The value of p so that both the roots of the | vedclass

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(A) Let $f(x) = (p - 5)x^2 - 2px + (p - 4)$.For the roots to be positive,one less than $2$ and the other between $2$ and $3$,the parabola must open up

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 positive,one less than $2$ and the other between $2$ and $3$,the parabola must open upwards,so $p - 5 > 0 \Rightarrow p > 5$.Given the conditions,we must have $f(0) > 0$,$f(2)  0$.$1$) $f(0) = p - 4 > 0 \Rightarrow p > 4$.$2$) $f(2) = (p - 5)(4) - 2p(2) + (p - 4) = 4p - 20 - 4p + p - 4 = p - 24 $3$) $f(3) = (p - 5)(9) - 2p(3) + (p - 4) = 9p - 45 - 6p + p - 4 = 4p - 49 > 0 \Rightarrow p > \frac{49}{4}$.Combining $p > 5$,$p > 4$,$p  \frac{49}{4}$,we get the interval $\left( \frac{49}{4}, 24 \right)$.

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

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