Question

The range of the function ${\sin }^{2}x-5\sin x-6$ is :

• A. $[-10,0]$
• B. $[-1,1]$
• C. $[0,\pi ]$
• D. $[-\frac{49}{4},0]$

Answer 1

The correct option is A $[-10,0]$

$f\left(x\right)=si{n}^{2}x-5\sin x-6$

$f\left(x\right)={({\sin }^{2}x-\frac{5}{2})}^2-6-\frac{25}{4}$

$f\left(x\right)={({\sin }^{2}x-\frac{5}{2})}^2-\frac{49}{4}$

We know $-1\le \sin x\le 1$

When $\sin x=-1$

$f\left(x\right)=\frac{49}{4}-\frac{49}{4}=0$

When, $\sin x=1$

$f\left(x\right)=\frac{9}{4}-\frac{49}{4}=\frac{-40}{4}=-10$

When, $\sin x=0$

$f\left(x\right)=\frac{25}{4}-\frac{49}{4}=-6$

So, here we can see minimum value is $-10$ and maximum value is $0.$

$\therefore$ Required range is $[-10,0]$