If the function f ( x ) = 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 [2026]

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Solution

Q.
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 [2026]

1 $\frac{2}{3}$

2 $\frac{3}{2}$

3 $2$

4 $\frac{1}{2}$

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