Question

Let $a>0,b>0$ and $f(x)=\left|\begin{array}{ccc}x& a& a\\ b& x& a\\ b& b& x\end{array}\right|,$ 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, $f(x)=\left|\begin{array}{ccc}x& a& a\\ b& x& a\\ b& b& x\end{array}\right|$
$=x(x^2-ab)-a(bx-ab)+a(b^2-bx)$
$\Rightarrow$ $f(x)=x^3-3abx+a^{2}b+a{b}^2$
On differentiating w. r. t. x, we get
$f^{\prime }(x)=3x^2-3ab$
Put $f^{\prime }(x)=0,$ $3x^2-3ab=0$
$\Rightarrow$ $x=\pm \sqrt{ab}$
Now, $f^{\prime }{}^{\prime }(x)=9x$
At $x=\sqrt{ab},$
$f^{\prime }{}^{\prime }(x)=9\sqrt{ab}>0,$ (minima)
Hence, $f(x)$ has a local minimum at
$x=\sqrt{ab}.$