Question

Consider the quadratic equation $(c-5)x^2-2cx+(c-4)=0$. 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 another root lies in the interval $(2,3)$. The number of elements in $S$ is

• A. 11.0
• B. 11.00
• C. 11

Answer 1

Answer: (A), (B), (C)

$(c-5)x^2-2cx+(c-4)=0$
For the equation to be quadratic in nature,
$c-5\ne 0\Rightarrow c\ne 5$
Now,
$(c-5)x^2-2cx+(c-4)=0\Rightarrow x^2-\frac{2c}{(c-5)}x+\frac{(c-4)}{(c-5)}=0$


The required conditions are,
$\left(i\right)f\left(0\right)\cdot f\left(2\right)<0\Rightarrow \frac{(c-4)}{(c-5)}\times (4-\frac{4c}{(c-5)}+\frac{(c-4)}{(c-5)})<0\Rightarrow \frac{(c-4)(c-24)}{(c-5{)}^2}<0\Rightarrow c\in (4,24)-\left\{5\right\}\cdots \left(1\right)\left(ii\right)f\left(2\right)\cdot f\left(3\right)<0\Rightarrow \frac{(c-24)(4c-49)}{(c-5{)}^2}<0\Rightarrow c\in (\frac{49}{4},24)\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
​​​​​​$c\in (\frac{49}{4},24)$
$\therefore S=\{13,14,\cdots ,23\}$

Hence, the number of elements in the set $S$ is
$n\left(S\right)=11$