Answer:
Option D
Explanation:
Let f(x)= $2x^{3}+3x+k$
On differentiating w.r,t x , we get
$f'(x)= 6x^{2}+3> 0,\forall x\epsilon R$
$\Rightarrow$ f(x) is strictly increasing function.
$\Rightarrow$ f(x)= 0 has only one real root, so two roots are not possible
