Multiple choice

The set of values of $p$ such that both the roots of the equation $\left ( p-5 \right )x^{2}-2px+\left ( p-4 \right )=0$ are positive and one of the roots is less than $2$ and the other root lies between $2$ & $3$ is

  1. $\left ( \displaystyle \frac{49}{4},24 \right )$
  2. $\left ( 5,\infty \right )$
  3. $\left ( -\infty ,4 \right )\cup \left ( \displaystyle \frac{49}{4},\infty \right )$
  4. $\left ( 5,\displaystyle \frac{49}{4} \right )$
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Explanation

For roots to be positive and satisfy the location conditions, we analyze the function f(x) = (p-5)x^2 - 2px + (p-4). Conditions include f(0) > 0, f(2) < 0, f(3) > 0, and discriminant > 0. Solving these inequalities leads to the range (49/4, 24).

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