Question

Let $f : R \rightarrow\left(\frac{-\pi}{2}, 0\right]$ be a function defined by $f ( x )=\tan ^{-1}\left(2 x - x ^2+\lambda\right)$. If $f$ is onto, then $\lambda$ lies in the interval

• A. $(-2,0)$
• B. $(0,2)$
• C. $(-1,1)$
• D. none

Answer 1

Answer: (A)

$\because f$ is onto
$\therefore$ Range of $\tan ^{-1}\left(2 x - x ^2+\lambda\right)$ should be $\left(\frac{-\pi}{2}, 0\right] $
$\Rightarrow$ Range of $2 x-x^2+\lambda$ should be $(-\infty, 0]$; hence $D=0 \Rightarrow 4+4 \lambda=0$, hence $\lambda=-1$