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If \( \Large A = \sin^{2} \theta + \cos^{4} \theta \), then for all real values of \( \Large \theta \)

A) \( \Large 1 \le A \le 2 \)
B) \( \Large \frac{3}{4} \le A \le 1 \)
C) \( \Large \frac{13}{16} \le A \le 1 \)
D) \( \Large \frac{3}{4} \le A \le \frac{13}{16} \)

Correct Answer:


B) \( \Large \frac{3}{4} \le A \le 1 \)

Description for Correct answer:

We have, \( \Large A = \sin^{2} \theta + \cos^{4} \theta \)

= \( \Large \sin^{2} \theta + \cos^{2} \theta \cos^{2} \theta \le \sin^{2} \theta + \cos^{2} \theta \) (Since, \( \Large \cos^{2} \theta \le 1 \))

=> \( \Large \sin^{2} \theta + \cos^{4} \theta \le 1 => A \le 1 \)

Again, \( \Large \sin^{2} \theta + \cos^{4} \theta =1 - \cos^{2} \theta + \cos^{4} \theta \)

= \( \Large \cos^{4} \theta - \cos^{2} \theta + 1 \)

= \( \Large \left(\cos^{2} \theta - \frac{1}{2}\right)^{2}+\frac{3}{4}\ge \frac{3}{4} \)

Hence \( \Large \frac{3}{4} \le A \le 1 \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

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