(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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$\frac{\pi}{12}$ 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 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

Difficult

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

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

2 Complex Numbers and Quadratic Equations: Quadratic Equations

Subtopic

Solution of quadratic equations and Nature of roots

Previous Year

KVPY 2021 Paper All KVPY years →

The roots of $|x - 2|^2 + |x - 2| - 6 = 0$ are

Medium

View Solution

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then $x^2(x-4)^2$ is equal to :

DifficultAP EAMCET 2006

View Solution

If both roots of the equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$ are negative,then in which interval does $p$ lie?

Medium

View Solution

If the roots of the equation $x^2+2ax+b=0$ are real,distinct and differ at most by $2m$,then $b$ lies in the interval

MediumAP EAMCET 2025

View Solution

The product of all the rational roots of the equation $(x^2-9x+11)^2-(x-4)(x-5)=3$ is equal to:

DifficultJEE MAIN 2025

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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

Similar Questions

The roots of $|x - 2|^2 + |x - 2| - 6 = 0$ are

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then $x^2(x-4)^2$ is equal to :

If both roots of the equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$ are negative,then in which interval does $p$ lie?

If the roots of the equation $x^2+2ax+b=0$ are real,distinct and differ at most by $2m$,then $b$ lies in the interval

The product of all the rational roots of the equation $(x^2-9x+11)^2-(x-4)(x-5)=3$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module