Question

For what real value of $a$ does the range of the function $y=\frac{x+1}{a+x^2}$ contain the interval $\left[0.1\right]$?

Answer 1

$y=\frac{x+1}{a+x^2}$ not contain value in $[0,1]$

$\Rightarrow ay+x^{2}y=x+1$

$\Rightarrow y{x}^2+x+ay+1=0$

Discriminant must be greater or equal to zero,

$D\equiv 1^2-4(ay+1)\left(y\right)\ge 0$

$\Rightarrow 1-4a{y}^2-4y\ge 0$

$\Rightarrow 4a{y}^2+4y-1\le 0$

$\Rightarrow y-(-\frac{1}{2a}\pm \frac{1}{2a}\sqrt{a+1})\le 0$

$\Rightarrow y\in [-\frac{1}{2a}-\frac{1}{2a}\sqrt{a+1},-\frac{1}{2a}+\frac{1}{2a}\sqrt{a+1}]$ -Equation 1

For the condition stated in question,

$-\frac{-1}{2a}-\frac{\sqrt{a+1}}{2a}>1$

$\Rightarrow a+1<4a^2-4a+1$(On squaring)

$\Rightarrow 4a^2-3a>0$

$\Rightarrow a(a-\frac{3}{4})>0$

$a\in (-\infty ,0)\cup (\frac{3}{4},\infty )$ -Equation 2

Or $-\frac{1}{2a}+\frac{\sqrt{a+1}}{2a}<0$

$\Rightarrow a+1<4a^2$(On squaring)

$\Rightarrow 4a^2-a-1>0$

$\Rightarrow (a-\frac{1+\sqrt{1}7}{8})(a-\frac{1-\sqrt{1}7}{8})>0$

$a\in (-\infty ,\frac{1-\sqrt{1}7}{8})\cup (\frac{1+\sqrt{1}7}{8},\infty )$ -Equation 3

Also, for definition of $\sqrt{a+1},a\ge -1\Rightarrow a\in [-1,\infty )$ - Equation 4

From, Equation 2,3 & 4

$a\in [-1,0)\cup (\frac{1+\sqrt{1}7}{8},\infty )$