Question

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

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

Answer 1

Answer: (D)

$\alpha =\underset{x\to \frac{\pi }{4}}{\lim }\frac{{\tan }^{3}x-\tan x}{\cos \left(x+\frac{\pi }{4}\right)};\frac{0}{0}$ form
Using L Hopital rule
$\alpha =\underset{x\to \frac{\pi }{4}}{\lim }\frac{3{\tan }^{2}x{\sec }^{2}x-{\sec }^{2}x}{-\sin \left(x+\frac{\pi }{4}\right)}$
$\Rightarrow \alpha =-4$
$\beta =\underset{x\to 0}{\lim }(\cos x{)}^{\cot x}=e^{{\lim }_{x\to 0}\frac{(\cos x-1)}{\tan x}}$
$\beta =e^{\underset{x\to 0}{\lim }\frac{-(1-\cos x)}{x^2}\cdot \frac{x^2}{{\left(\frac{\tan x}{x}\right)}^x}}$
$\beta =e^{\underset{x\to 0}{\lim }\left(\frac{-1}{2}\right)\cdot \frac{x}{1}}=x_1$
$\alpha =-4;\beta =1$
If $a{x}^2+bx-4=0$ are the roots then
$16a-4b-4=0\&a+b-4=0$
$\Rightarrow a=1\&b=3$