Consider the quadratic equation (c-5)x^2-2cx+(c-4)=0,c≠ 5. Let S be the set of all integral values of c for which one root of the equation lies in…

Case $-I$−I

$$c-5>0$$c−5>0

$$\Rightarrow c>5...(i)$$⇒c>5...(i)

Then, the graph of the quadratic is shown below

Hence, we have $f(0)>0$f(0)>0

$$\Rightarrow c-4>0$$⇒c−4>0

$$\Rightarrow c>4...(ii)$$⇒c>4...(ii)

And $f(2)<0$f(2)<0

$$\Rightarrow 4(c-5)-4c+c-4<0$$⇒4(c−5)−4c+c−4<0

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

$$\Rightarrow c<24...(iii)$$⇒c<24...(iii)

And $f(3)>0$f(3)>0

$$\Rightarrow 9(c-5)-6c+c-4>0$$⇒9(c−5)−6c+c−4>0

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

$$\Rightarrow 4c-49>0$$⇒4c−49>0

$$\Rightarrow c>\frac{49}{4}...(iv)$$⇒c>449​...(iv)

Hence, on taking the intersection of all the above inequalities, we get $c\in (\frac{49}{4}, 24).$c∈(449​,24).

Case $-II$−II

$$c-5<0$$c−5<0

$$\Rightarrow c<5...(v)$$⇒c<5...(v)

Then, the graph of the quadratic is shown below

Hence, we have $f(0)<0$f(0)<0

$$\Rightarrow c<4...(vi)$$⇒c<4...(vi)

And, $f(2)>0$f(2)>0

$$\Rightarrow c>24...(vii)$$⇒c>24...(vii)

Also, $f(3)<0$f(3)<0

$$\Rightarrow c<\frac{49}{4}...(viii)$$⇒c<449​...(viii)

Hence, on taking the intersection of all the above inequalities, we get $\Rightarrow c\in \phi$⇒c∈ϕ

Hence, $c\in (\frac{49}{4}, 24)$c∈(449​,24)

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