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 = \(\pi\over3\)