Question

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

• A. $-1$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

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