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Find the maximum value of the function f (x) = 1 + 2 sin x + 3 cos2 x , 0 = x = 2p/3.

Find the maximum value of the function f (x) = 1 + 2 sin x + 3 cos2 x , 0 = x = 2p/3.

Grade:11

6 Answers

Robin Chourasia
askIITians Faculty 9 Points
7 years ago
y= 1 + 2 sin x + 3 cos2x
y'= 2 cos x -6 sin 2x
y''= -2sinx-12cos2x

from y' we get,

2 cos x(1-6sinx)=0

=> x=90 or sin x= 1/6

y'' at x= 90 we get y'' = -2 - 12 (-1) = 10
at sinx = 1/6 we get y'' = -2 *1/6 -12 (1- 1/18)= -1/3- 34/3 =-35/3

Substituting both values in Y we get, @ x = 90 y = 0 @ sinx = 1/6 y = 1+ 1/3 + 3 (1-1/18)= 25/6 is maximum value..

Regards,

Robin
askiitian faculty
Parvez ali
askIITians Faculty 47 Points
7 years ago

147-1177_140303-105816.jpg

Thanks & Regards
Parvez Ali
askIITians Faculty




Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
f'(x)=2cosx-6cosxsinx=0
2cosx(1-3sinx)=0
cosx=0 or sin(x)=1 f(x)=1+2=3
sinx=1/3 imply cos^2(x)=1-1/9=8/9

f(x)=1+2/3+3*8/9=1+10/3=4.33
hence answer is 4.33.

Sher Mohammad
B.Tech, IIT Delhi.
bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
f(x)= -3sin2x + 2sinx + 4
= -3(sin2x-2/3 sinx)+
=-3(sinx-1/3)^2+3/9+4
=-3(sinx - 1/3)+13/3

Now, for f(x) to be max, the first term should be minimum.
Hence it is at sin x =0
Hence, max. value of f(x) = 16/3
Thanks & Regards
Bharat Bajaj
askIITians Faculty
IIT Delhi


bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
f(x)= -3sin2x + 2sinx + 4
= -3(sin2x-2/3 sinx)+
=-3(sinx-1/3)^2+3/9+4
=-3(sinx - 1/3)+13/3

Now, for f(x) to be max, the first term should be minimum.
Hence it is at sin x =0
Hence, max. value of f(x) = 16/3
Thanks & Regards
Bharat Bajaj
askIITians Faculty
IIT Delhi

Sunil Kumar FP
askIITians Faculty 183 Points
7 years ago

204-1662_20140303_130218.jpg

thanks and regards
sunil kr
askIItian faculty

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