Question

Let $A=\{a:0\le a<\frac{\pi }{2}\}$ and $f:R\to A$ be an onto function given by $f\left(x\right)={\tan }^{-1}(x^2+x+\lambda ),$ where $\lambda$ is a constant. Then $\left[\lambda \right]$ is,
(where $[.]$ represents greatest integer function)

• A. $-1$
• B. $1$
• C. $0$
• D. None of the above

Answer 1

The correct option is C $0$

Given $f$ is onto $\Rightarrow$ Range of $f=A$
$\Rightarrow 0\le f\left(x\right)<\frac{\pi }{2}$
$\Rightarrow 0\le {\tan }^{-1}(x^2+x+\lambda )<\frac{\pi }{2}$
$\Rightarrow 0\le x^2+x+\lambda <\infty$
$x^2+x+\lambda \in [0,\infty )$
which is possible iff $D=0$
$\Rightarrow 1-4\lambda =0$
$\Rightarrow \lambda =\frac{1}{4}$
$\therefore \left[\lambda \right]=0$

Alternate Solution:
$x^2+x+\lambda \in [0,\infty )$
$\Rightarrow {(x+\frac{1}{2})}^2+\lambda -\frac{1}{4}\ge 0$
Since, ${(x+\frac{1}{2})}^2\ge 0\forall x\in R$
Hence, to satisfy above one, we must have
$\lambda -\frac{1}{4}=0$
$\Rightarrow \lambda =\frac{1}{4}$