=0

, so continuous everywhere; derivatives at

0

-5

(right),

+5

(left)

\Rightarrow

non-differentiable only at

x=0

. Step 2:

|f(x)|

\lim_{x\to0^-}e^{x-1}=e^{-1}

f(0)=6\Rightarrow

jump, so discontinuous at

x=0

. On

x\ge0

f(x)=(x-2)(x-3)=0

x=2,3

with sign change

\Rightarrow

non-differentiable at

x=2,3

(and

x=0

). Step 3:

g(x)=f(|x|)+|f(x)|

. Discontinuous: only

x=0\Rightarrow \alpha=1

. Non-differentiable:

\{0\}\cup\{0,2,3\}=\{0,2,3\}\Rightarrow \beta=3

. Step 4:

\alpha+\beta=1+3=4.

Correct answer: 4

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