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Q.
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
Relations and Functions - Part 2
Solution:
We have $y=\frac{x^2+a}{x+1}$
$\Rightarrow x ^2- yx +( a - y )=0$
As $x \in R$, so discriminant $\geq 0$
$\Rightarrow y^2-4(a-y) \geq 0, \forall y \in R $
$ \Rightarrow y^2+4 y-4 a \geq 0 \forall y \in R$
$\Rightarrow \text { It is possible if } 16+16 a \leq 0$
$ \Rightarrow 1+ a \leq 0$
$\Rightarrow a \leq-1 \text { but } a \neq-1 $
$\therefore a \in(-\infty,-1) $