Grade 12Differential Calculus
que-3
The minimum value of the function 2cos2x-cos4x in 0<x<pie is
2 Answers
We have
f(x) = 2cos2x - cos4x
f ''(x) = -4sin2x + 4sin4x
= 4(sin4x - sin2x)
= 8cos3x.sinx
Now,
f ''(x) = 0 
8cos3xsinx= 0
cos3xsinx = 0
cos3x = 0 or sinx = 0
or x = 0, 
.
or x = 0, 
. 
(As 0 
x 
)
Now,
f(0) = 2cos0 - cos0 = 2 - 1 = 1 
f 
= 2cos
- cos2
= -2 - 1 = -3 
and f(
) = 2cos2
- cos4
= 2 - 1 = 1
Thus, the maximum value of f(x) is 
and the minimum value of f(x) is -3.
--
Regards
Ramesh
Hi pallavi
2cos2x-cos4x
=2 cos2x -2cos22x +1
=-2(cos2x-1/2)2 +3/2
minimum value occur when cos2x-1/2 will be maximum
=-2(-1-1/2)2 +3/2
=-3
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