Question

The number of intergral values of $a$ for which $y=\frac{x^2-ax+11}{x^2-5x+4}$ can take all real values is

Answer 1

Given $y=\frac{x^2-ax+11}{x^2-5x+4}$
Finding the common root, we get
$x^2-ax+11=0x^2-5x+4=0\Rightarrow (5-a)x+7=0$
Let $a\ne 5$, then
$\Rightarrow x=\frac{7}{a-5}$

Now, $f\left(x\right)=x^2-ax+11$
$f\left(\frac{7}{a-5}\right)<0\Rightarrow \frac{49}{(a-5{)}^2}-\frac{7a}{(a-5)}+11<0\Rightarrow 49-7a(a-5)+11(a-5{)}^2<0(\because a\ne 5)\Rightarrow 4a^2-75a+324a<0\Rightarrow 4a^2-48a-27a+324a<0\Rightarrow (a-12\left)\right(4a-27)<0\Rightarrow a\in (\frac{27}{4},12)\therefore a=7,8,9,10,11$

When $a=5$, we get
$y=\frac{x^2-5x+11}{x^2-5x+4}\Rightarrow y=1+\frac{7}{x^2-5x+4}$
Which cannot take all real values

Hence, there are $5$ possible values of $a$.