A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{5}{3}\]
Correct Answer: B
Solution :
[b] Let \[y=\frac{1}{40}(3{{x}^{4}}+8{{x}^{3}}-18{{x}^{2}}+60)\] \[\Rightarrow \frac{dy}{dx}=\frac{1}{40}(12{{x}^{3}}+24{{x}^{2}}-36x)\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{40}(36{{x}^{2}}+48x-36)\] Now \[\frac{dy}{dx}=0\Rightarrow {{x}^{3}}+2{{x}^{2}}-3x=0\] or \[x(x-1)(x+3)=0\] or \[x=0,\,\,1,\,\,-3\] At \[x=0,\,\,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=-36<0\therefore y\] is maximum at \[x=0\] \[\Rightarrow \] The given function i.e., \[\frac{1}{y}\] is minimum at \[x=0\] \[\therefore \] minimum value of the function \[=\frac{40}{60}=\frac{2}{3}\]
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