(B) For the quadratic equation $x^2 + 4x \cos \theta + \cot \theta = 0$ to have equal roots,its discriminant $D$ must be zero.$D = b^2 - 4ac = (4 \cos

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$\frac{\pi}{12}$ or $\frac{5 \pi}{12}$

(B) For the quadratic equation $x^2 + 4x \cos \theta + \cot \theta = 0$ to have equal roots,its discriminant $D$ must be zero.$D = b^2 - 4ac = (4 \cos \theta)^2 - 4(1)(\cot \theta) = 0$$16 \cos^2 \theta - 4 \cot \theta = 0$$4 \cos^2 \theta = \cot \theta$$4 \cos^2 \theta = \frac{\cos \theta}{\sin \theta}$Since $0 $4 \cos \theta \sin \theta = 1$$2 \sin 2 \theta = 1$$\sin 2 \theta = \frac{1}{2}$Since $0 Therefore,$\theta = \frac{\pi}{12}$ or $\theta = \frac{5 \pi}{12}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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Let $\theta$,$0 < \theta < \pi / 2$,be an angle such that the equation $x^2 + 4x \cos \theta + \cot \theta = 0$ has equal roots for $x$. Then $\theta$ in radians is

KVPY 2021 All KVPY

A
$\frac{\pi}{6}$ only
B
$\frac{\pi}{12}$ or $\frac{5 \pi}{12}$
C
$\frac{\pi}{6}$ or $\frac{5 \pi}{12}$
D
$\frac{\pi}{12}$ only

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