Maxima and MinimaHard
Question
f(x) = 2sinx + cos2x, 0 < x < 2π is maximum at-
Options
A.x = π/2
B.x = 3π/2
C.x = π/6
D.No where
Solution
f(x) = 2 sin x + cos 2x, 0 ≤ x ≤ 2 π
f′(x( = 2 cos x - 2 sin 2x
⇒ f′(x) = 0 ⇒ 2cos x (1 - 2 sin x ) = 0
⇒ x =
Now , f″(x) = -2 sin x - 4cos 2x
⇒ f″ = - 2,
= - 2. 
- 4. 
= -3 < 0
⇒ f″
= -2 - 4 (- 1) = 2 > 0
Hence f(x) is maixmum at x = π/6
f′(x( = 2 cos x - 2 sin 2x
⇒ f′(x) = 0 ⇒ 2cos x (1 - 2 sin x ) = 0
⇒ x =
Now , f″(x) = -2 sin x - 4cos 2x
⇒ f″ = - 2,
= - 2. - 4. = -3 < 0⇒ f″
= -2 - 4 (- 1) = 2 > 0Hence f(x) is maixmum at x = π/6
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