f(x) = 1 + 2 sinx + 3cos^2x, 0 ≤ x ≤ 2π/3 is : A. Minimum at x = π/2 B. Maximum at x = sin^(-1)(1/3) ​

Question

f(x) = 1 + 2 sinx + 3cos²x, 0 ≤ x ≤ \(\frac{2\pi}{3}\) is :

A. Minimum at x = \(\frac{\pi}{2}\)

B. Maximum at x = sin^-1(\(\frac{1}{3}\))

C. Minimum at x = \(\frac{\pi}{6}\)

D. Maximum at  sin^-1(\(\frac{1}{6}\))

Answer 1

f(x) = 1 + 2sinx + 3cos²x, 

x ∈ [0,\(\frac{2\pi}{3}\)]

Differentiating f(x) with respect to x, we get

f’(x)=2cosx-6cosxsinx 

Also, 

Differentiating f’(x) with respect to x, we get

f’’(x) = - 2sinx - 6cos²x + 6sin²x 

=- 2sinx + 12sin²x - 6 (Since sin²x + cos²x = 1)

For extreme points, 

f’(x) = 0 

⇒ 2cosx(1 - 3sinx) = 0 

or x = \(\frac{\pi}{2}\)

or sin^-1(\(\frac{1}{3}\))

f"(\(\frac{\pi}{2}\)) = 4 > 0 and f"( sin^-1(\(\frac{1}{3}\)))

= \(-\frac{16}{3}\) < 0

Hence,

x = \(\frac{\pi}{2}\) is a point of minima and x = sin^-1(\(\frac{1}{3}\)) s a point of maxima.