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 is

Step-by-step solution

Given, f ( x ) = 2 x 3 + ( 2 p - 7 ) x 2 + 3 ( 2 p - 9 ) x - 6 On taking differentiation w.r.t x , we get f ' ( x ) = 6 x 2 + 2 ( 2 p - 7 ) x + 3 ( 2 p - 9 ) ⇒ f ' ( 0 ) = 6 ( 0 ) 2 + 2 ( 2 p - 7 ) ( 0 ) + 3 ( 2 p - 9 ) ⇒ f ' ( 0 ) = 3 ( 2 p - 9 ) Here, given that for x < 0 the function has maxima and for x > 0 the function has minima so the function is decreasing function about x = 0 , Hence f ' ( 0 ) < 0 So, 3 ( 2 p - 9 ) < 0 ⇒ p < 9 2 Therefore, p ∈ - &