Question

Let $f:(-\infty ,-1]\to \left(\frac{\pi }{2},\pi \right]$ 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 $(-\infty ,-1]\forall x\in (-\infty ,-1]$
$\therefore y=-\left(x^2-x-a\right)=-\left[{\left(x-\frac{1}{2}\right)}^2-\left(a+\frac{1}{4}\right)\right]$
$\Rightarrow y=a+\frac{1}{4}-{\left(x-\frac{1}{2}\right)}^2\Rightarrow y\in (-\infty ,a-2]\text{ [Maximum value occurs at }x=-1$
$\therefore a-2=-1\Rightarrow a=1$