Prove the following


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 :

  1. $(1,-3)$

  2. $(-1,3)$

  3. $(-1,-3)$

  4. $(1,3)$

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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