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The maximum and minimum values of the function 1+2sinx+2cos 2 x in(0,pi/2) where() are closed brackets.

The maximum and minimum values of the function 1+2sinx+2cos2x in(0,pi/2) where() are closed brackets.

Grade:11

2 Answers

Rinkoo Gupta
askIITians Faculty 81 Points
9 years ago
f(x)=1+2sinx+2cos2x
f’(x)=2cosx-4cosxsinx=0
2cosx(1-2sinx)=0
cosx=0 or sinx=1/2
x=pi/2 or s=pi/6
pi/6 belongs to (0,pi/2) so we check this point.
f”(x)=-2sinx-4cos2x
f”(pi/6)=-2sin(pi/6)-4cos(pi/3)
=-2(1/2)-4(1/2)
=-1-2=-3 which is negative
so function is maximum at x=pi/6
so f(pi/6)=1+2sin(pi/6)+2cos2(pi/6)
=1+2(1/2)+2(sqrt3/2)2
=1+1+2(3/4)
=2+3/2=7/2
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
f(x)=1+2sinx+2cos^{2}x
Lets find the critical points
f^{'}(x)=0
f^{'}(x)=2cosx+4cosx(-sinx)=0
2cosx(1-2sinx)=0
cosx=0, sinx = \frac{1}{2}
x = \frac{\pi }{6}, \frac{\pi }{2}
To find local minima & maxima, we will find second derivative at critical points
f^{'}(x) = 2cosx-2sin2x
f^{''}(x) = -2sinx-4cos2x
f^{''}(\frac{\pi }{6}) = -2sin(\frac{\pi }{6})-4cos2(\frac{\pi }{6}) = -3
f^{''}(\frac{\pi }{2}) = -2sin(\frac{\pi }{2})-4cos2(\frac{\pi }{2}) = 2
x = \frac{\pi }{2}
is the minima
x = \frac{\pi }{6}
is the maxima
Maximum value of function:
f(\frac{\pi }{6}) = \frac{7}{2}
Minimumvalue of function:
f(\frac{\pi }{2}) = 3
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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