Let f(x)= cases (1+ax)^ 1 / x & , x 0 1+b & , x=0 (x+4)^ 1 / 2 -2 (x+c)^ 1 / 3 -2 & , cases be continuous at x=0 . Then e^a b c is equal to

Correct answer: C. 48 aligned & f(0^ - )=e^ _ x arrow 0 ax x =e^ a & f(0)=1+b & f(0^ + )= 1 2 x+4 (1)/(3)(x+c)^ -(2)/(3) = (1)/(2(2)) (1)/(3) · c^ -(2)/(3) & =(3)/(4) c^ 2 / 3 aligned Also at x=0 ; c^ 1 / 3 =2 c=8 So f(0^ + )=(3)/(4)(8)^ 2 / 3 =3 Now, e^ a =b+1=3 e^ a . b · c=3 · 2 · 8=48

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Continuity and Differentiability

Let f(x)= cases (1+ax)^ 1 / x & , x 0 1+b & , x=0 (x+4)^ 1 / 2 -2...

Let

f(\mathrm{x})=

\begin{cases}(1+\mathrm{ax})^{1 / \mathrm{x}} & , \quad \mathrm{x} \lt 0 \\ 1+\mathrm{b} & , \quad \mathrm{x}=0 \\ \frac{(\mathrm{x}+4)^{1 / 2}-2}{(\mathrm{x}+\mathrm{c})^{1 / 3}-2} & ,\end{cases}$

<br/>be continuous at

x=0

. Then

e^a b c$ is equal to

Options

A $64$
B $72$
C $48$
D $36$

Correct answer

C. $48$

Step-by-step solution

\begin{aligned}<br/>& f\left(0^{-}\right)=\mathrm{e}^{\lim _{x \rightarrow 0} \frac{\mathrm{ax}}{\mathrm{x}}}=\mathrm{e}^{\mathrm{a}} \\ & \mathrm{f}(0)=1+\mathrm{b} \\ & \mathrm{f}\left(0^{+}\right)=\frac{\frac{1}{2 \sqrt{\mathrm{x}+4}}}{\frac{1}{3}(\mathrm{x}+\mathrm{c})^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3} \cdot \mathrm{c}^{-\frac{2}{3}}} \\ & =\frac{3}{4} \mathrm{c}^{2 / 3}<br/>\end{aligned}

Also at

\mathrm{x}=0

;

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

\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} \cdot \mathrm{c}=3 \cdot 2 \cdot 8=48

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