# Value of $a$ such that range contains the interval $[0,1]$

Find the number of integral values of $a$ in the interval $[0,100]$ so that the range of the function $y= \frac{x+a}{x^2-1}$, $x\in R$ contains the interval $[0,1]$?

After rearranging $y= \frac{x+a}{x^2-1}$, we get

$yx^2-x-(y+a)$ As $x \in R$, hence $Discriminant \geq0$ which gives

$1+4y(y+a)\geq0$

Could someone hint me as how to proceed from here?

• For $a>1$ the interval $(-a,-1)$ for $x$ maps onto the interval $[0,\infty)]$. For $a=1$ the function does not assume the value 0 (the value at $x=-1$ is $-2$ by continuity). For $a=0$ the interval $(-1,0]$ maps onto the interval $[0,\infty)$. So answer 100. Apr 27, 2016 at 21:18
• You can refer math.stackexchange.com/questions/1260052/… Apr 28, 2016 at 11:39

## 1 Answer

$f(x)= \frac{x+a}{x^2-1}$

$a \in {1,2,3,.....,100}$

When you take $a=1$, then $f(x)$ will not be able to achieve $0$ as denominator also becomes $0$. In other words $x= \pm 1$ are not in domain of function.

For all other values of $a$, $f(x)=0$ for $x=-a$

Now set $\frac{x+a}{x^2-1}=1$ and obtain quadratic in $x$

Try to use the fact that $Ax^2+Bx+C=0$ $(A,B,C \in R)$ has real roots when $AC \leq0$ .

I hope you can take it from here.