A) \[f(x)\] has a local minimum at \[x=\sqrt{ab}.\]
B) \[f(x)\] has a local maximum at \[x=\sqrt{ab}.\]
C) \[f(x)\]has a neither local minimum at \[x=\sqrt{ab}.\] nor local maximum at \[x=\sqrt{ab}.\].
D) None of the above
Correct Answer: A
Solution :
Given, \[f(x)=\left| \begin{matrix} x & a & a \\ b & x & a \\ b & b & x \\ \end{matrix} \right|\] \[=x({{x}^{2}}-ab)-a(bx-ab)+a({{b}^{2}}-bx)\] \[\Rightarrow \] \[f(x)={{x}^{3}}-3abx+{{a}^{2}}b+a{{b}^{2}}\] On differentiating w. r. t. x, we get \[f'(x)=3{{x}^{2}}-3ab\] Put \[f'(x)=0,\] \[3{{x}^{2}}-3ab=0\] \[\Rightarrow \] \[x=\pm \sqrt{ab}\] Now, \[f'\,'(x)=9x\] At \[x=\sqrt{ab},\] \[f'\,'(x)=9\sqrt{ab}>0,\] (minima) Hence, \[f(x)\] has a local minimum at \[x=\sqrt{ab}.\]
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