Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the absolute maximum and minimum values of the function f given by f (x) = cos ^ 2 x + sin x, x belongs to 0 , pi

14) Find the absolute maximum and minimum values of the function f given by
f (x) = \cos ^2 x + \sin x , x \epsilon [ 0 , \pi ]

Answers (1)

best_answer

Given function is 
f (x) = \cos ^2 x + \sin x
f^{'} (x) = 2\cos x(-\sin x) + \cos x \\ f^{'}(x) = 0\\ -2\cos x\sin x + \cos x=0\\ \cos x(1-2\sin x) = 0\\ either\\ \cos x = 0 \ \ \ \ \ \ and \ \ \ \ \ \ \ \sin x = \frac{1}{2} \\ x = \frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{6} \ \ \ \ \ as \ x \ \epsilon [0,\pi]
Now,
f^{''} (x) = -2(-\sin x)\sin x - 2\cos x\cos x + (-\sin x) \\ f^{''}(x)= 2\sin^2x - 2\cos^2x - \sin x\\ f^{''}(\frac{\pi}{6}) = \frac{-3}{2} < 0
Hence, the point  x = \frac{\pi}{6}  is the point of maxima and the maximum value is
f (\frac{\pi}{6}) = \cos ^2 \frac{\pi}{6} + \sin \frac{\pi}{6} = \frac{3}{4}+\frac{1}{2} = \frac{5}{4}
And
f^{''}(\frac{\pi}{2}) = 1 > 0
Hence, the point  x = \frac{\pi}{2} is the point of minima and the minimum value is
f (\frac{\pi}{2}) = \cos ^2 \frac{\pi}{2} + \sin \frac{\pi}{2} =0 + 1 = 1

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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question