Consider the quadratic equation c - 5 x 2 - 2 c x + c - 4 = 0 ,   c ≠ 5 . Let S be the set of all…

Case $-I$

$c-5>0$

$\Rightarrow c>5 ...\left(i\right)$

Then, the graph of the quadratic is shown below

Hence, we have $f\left(0\right)>0$

$\Rightarrow c-4>0$

$\Rightarrow c>4 ...\left(ii\right)$

And $f\left(2\right)<0$

$\Rightarrow 4\left(c-5\right)-4c+c-4<0$

$\Rightarrow 4c-20-4c+c-4<0$

$\Rightarrow c<24 ...\left(iii\right)$

And $f\left(3\right)>0$

$\Rightarrow 9\left(c-5\right)-6c+c-4>0$

$\Rightarrow 9c-45-6c+c-4>0$

$\Rightarrow 4c-49>0$

$\Rightarrow c>\frac{49}{4} ...\left(iv\right)$

Hence, on taking the intersection of all the above inequalities, we get $c\in \left(\frac{49}{4}, 24\right).$

Case $-II$

$c-5<0$

$\Rightarrow c<5 ...\left(v\right)$

Then, the graph of the quadratic is shown below

Hence, we have $f\left(0\right)<0$

$\Rightarrow c<4 ...\left(vi\right)$

And, $f\left(2\right)>0$

$\Rightarrow c>24 ...\left(vii\right)$

Also, $f\left(3\right)<0$

$\Rightarrow c<\frac{49}{4} ...\left(viii\right)$

Hence, on taking the intersection of all the above inequalities, we get $\Rightarrow c\in \varphi$

Hence, $c\in \left(\frac{49}{4}, 24\right)$

Hence, the total number of element in $S$ are $11.$