Question

If $y=4{\sin }^2\theta -\cos 2\theta$, then the interval in which $y$ lies is

• A. $(-1,5)$
• B. $[-1,5]$
• C. $(-\infty ,-1)\cup (5,\infty )$
• D. $noneofthese$

Answer 1

Answer: (B)

$[-1,5]$

We have, $y=4{\sin }^2\theta -\cos 2\theta$

$=4{\sin }^2\theta -(1-2{\sin }^2\theta )=6{\sin }^2\theta -1$

$y=6{\sin }^2\theta -1$

$\Rightarrow y+1=6{\sin }^2\theta$

$\therefore \frac{y+1}{6}={\sin }^2\theta$

But $\sin \theta$lies in the interval $[0,1]$

$\Rightarrow 0\le {\sin }^2\theta \le 1$

$\therefore 0\le \frac{y+1}{6}\le 1$

$\Rightarrow 0\le y+1\le 6$

$\Rightarrow 0-1\le y\le 6-1$

$\Rightarrow -1\le y\le 5$

$\Rightarrow y\in [-1,5]$