Consider the quadratic equation (c - 5)x^2 - | vedclass

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(D) Let $f(x) = (c - 5)x^2 - 2cx + (c - 4)$.For one root to lie in $(0, 2)$ and the other in $(2, 3)$,the value of $f(2)$ must have a sign opposite to

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$11$

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(D) Let $f(x) = (c - 5)x^2 - 2cx + (c - 4)$.For one root to lie in $(0, 2)$ and the other in $(2, 3)$,the value of $f(2)$ must have a sign opposite to the leading coefficient $(c - 5)$.Case $I$: If $c - 5 > 0$ (i.e.,$c > 5$),then $f(2) $f(2) = (c - 5)(2)^2 - 2c(2) + (c - 4) = 4c - 20 - 4c + c - 4 = c - 24$.So,$c - 24 Also,we need $f(0) > 0$ and $f(3) > 0$ for the roots to be in the specified intervals.$f(0) = c - 4 > 0 \Rightarrow c > 4$.$f(3) = (c - 5)(9) - 6c + (c - 4) = 9c - 45 - 6c + c - 4 = 4c - 49 > 0 \Rightarrow c > 12.25$.Combining $c > 5$,$c  4$,and $c > 12.25$,we get $c \in (12.25, 24)$.Integral values of $c$ are ${13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}$.Number of elements = $11$.Case $II$: If $c - 5  0$.$c - 24 > 0 \Rightarrow c > 24$,which contradicts $c Therefore,$S = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}$,and the number of elements is $11$.

Consider the quadratic equation $(c - 5)x^2 - 2cx + (c - 4) = 0$,where $c \ne 5$. 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

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