Let the function f ( x ) = 2 x ^ 3 + ( 2p - 7 ) x ^ 2 + 3 ( 2p - 9...

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

On taking differentiation w.r.t $x$, we get

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

$⇒ {f}^{'} ( 0 ) = 6 ( 0{)}^{2} + 2 ( 2p - 7 ) ( 0 ) + 3 ( 2p - 9 )$

$⇒ {f}^{'} ( 0 ) = 3 ( 2p - 9 )$

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, $3 ( 2p - 9 ) < 0$

$⇒ p < \dfrac{9}{2}$

Therefore, $p \in \left(- \infty , \dfrac{9}{2}\right)$