Question

Let $ a>0,b>0 $ and $ f(x)=\left| \begin{matrix} x & a & a \\ b & x & a \\ b & b & x \\ \end{matrix} \right|, $ then which of the following statement is true?

• 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

Answer 1

Answer: (A)

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}. $