If the function f(x)= e^ x (e^ x-x -1)+ _ e ( x+ x)-x x-x is continuous at x=0 , then the value of f(0) is equal to

Correct answer: C. (3)/(2) f(0) = _ x → 0 e^ x - e^x + ( x + x) - x x - x Applying L'hospital rule f(0) = _ x → 0 e^ x · ^2 x - e^x + x - 1 ^2 x - 1 f(0) = _ x → 0 e^ x ( ^2 x - 1) + (e^ x - e^x) + x - 1 ^2 x f(0) = _ x → 0 ( e^ x + e^x ( e^ x - x - 1 ) ^2 x + (1)/( x + 1) ) f(0) = 1 + 0 + (1)/(2) = (3)/(2)

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Continuity and Differentiability

If the function f(x)= e^ x (e^ x-x -1)+ _ e ( x+ x)-x x-x is...

If the function

f(x)=\frac{e^{x}\left(e^{\tan x-x}-1\right)+\log _{e}(\sec x+\tan x)-x}{\tan x-x}

is continuous at

x=0

, then the value of

f(0)

is equal to

Options

A $\frac{2}{3}$
B2
C $\frac{3}{2}$
D $\frac{1}{2}$

Correct answer

C. $\frac{3}{2}$

Step-by-step solution

f(0) = \lim_{x \to 0} \dfrac{e^{\tan x} - e^x + \ln(\sec x + \tan x) - x}{\tan x - x}

Applying L'hospital rule

\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x} \cdot \sec^2 x - e^x + \sec x - 1}{\sec^2 x - 1}

\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x}(\sec^2 x - 1) + (e^{\tan x} - e^x) + \sec x - 1}{\tan^2 x}

\Rightarrow f(0) = \lim_{x \to 0} \left( e^{\tan x} + \dfrac{e^x \left( e^{\tan x - x} - 1 \right)}{\tan^2 x} + \dfrac{1}{\sec x + 1} \right)

\Rightarrow f(0) = 1 + 0 + \dfrac{1}{2} = \dfrac{3}{2}

Practice Continuity and Differentiability on Quantrex Academy →

More from Continuity and Differentiability