Question

Let $f\left(x\right)={\sin }^{2}x-(2a+1)\sin x+(a-3).$ If $f\left(x\right)\le 0$ for all $x\epsilon [0,\frac{\pi }{2}],$ then range of values of $a$ is

• A. $[-3,0]$
• B. $[-3,\infty )$
• C. $[-3,3]$
• D. $(-\infty ,3]$

Answer 1

Answer: (C)

$[-3,3]$

The value of function will be calculated for all angles between $0$ and $\frac{\pi }{2}$
So, $f\left(0\right)\le 0$
$=>si{n}^{2}0-(2a+1)sin0+(a-3)\le 0$
$=>0-0+a-3\le 0$
$=>a\le 3$
And $f\left(\frac{\pi }{2}\right)\le 0$
$=>si{n}^2\frac{\pi }{2}-(2a+1)sin\frac{\pi }{2}+(a-3)\le 0$
$=>1-2a-1+a-3\le 0$
$=>-a-3\le 0$
$=>a\ge 3$
So, $a$ belongs to $[-3,3]$