Question

If $y=\cot \theta ({\sin }^2\theta +\sin \theta \cos \theta )$, then

• A. $y_{min}=\frac{1-\sqrt{2}}{4}$
• B. $y_{min}=\frac{1-\sqrt{2}}{2}$
• C. $y_{max}=\frac{1+\sqrt{2}}{4}$
• D. $y_{max}=\frac{1+\sqrt{2}}{2}$

Answer 1

Answer: (B), (D)

The correct options are
B $y_{min}=\frac{1-\sqrt{2}}{2}$
D $y_{max}=\frac{1+\sqrt{2}}{2}$

$y=\cot \theta ({\sin }^2\theta +\sin \theta \cos \theta )=\frac{\cos \theta }{\sin \theta }({\sin }^2\theta +\sin \theta \cos \theta )=\cos \theta (\sin \theta +\cos \theta )=\cos \theta \sin \theta +{\cos }^2\theta =\frac{\sin 2\theta }{2}+\frac{1+\cos 2\theta }{2}=\frac{1}{2}+\frac{\sin 2\theta +\cos 2\theta }{2}$

Now, $-\sqrt{2}\le \sin 2\theta +\cos 2\theta \le \sqrt{2}$
$\Rightarrow \frac{1-\sqrt{2}}{2}\le \frac{1}{2}+\frac{\sin 2\theta +\cos 2\theta }{2}\le \frac{1+\sqrt{2}}{2}$