Question

If $\underset{x\to 0}{\lim }{\left(\frac{asin2x-x^{x+1}}{ln\left(2x+1\right)}\right)}^{\frac{2x}{x^2+x}}=\frac{9}{4},a\in R,$ then sum of all possible values of $\_{}^{{}^{\prime }}{a}_{}^{{}^{\prime }}$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

We have, $\underset{x\to 0}{\lim }{\left(\frac{asin2x-x^{x+1}}{ln\left(2x+1\right)}\right)}^{\frac{2x}{x^2+x}}=\frac{9}{4},a\in R$ $\Rightarrow \underset{x\to 0}{\lim }{\left(\frac{a\frac{sin2x}{2x}-x^x\cdot \frac{x}{2x}}{\frac{ln\left(1+2x\right)}{2x}}\right)}^{\frac{2}{x+1}}=\frac{9}{4}$
We know,
$\underset{x\to 0}{\lim }\frac{sinx}{x}=1,\underset{x\to 0}{\lim }\frac{ln\left(1+x\right)}{x}=1,\underset{x\to 0}{\lim }{x}^x=1$
$\Rightarrow {\left(\frac{a-\frac{1}{2}}{1}\right)}^2=\frac{9}{4}$
$\Rightarrow a=\frac{1}{2}\pm \frac{3}{2}$
$\Rightarrow a=2\&a=-1$
Then, the sum of all possible value of $a=2+\left(-1\right)=1.$