Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Prove that y= 4 sin theta/ 2 + cos theta - theta is an increasing function of theta in 0 to pie by 2

9. Prove that   y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is an increasing function of    \theta\: \: in\: \: [ 0 , \pi /2 ]

Answers (1)

best_answer

Given function is,
f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta

f^{'}(x) = \frac{dy}{d\theta} = \frac{4 \cos \theta(2+\cos \theta) - (-\sin \theta)4\sin \theta) }{(2+ \cos \theta )^2} - 1
                         = \frac{8 \cos \theta+4\cos^2 \theta + 4\sin^2 \theta - (2+ \cos \theta )^2 }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4(\cos^2 \theta + \sin^2 \theta) - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4 - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}                                                 
                         = \frac{4 \cos \theta-\cos^2 \theta }{(2+ \cos \theta )^2}
Now,  for  \theta \ \epsilon \ [0,\frac{\pi}{2}]
\\ 4 \cos \theta \geq \cos^2 \theta\\ 4 \cos \theta - \cos^2 \geq 0\\ and \ (2+\cos \theta)^2 > 0
So, f^{'}(x) > 0 \ for \ \theta \ in \ [0,\frac{\pi}{2}]
Hence, f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is increasing function in  \theta \ \epsilon \ [0,\frac{\pi}{2}]
    

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question