Question

Let $f:(-\infty,-1] \rightarrow\left(\frac{\pi}{2}, \pi\right]$ be defined as $f(x)=\sec ^{-1}\left(-x^2+x+a\right)$. If $f(x)$ is surjective, then the range of $a$ is

• A. $\{1\}$
• B. $\left\{\frac{-5}{4}\right\}$
• C. $\left(-\infty, \frac{-5}{4}\right]$
• D. $(-\infty, 1]$

Answer 1

Answer: (A)

For $f ( x )$ to be surjective.
Range of $f(x)$ must be $\left(\frac{\pi}{2}, \pi\right]$ and hence range of $y=-x^2+x+$ a must be $(-\infty,-1] \forall x \in(-\infty,-1]$
$\therefore y=-\left(x^2-x-a\right)=-\left[\left(x-\frac{1}{2}\right)^2-\left(a+\frac{1}{4}\right)\right]$
$\Rightarrow y = a +\frac{1}{4}-\left( x -\frac{1}{2}\right)^2 \Rightarrow y \in(-\infty, a -2] \text { [Maximum value occurs at } x =-1 $
$\therefore a -2=-1 \Rightarrow a =1$