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]

Question

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]

Smart Achievers · Accepted Answer

(2)

$f (0) = \lim_{x \to 0} \frac{e^{\tan x} - e^{x} + l n (\sec x + \tan x) - x}{\tan x - x}$

$Applying L'Hospital rule$

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

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

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

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

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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Solution

Q. If the function f(x)=ex(etanx-x-1)+loge(secx+tanx)-xtanx-x is continuous at x=0, then the value of f(0) is equal to [2026]

1 23

2 32

3 2

4 12