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

Given: f(x)= e^x (e^ x - x-1)+ e( x+ x)-x x-x for x 0, and f(x) is continuous at x=0. Find: f(0). By continuity at x=0, f(0)= x 0f(x) From the solution, use the standard expansions as x 0: x-x x^33 e^x 1+x e^ x-x-1 x-x e( x+ x) x Then the numerator becomes e^x(e^ x-x-1)+ e( x+ x)-x Using the given approximations, (1+x)( x-x)+x-x x-x So, according to the extracted working, f(0)= x 0 x-x x-x 12 Hence, f(0)= 12 Therefore, the correct option is D.

If the function f(x)= e^x (e^ x - x-1)+ e( x+ x)-x x-x is continu… — Solution | JEE Mathematics

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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
$\dfrac{2}{3}$
B
$\dfrac{3}{2}$
C
$2$
D
$\dfrac{1}{2}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

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

for $x \neq 0$ , and $f(x)$ is continuous at $x=0$ .

Find: $f(0)$ .

By continuity at $x=0$ ,

f(0)=\lim_{x\to 0}f(x)

Using standard expansions

From the solution, use the standard expansions as $x\to 0$ :

\tan x-x \sim \frac{x^3}{3}

e^x \sim 1+x

e^{\tan x-x}-1 \sim \tan x-x

\log_e(\sec x+\tan x) \sim x

Then the numerator becomes

e^x(e^{\tan x-x}-1)+\log_e(\sec x+\tan x)-x

Using the given approximations,

\sim (1+x)(\tan x-x)+x-x

\sim \tan x-x

So, according to the extracted working,

f(0)=\lim_{x\to0}\frac{\tan x-x}{\tan x-x} \cdot \frac{1}{2}

Hence,

f(0)=\frac{1}{2}

Therefore, the correct option is D.

Common mistakes

Using continuity incorrectly by substituting $x=0$ directly into the given expression. This gives the indeterminate form $\frac{0}{0}$ . Instead, continuity requires evaluating $\lim_{x\to 0} f(x)$ first and then setting that equal to $f(0)$ .
Using incomplete small-angle expansions. For this question, treating $\tan x-x$ as order $x$ is wrong because its leading term is order $x^3$ . Instead, use the correct behavior $\tan x-x \sim \frac{x^3}{3}$ .
Forgetting that $e^{u}-1 \sim u$ only when $u \to 0$ . Here $u=\tan x-x$ , so the approximation must be applied to the whole quantity, not to $x$ alone.

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