Question

The set of all real values of a so that the range of the function $y=\frac{x^2+a}{x+1}$ is $R$, is

• A. $[1,\infty )$
• B. $(-\infty ,-1)$
• C. $(1,\infty )$
• D. $(-\infty ,-1]$

Answer 1

Answer: (B)

We have $y=\frac{x^2+a}{x+1}$
$\Rightarrow x^2-yx+(a-y)=0$
As $x\in R$, so discriminant $\ge 0$
$\Rightarrow y^2-4(a-y)\ge 0,\forall y\in R$
$\Rightarrow y^2+4y-4a\ge 0\forall y\in R$
$\Rightarrow \text{ It is possible if }16+16a\le 0$
$\Rightarrow 1+a\le 0$
$\Rightarrow a\le -1\text{ but }a\ne -1$
$\therefore a\in (-\infty ,-1)$