Given,f(x)=2x3+(2p−7)x2+3(2p−9)x−6
On taking differentiation w.r.t x, we get
f′(x)=6x2+2(2p−7)x+3(2p−9)
⇒f′(0)=6(0)2+2(2p−7)(0)+3(2p−9)
⇒f′(0)=3(2p−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(2p−9)<0
⇒p<29
Therefore, p∈(−∞,29)