question_answer

If the equation \[4{{x}^{3}}+5x+k=0(k\in R)\] has a negative real root then

A) \[k=0\]                                

B) \[-\infty <k<0\]

C) \[0<k<\infty \]                 

D) \[-\infty <k<\infty \]

Correct Answer: C

Solution :

\[f(x)=4{{x}^{3}}+5x+k\] \[f'(x)=12{{x}^{2}}+5>0\forall x\in R\] \[\therefore \]\[f(x)\] is strictly increasing on R. So, for \[f(x)=0\] to have a negative real root, \[f(0)>0\Rightarrow k>0\]