Question

If $L={lim}_{x\to 0}\frac{\sin x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}$ exists, then the solution set of the inequality $||x-c|-2a|<4b$ is

• A. $(-1,1)$
• B. $(-2,2)$
• C. $[-1,1]$
• D. $[-2,2]$

Answer 1

The correct option is A $(-1,1)$

$L=\underset{x\to 0}{lim}\frac{\sin x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}$

Its limit exists, which implies that it is of form $\frac{0}{0}$

$\therefore a+b=0$ ...(1)

Now, we can apply L'Hospital's Rule

$L=\underset{x\to 0}{lim}\frac{\cos x+a{e}^x-b{e}^{-x}+\frac{c}{1+x}}{3x^2}$

From this we get,

$a-b+c=-1$ ...(2)

Again applying L'Hospital's Rule, we get

$L=\underset{x\to 0}{lim}\frac{-\sin x+a{e}^x+b{e}^{-x}-\frac{c}{(1+x{)}^2}}{6x}$

From this we get,

$a+b=c$ ...(3)

From these three equation, we can say

$a=-\frac{1}{2},b=\frac{1}{2},c=0$

Now, solution set for $||x-c|-2a|<4b$

Substituting required values, we get

$||x|+1|<2$

$\Rightarrow -2<|x|+1<2$

$\Rightarrow -3<|x|<1$

$\Rightarrow |x|<1$ $\dots [\because |x|$ is always positive $]$

$\Rightarrow -1<x<1$

$\therefore x\in (-1,1)$

Hence, option $A.$