Question

Let $f:R\to A=\{y:0\le y<\frac{\pi }{2}\}$ be a function such that $f\left(x\right)={\tan }^{-1}(x^2+x+k),$ where $k$ is a constant. The value of $k$ for which $f$ is an onto function is

• A. $1$
• B. $0$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

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

For $f$ to be an onto function, Range and co-domain of $f$ should be equal

$\Rightarrow f\left(x\right)\ge 0\forall x\in R$

$\Rightarrow {\tan }^{-1}(x^2+x+k)\ge 0$

For the above equation to be valid for all $x$

We must have, the discriminant of $x^2+x+k=0$, is zero

$b^2-4ac=0$

$\Rightarrow 1-4k=0\Rightarrow k=\frac{1}{4}$