Question

Let $f:R\to [0,\frac{\pi }{2})$ be defined by $f\left(x\right)=ta{n}^{-1}(x^2+x+a)$. Then the set of values of '$a$' for which $f$ is onto is

• A. $[0,\infty )$
• B. $[2,1]$
• C. $\left\{\frac{1}{4}\right\}$
• D. $[\frac{1}{4},\infty )$

Answer 1

The correct option is C $\left\{\frac{1}{4}\right\}$

$f\left(x\right)={\tan }^{-1}(x^2+x+a)$
For, $f\left(x\right)$ to be onto, codomain should be exactly equal to range.
That is, range of funnction $f\left(x\right)=[0,\frac{\pi }{2})$
So, $0⩽{\tan }^{-1}(x^2+x+a)<\frac{\pi }{2}$
Now, $x^2+x+a={(x+\frac{1}{2})}^2+a-\frac{1}{4}$
The above expression will take all real values from $[0,\infty )$, only if $a=\frac{1}{4}$
Hence, only for $a=\frac{1}{4}$, the function is onto.