Q39 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
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
- A$\frac{2}{3}$
- B2
- Correct answerC$\frac{3}{2}$
- D$\frac{1}{2}$
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
$\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x} \cdot \sec^2 x - e^x + \sec x - 1}{\sec^2 x - 1}$
$\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x}(\sec^2 x - 1) + (e^{\tan x} - e^x) + \sec x - 1}{\tan^2 x}$
$\Rightarrow f(0) = \lim_{x \to 0} \left( e^{\tan x} + \dfrac{e^x \left( e^{\tan x - x} - 1 \right)}{\tan^2 x} + \dfrac{1}{\sec x + 1} \right)$
$\Rightarrow f(0) = 1 + 0 + \dfrac{1}{2} = \dfrac{3}{2}$
Applying L'hospital rule
$\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x} \cdot \sec^2 x - e^x + \sec x - 1}{\sec^2 x - 1}$
$\Rightarrow f(0) = \lim_{x \to 0} \dfrac{e^{\tan x}(\sec^2 x - 1) + (e^{\tan x} - e^x) + \sec x - 1}{\tan^2 x}$
$\Rightarrow f(0) = \lim_{x \to 0} \left( e^{\tan x} + \dfrac{e^x \left( e^{\tan x - x} - 1 \right)}{\tan^2 x} + \dfrac{1}{\sec x + 1} \right)$
$\Rightarrow f(0) = 1 + 0 + \dfrac{1}{2} = \dfrac{3}{2}$
Hence, the correct answer is option: 'C'.