Let the function $f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x-6$ have a maxima for some value of $\mathrm{x} 0$. Then, the set of all values of $p$ is

(C) $\left(-\infty, \frac{9}{2}\right)$

(A) $\left(\frac{9}{2}, \infty\right)$

(B) $\left(0, \frac{9}{2}\right)$

(D) $\left(-\frac{9}{2}, \frac{9}{2}\right)$

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Application of Derivatives

Let the function

f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6

have a maxima for some value of

x < 0

and a minima for some value of

x > 0

. Then, the set of all values of

p

Answer: (C)

The derivative

f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)

is an upward parabola, so its smaller root gives the maximum and its larger root the minimum. A maximum at a negative

x

and a minimum at a positive

x

means the two roots straddle zero, which happens exactly when

f^{'} (0) < 0

. Since

f^{'} (0) = 3 (2 p - 9)

, this needs

2 p - 9 < 0

, that is

p < \frac{9}{2}

. The set of values is

(- \infty, \frac{9}{2})

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option