A) \[1/2<x<1\]
B) \[1/2<x<2\]
C) \[3<x<59/4\]
D) \[-\infty <x<\infty \]
Correct Answer: D
Solution :
Given, \[f(x)=2{{x}^{3}}-3{{x}^{2}}+90x+174\] \[\therefore \] \[f'(x)=6{{x}^{2}}-6x+90\] For increasing function, \[f'(x)>0\] \[\Rightarrow \] \[6{{x}^{2}}-6x+90-0\] \[\Rightarrow \] \[{{x}^{2}}-x+15>0\] Clearly, \[f'(x)>0\forall x\] So, \[f(x)\] increases in \[(-\infty ,\infty )\].
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