1. Tardigrade
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  3. Mathematics
  4. Let f(x)=|x 1 (-3/2) 2 2 1 (1/x-1) 0 (1/2)| The minimum value of f(x) (given x > 1 ), is

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Q. Let $f(x)=\begin{vmatrix}x & 1 & \frac{-3}{2} \\ 2 & 2 & 1 \\ \frac{1}{x-1} & 0 & \frac{1}{2}\end{vmatrix}$ The minimum value of $f(x)$ (given $x > 1$ ), is

Determinants

Solution:

$y=\begin{vmatrix}x & 1 & \frac{-3}{2} \\ 2 & 2 & 1 \\ \frac{1}{x-1} & 0 & \frac{1}{2}\end{vmatrix}$
$=x(1-0)-\left(1-\frac{1}{x-1}\right)-\frac{3}{2}\left(0-\frac{2}{x-1}\right)$
$=x-1+\frac{1}{x-1}+\frac{3}{x-1}$
$=(x-1)+\frac{4}{(x-1)}$
By applying A.M.-G.M. inequality, we have
$y=(x-1)+\frac{4}{x-1} \geq 2 \sqrt{(x-1) \frac{4}{x-1}}=4$ (As $x > 1$)