Question

If $\alpha=\lim _{x \rightarrow \pi / 4} \frac{\tan ^{3} x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}$ and $\beta=\lim _{x \rightarrow 0}(\cos x)^{\cot x}$ are the roots of the equation, $a x^{2}+b x-4=0$, then the ordered pair $(a, b)$ is :

• A. $(1,-3)$
• B. $(-1,3)$
• C. $(-1,-3)$
• D. $(1,3)$

Answer 1

Answer: (D)

$\alpha=\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^{3} x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} ; \frac{0}{0}$ form
Using L Hopital rule
$\alpha=\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{3 \tan ^{2} x \sec ^{2} x-\sec ^{2} x}{-\sin \left(x+\frac{\pi}{4}\right)} $
$\Rightarrow \alpha=-4$
$\beta=\displaystyle\lim _{x \rightarrow 0}(\cos x)^{\cot x}=e^{\lim _{x \rightarrow 0} \frac{(\cos x-1)}{\tan x}} $
$\beta=e^{\displaystyle\lim _{x \rightarrow 0} \frac{-(1-\cos x)}{x^{2}} \cdot \frac{x^{2}}{\left(\frac{\tan x}{x}\right)^{x}}}$
$\beta=e^{\displaystyle\lim _{x \rightarrow 0}\left(\frac{-1}{2}\right) \cdot \frac{x}{1}}=x_{1}$
$\alpha=-4 ; \beta=1$
If $a x^{2}+b x-4=0$ are the roots then
$16 a-4 b-4=0 \& a+b-4=0 $
$\Rightarrow a=1 \& b=3$