SATHEE: Trigonometrical Equations 3 Question 6

Trigonometrical Equations 3 Question 6

7. The equation $(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$ in the variable $x$ , has real roots. Then, $p$ can take any value in the interval

(1990, 2M)

(a) $(0, 2 π)$

(b) $(- π, 0)$

(d) $(0, π)$

Show Answer

Answer:

Correct Answer: 7. (d)

Solution:

Since, the given quadratic equation

$(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$

has real roots.

$∴$ Discriminant, $\cos^{2} p - 4 \sin p (\cos p - 1) \geq 0$

$\Rightarrow (\cos p - 2 \sin p)^{2} - 4 \sin^{2} p + 4 \sin p \geq 0$

$\Rightarrow (\cos p - 2 \sin p)^{2} + 4 \sin p (1 - \sin p) \geq 0$

$∵ 4 \sin p (1 - \sin p) > 0$ for $0 < p < π$

and $(\cos p - 2 \sin p)^{2} \geq 0$

Thus, $(\cos p - 2 \sin p)^{2} + 4 \sin p (1 - \sin p) \geq 0$

for

$0 < p < π$ .

Hence, the equation has real roots for $0 < p < π$ .

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Trigonometrical Equations 3 Question 6

7. The equation (cos⁡p−1)x2+(cos⁡p)x+sin⁡p=0 in the variable x, has real roots. Then, p can take any value in the interval

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

7. The equation $(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$ in the variable $x$ , has real roots. Then, $p$ can take any value in the interval