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Find the maximum value of the function f(x) = 2sin x + sin 2x - 1 is at 
1. π 
2. \(\pi\over3\)
3. \(\pi\over6\)
4. \(\pi\over2\)

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Correct Answer - Option 2 : \(\pi\over3\)

Concept:

Following steps to finding minima using derivatives.

  • Find the derivative of the function.
  • Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
  • Now we have to find the second derivative: If f"(x) Is greater than 0 then the function is said to be minima


Calculation:

f(x) = 2sin x + sin 2x - 1

Differentiating wrt x

f'(x) = 2 cos x + 2 cos 2x

Putting f'(x) = 0

2 cos x + 2(2 cos2 x - 1) = 0

cos x + 2 cos2 x - 1 = 0

Let cos x = a

2a2 + a - 1 = 0

\(\rm a = {-1 \pm \sqrt{1^2-4(2)(-1)} \over 2(2)}\)

\(\rm a = {-1 \pm 3 \over 4}\)

a = -1, 0.5

If cos x = -1 ⇒ x = π

If cos x = 0.5 ⇒ x = \(\pi\over3\)

Now f''(x) = -2 sin x + 2 (-2 sin 2x)

At  x = \(\pi\over3\),

f''(x) = -2(sin \(\pi\over3\) + 2 sin \(2\pi\over3\)) = \(-3\sqrt3\) < 0

∴ f(x) is maximum at  x = \(\pi\over3\)

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