Let f( x )= \ begin array lll (1+ ax ) 1 / x & , & x <0\\ 1+ b & , & x =0\\ frac ( x +4) 1 / 2 -2 ( x + c ) 1 / 3 -2 & , & x 0end array . be continuous at x =0 . Then e a bc is equal to

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 3 April, Morning Shift — Question 38

Let $f(\mathrm{x})=\left\{\begin{array}{lll}(1+\mathrm{ax})^{1 / \mathrm{x}} & , & \mathrm{x}<0\\ 1+\mathrm{b} & , & \mathrm{x}=0\\ \frac{(\mathrm{x}+4)^{1 / 2}-2}{(\mathrm{x}+\mathrm{c})^{1 / 3}-2} & , & \mathrm{x}>0\end{array}\right.$ be continuous at $\mathrm{x}=0$ . Then $\mathrm{e}^{\mathrm{a}} \mathrm{bc}$ is equal to

Option A:
64
Option B:
72
Option C:
48
Correct
Option D:
36

Answer: C

Step-by-step solution

$f\left(0^{-}\right)=e^{\lim _{x \rightarrow 0} \frac{a x}{x}}=e^{a}$

$f(0)=1+b$

$f\left(0^{+}\right)=\frac{\frac{1}{2 \sqrt{x+4}}}{\frac{1}{3}(x+c)^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3} \cdot c^{-\frac{2}{3}}}$

$=\frac{3}{4} \mathrm{c}^{2 / 3}$

Also at $\mathrm{x}=0$ ;

$c^{1 / 3}=2 \Rightarrow c=8$

So $\mathrm{f}\left(0^{+}\right)=\frac{3}{4}(8)^{2 / 3}=3$

Now, $\mathrm{e}^{\mathrm{a}}=\mathrm{b}+1=3$

$\mathrm{e}^{\mathrm{a}} . \mathrm{b} . \mathrm{c}=3.2 .8=48$

Answer key and solution verified before publishing.

Practise Limits, Continuity and Differentiability

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity