If expression left( {mx - 1 + frac{1}{x}} right) is always non-negative, then minimum value of m mus

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4-2.Quadratic Equations and Inequations

hard

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be

A

$ - \frac{1}{2}$

B

$0$

C

$\frac{1}{4}$

D

$\frac{1}{2}$

Solution

(c) We know that $a{x^2} + bx + c \ge 0$ if $a > 0$

and ${b^2} – 4ac \le 0$.

? $mx – 1 + \frac{1}{x} = 0\,\,\, \Rightarrow \,\,\frac{{m{x^2} – x + 1}}{x} \ge 0$

==> $m{x^2} – x + 1 \ge 0$ and $x > 0$

Now $m{x^2} – x + 1 \ge 0$ if $m > 0$ and $1 – 4m \le 0$

or if $m > 0$ and $m \ge \frac{1}{4}$

Thus the minimum value of m is $\frac{1}{4}$.

Standard 11

Mathematics

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