Question

$\text{If}\underset{x\to -1}{lim}{\left(\frac{\tan (x+1)+a(e.e^x+x)}{x-a^2{\sin }^{-1}(x+1)+1}\right)}^{\frac{x+1}{\sqrt{x+2}-1}}=1,$
$\text{then possible values of}{}^{\prime }{a}^{\prime }\text{is/are}$

• A. $-2$
• B. $0$
• C. $\sqrt{3}+1$
• D. $\sqrt{3}-1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $-2$
B $0$
C $\sqrt{3}+1$

$\underset{x\to -1}{lim}{\left(\frac{\tan (x+1)+a{e}^{x+1}+ax+a-a}{x+1-a^2{\sin }^{-1}(x+1)}\right)}^{\frac{(x+1)(\sqrt{x+2}+1)}{x+1}}=1$
$\Rightarrow \underset{x\to -1}{lim}{\left(\frac{\tan (x+1)+a(e^{x+1}-1)+a(x+1)}{x+1-a^2{\sin }^{-1}(x+1)}\right)}^{\sqrt{x+2}+1}=1$
$\Rightarrow \underset{x\to -1}{lim}{\left(\frac{\frac{\tan (x+1)}{x+1}+a\frac{(e^{x+1}-1)}{x+1}+a}{1-a^2\frac{{\sin }^{-1}(x+1)}{x+1}}\right)}^{\sqrt{x+2}+1}=1$
$\Rightarrow {\left(\frac{1+a+a}{1-a^2}\right)}^2=1$
$\Rightarrow \left(\frac{1+2a}{1-a^2}\right)=\pm 1$

$\Rightarrow a^2+2a=0\text{or}{a}^2-2a-2=0$
$\Rightarrow a=0,-2\text{or}a=1\pm \sqrt{3}$