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