If α=limx→π4 tan3⁡x−tan⁡xcos⁡x+π4 and β=limx→0 (cos⁡x)cot⁡x are the roots of the equation, ax2+bx−4=0, then the ordered pair (a, b) is

α=limx→π4 tan3⁡x−tan⁡xcos⁡x+π4 =limx→π4 tan⁡x(tan⁡x+1)(tan⁡x−1)cos⁡x+π4 =limx→π4 sin⁡xcos⁡x⋅sin⁡x−cos⁡xcos⁡xsin⁡x+cos⁡xcos⁡x12(cos⁡x−sin⁡x) =limx→π4 −2sin⁡x(sin⁡x+cos⁡x)cos3⁡x =−2×12×2122=−4and β=limx→0 (cos⁡x)cot⁡x =elimx→0 cos⁡x−1tan⁡x=elimx→0 −sin⁡xsec2⁡x=e0=1Equation whose roots are α and β, is x2+3x−4=0∴ a=1, b=3

If α=limx→π4 tan3⁡x−tan⁡xcos⁡x+π4 and β=limx→0 (cos⁡x)cot⁡x are the roots of the equation, ax2+bx−4=0, then the ordered pair (a, b) is

α=limx→π4 tan3⁡x−tan⁡xcos⁡x+π4 =limx→π4 tan⁡x(tan⁡x+1)(tan⁡x−1)cos⁡x+π4 =limx→π4 sin⁡xcos⁡x⋅sin⁡x−cos⁡xcos⁡xsin⁡x+cos⁡xcos⁡x12(cos⁡x−sin⁡x) =limx→π4 −2sin⁡x(sin⁡x+cos⁡x)cos3⁡x =−2×12×2122=−4and β=limx→0 (cos⁡x)cot⁡x =elimx→0 cos⁡x−1tan⁡x=elimx→0 −sin⁡xsec2⁡x=e0=1Equation whose roots are α and β, is x2+3x−4=0∴ a=1, b=3

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If $α = lim_{x \to \frac{π}{4}} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$ and $β = lim_{x \to 0} (\cos x)^{\cot x}$ are the roots of the equation, ${ax}^{2} + bx - 4 = 0$ , then the ordered pair (a, b) is

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(– 1, 3)

(1, 3)

(1, – 3)

(–1, – 3)

answer is D.

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Detailed Solution

$\begin{array}{l} α = lim_{x \to \frac{π}{4}} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})} \\ = lim_{x \to \frac{π}{4}} \frac{\tan x (\tan x + 1) (\tan x - 1)}{\cos (x + \frac{π}{4})} \\ = lim_{x \to \frac{π}{4}} \frac{\frac{\sin x}{\cos x} \cdot (\frac{\sin x - \cos x}{\cos x}) (\frac{\sin x + \cos x}{\cos x})}{\frac{1}{\sqrt{2}} (\cos x - \sin x)} \\ = lim_{x \to \frac{π}{4}} \frac{- \sqrt{2} \sin x (\sin x + \cos x)}{\cos^{3} x} \\ = \frac{- \sqrt{2} \times \frac{1}{\sqrt{2}} \times \sqrt{2}}{\frac{1}{2 \sqrt{2}}} = - 4 \end{array}$

$\begin{array}{l} and β = lim_{x \to 0} (\cos x)^{\cot x} \\ = e^{lim_{x \to 0} \frac{\cos x - 1}{\tan x}} = e^{lim_{x \to 0} - \frac{\sin x}{\sec^{2} x}} = e^{0} = 1 \end{array}$