Question

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

• A. $R-\{-1\}$
• B. $(-\infty ,-1)$
• C. $(1,\infty )$
• D. $(-\infty ,-1]$

Answer 1

The correct option is B $(-\infty ,-1)$

$y=\frac{x^2+a}{x+1}\Rightarrow xy+y=x^2+a\Rightarrow x^2-yx+(a-y)=0$
As $x\in R-\{-1\}$,
$D\ge 0\Rightarrow y^2-4(a-y)\ge 0\forall y\in R\Rightarrow y^2+4y-4a\ge 0\forall y\in RD\le 0$
$16+16a\le 0\Rightarrow a\le -1\therefore a\in (-\infty ,-1]$

Now, checking on the boundary values, i.e. at $a=-1$
$y=\frac{x^2-1}{x+1}=x-1$
When $x=-1,y=-2$
So, the range of $\frac{x^2-1}{x+1}$ in this case is $R-\{-2\}$
So, $a\ne -1$

Hence, $a\in (-\infty ,-1)$