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?
$f(x)= \frac{x+a}{x^2-1}$
$a \in {1,2,3,.....,100}$
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.