Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the intervals in which the function f given by f x= 4 sin x - 2x - x cos x / 2+cos x is increasing

6) Find the intervals in which the function f given by f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x } is  

(i) increasing

Answers (1)

best_answer

Given function is
f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }
f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}
             =\frac{4\cos x - \cos^2x}{2+\cos x}
f^{'}(x)=0\\ \frac{4\cos x - \cos^2x}{2+\cos x} =0\\ \cos x(4-\cos x) = 0\\ \cos x = 0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \cos x =4
But    \cos x \neq 4
So,
\cos x = 0 \\ x = \frac{\pi}{2} \ and \ \frac{3\pi}{2}
Now three ranges are there \left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )
In interval \left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )  , f^{'}(x) > 0

Hence, the given  function  f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }  is increasing in the interval  \left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )
in interval ,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0  so function is decreasing in this inter
 

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 Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question