If α = _ x → π / 4 ^ 3 x - x (x + (π)/(4)) and β = _ x → 0 ( x) ^ x...

If, \beta = \lim_{x \to 0}{\left(\text{ \cos x }\right)}^{\cot x}$(${1}^{\infty}$form)</p><p>$\beta = {e}^{\lim_{x \to 0}\cot x\left[\cos x - 1\right]}$</p><p>$\beta = {e}^{\lim_{x \to 0}\dfrac{- \cos x}{\sin x}\left[1 - 1 + 2\sin^{2}\dfrac{x}{2}\right]}$</p><p>$\beta = {e}^{\lim_{x \to 0}\dfrac{- \cos x}{2sin\dfrac{x}{2} . cos\dfrac{x}{2}}\left[1 - 1 + 2\sin^{2}\dfrac{x}{2}\right]}$</p><p>$\beta = {e}^{\lim_{x \to 0}\dfrac{- \cos x . sin\dfrac{x}{2}}{cos\dfrac{x}{2}}}$</p><p>$\beta = {e}^{0}$</p><p>$\beta = 1$<br />$\alpha = \lim_{x \to \pi / 4}\dfrac{\tan^{3}x - \tan x}{\cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$= \lim_{x \to \pi / 4}\dfrac{\tan x ( \tan x + 1 ) ( \tan x - 1 )}{\cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$\alpha = \lim_{x \to \pi / 4}\tan x \times \lim_{x \to \pi / 4}\left(\tan x + 1\right) \times \lim_{x \to \pi / 4}\dfrac{( \tan x - 1 )}{\cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$\alpha = 1 \times \left(1 + 1\right) \times \lim_{x \to \pi / 4}\dfrac{( \tan x - 1 )}{\cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$= 2\lim_{x \to \pi / 4}\dfrac{\tan x - 1}{\cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$= 2\lim_{x \to \pi / 4}\dfrac{\sin x - \cos x}{\cos x . \cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$= - 2\sqrt{2}\lim_{x \to \pi / 4}\dfrac{\left(\dfrac{1}{\sqrt{2}}\cos x - \dfrac{1}{\sqrt{2}}\sin x\right)}{\cos x . \cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$\alpha = - 2\sqrt{2}\lim_{x \to \pi / 4}\dfrac{\cos \left(x + \dfrac{\pi}{4}\right)}{\cos x . \cos \left(x + \dfrac{\pi}{4}\right)}$</p><p>$\alpha = - 4$</p><p>So, the equation whose roots are$\alpha and $\beta$ is

${x}^{2} + 3x - 4 = 0$

Hence, $a = 1, b = 3$