Question

If $\underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\frac{2}{3}$, where $\alpha ,\beta ,\gamma \in R$, then which of the following is NOT correct ?

• A. ${\alpha }^2+{\beta }^2+{\gamma }^2=6$
• B. $\alpha \beta +\beta \gamma +\gamma \alpha +1=0$
• C. $\alpha {\beta }^2+\beta {\gamma }^2+\gamma {\alpha }^2+3=0$
• D. ${\alpha }^2-{\beta }^2+{\gamma }^2=4$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{\alpha \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)+\beta \left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots \right)+\gamma \left(x-\frac{x^3}{3!}+\dots \right)}{x^3)}$
constant terms should be zero
$\Rightarrow a+\beta =0$
coeff of $x$ should be zero
$\Rightarrow \alpha -\beta +\gamma =0$
coeff of $x^2$ should be zero
$\underset{x\to 0}{\lim }\frac{x^3\left(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)+x^4\left(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)}{x^3}=\frac{2}{3}$
$\Rightarrow \frac{\alpha }{2}+\frac{\beta }{2}=0$
$\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=2/3$
$\Rightarrow \alpha =1,\beta =-1,\gamma =-2$