The value of p such that both the roots of th | vedclass

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Question

(A) Let $f(x) = (p - 5)x^2 - 2px + (p - 4)$.For the roots to be real and satisfy the given conditions,we analyze the sign of $f(x)$ at specific points

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(x) = (p - 5)x^2 - 2px + (p - 4)$.For the roots to be real and satisfy the given conditions,we analyze the sign of $f(x)$ at specific points.Case $1$: If $p - 5 > 0$ (i.e.,$p > 5$),the parabola opens upward.For one root to be in $(0, 2)$ and the other in $(2, 3)$,we must have $f(0) > 0$,$f(2)  0$.$f(0) = p - 4 > 0 \implies p > 4$.$f(2) = (p - 5)(4) - 2p(2) + (p - 4) = 4p - 20 - 4p + p - 4 = p - 24 $f(3) = (p - 5)(9) - 2p(3) + (p - 4) = 9p - 45 - 6p + p - 4 = 4p - 49 > 0 \implies p > \frac{49}{4}$.Combining $p > 5$,$p  \frac{49}{4}$,we get $p \in \left( \frac{49}{4}, 24 \right)$.Case $2$: If $p - 5 For one root to be in $(0, 2)$ and the other in $(2, 3)$,we must have $f(0)  0$,and $f(3) $f(0) = p - 4 $f(2) = p - 24 > 0 \implies p > 24$. This is impossible since $p Thus,the interval is $\left( \frac{49}{4}, 24 \right)$.

The value of $p$ such that both the roots of the equation $(p - 5)x^2 - 2px + (p - 4) = 0$ are positive,one is less than $2$ and the other lies between $2$ and $3$,lies in the interval:

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