Consider the quadratic equation $(c-5)x^2-2cx+(c-4)=0,$ $c\ne 5$  Let $S$ be the set of all integral values of$c$ for which one root of the equation lies in the interval $(0, 2)$ and its other root lies in the interval $(2, 3)$. Then the number of elements in $S$ is

 : Let $f\left(x\right)=(c-5)x^2-2cx+c-4$

$\therefore f\left(0\right)=c-4$

      $\begin{array}{l}f\left(2\right)=(c-5)4-4c+c-4=c-24\\ f\left(3\right)=(c-5)9-6c+c-4=4c-49\end{array}$

Now, $f\left(0\right)f\left(2\right)<0\text{ and }f\left(2\right)f\left(3\right)<0$

$\begin{array}{l}\Rightarrow (c-4)(c-24)<0\text{ and }(c-24)(4c-49)<0\\ \Rightarrow 4<c<24\text{...(i) and }49/4<c<24\end{array}$

From $\text{ (i) and (ii), c∈}(49/4,24)$

$\therefore S=\{13,14,15,\dots ,23\}$

So, number of elements in S is 11.