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

1.	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» 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.

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»

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.