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Question:
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Which of the following statements is /are correct for $0 < \theta < \frac{\pi}{2} ?$
Options:
Solution:
2154 Upvotes
Verified Answer
The correct answers are:
$(\cos \theta)^{1 / 2} \leq \cos \frac{\theta}{2}$, $\cos \frac{5 \theta}{6} \geq(\cos \theta)^{5 / 6}$
$0 < 0 < \frac{\pi}{2}, cos \theta$ is monotonically decreasing function
Option
(a) When $\theta>\frac{\theta}{2},$ then $\cos \theta \leq \cos \frac{\theta}{2}$
$\therefore \quad \sqrt{\cos \theta} \leq \cos \frac{\theta}{2} \quad$ correct
(b) When $0>\frac{30}{4},$ then $\cos \theta \leq \cos \frac{30}{4}$
$\therefore \quad(\cos \theta)^{3 / 4} \leq \cos \frac{30}{4}$
incorrect|
(c) When $\theta>\frac{5 \theta}{6},$ then $\cos \theta \leq \cos \frac{50}{6}$
$\therefore \quad(\cos \theta)^{5 / 6} \leq \cos \frac{50}{6}$
[correct|
(d) When $\frac{7 \theta}{8} < \theta,$ then $\cos \theta \leq \cos \frac{7 \theta}{8}$
But $(\cos \theta)^{7 / 8} \leq \cos \frac{7 \theta}{8}$
|incorrect|
Option
(a) When $\theta>\frac{\theta}{2},$ then $\cos \theta \leq \cos \frac{\theta}{2}$
$\therefore \quad \sqrt{\cos \theta} \leq \cos \frac{\theta}{2} \quad$ correct
(b) When $0>\frac{30}{4},$ then $\cos \theta \leq \cos \frac{30}{4}$
$\therefore \quad(\cos \theta)^{3 / 4} \leq \cos \frac{30}{4}$
incorrect|
(c) When $\theta>\frac{5 \theta}{6},$ then $\cos \theta \leq \cos \frac{50}{6}$
$\therefore \quad(\cos \theta)^{5 / 6} \leq \cos \frac{50}{6}$
[correct|
(d) When $\frac{7 \theta}{8} < \theta,$ then $\cos \theta \leq \cos \frac{7 \theta}{8}$
But $(\cos \theta)^{7 / 8} \leq \cos \frac{7 \theta}{8}$
|incorrect|
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