Question

If $\alpha =\underset{x\to \pi /4}{lim}\frac{{\tan }^{3}x-\tan x}{\cos (x+\frac{\pi }{4})}$ and $\beta =\underset{x\to 0}{lim}{\left(\cos x\right)}^{\cot x}$
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

The correct option is C $(1,3)$

$\alpha =\underset{x\to \pi /4}{lim}\frac{{\tan }^{3}x-\tan x}{\cos (x+\frac{\pi }{4})}\frac{0}{0}$ form
Applying L'Hospital Rule
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{3{\tan }^{2}x\cdot {\sec }^{2}x-{\sec }^{2}x}{-\sin (x+\frac{\pi }{4})}$
$\alpha =\frac{3\times 1\times 2-2}{-1}=-4$
$=-2\underset{x\to \frac{\pi }{4}}{lim}\frac{cos2x}{cos(x+\frac{\pi }{4})}\Rightarrow \alpha =-4$

$\beta =\underset{x\to 0}{lim}\left(\cos x{)}^{\cot x}\right(1^{\infty }$ form$)$
$=e^{\underset{x\to 0}{lim}\left(\cos x-1\right)\cot x}$
$=e^{\underset{x\to 0}{lim}\frac{\cos x-1}{\tan x}}$ $\frac{0}{0}$ form
Applying L'Hospital Rule
$=e^{\underset{x\to 0}{lim}\frac{-\sin x}{{\sec }^{2}x}}$
$=e^0\Rightarrow \beta =1$

Quadratic equation having roots $(\alpha ,\beta )=(-4,1)$ is
$x^2+3x-4=0$
Clearly $a=1,b=3$