JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If the function \( f(x) = \frac{e^{x}(e^{\tan x-x}-1)+\log_{e}(\sec x+\tan x)-x}{\tan x-x} \) is continuous at \( x=0 \), then the value of \( f(0) \) is equal to
- A 2
- B \(\frac{2}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(f(0)=\operatorname{Lim}_{x \rightarrow 0} \frac{e^{\tan x}-e^x+\ell n(\sec x+\tan x)-x}{\tan x-x}\) Applying L'hospital rule \(\Rightarrow f(0)=\operatorname{Lim}_{x \rightarrow 0} \frac{e^{\tan x} \cdot \sec ^2 x-e^x+\sec x-1}{\sec ^2 x-1}\)…
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