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Question: If the expression \(\left( mx - 1 + \frac{1}{x} \right)\) is non-negative for all positive real x, t...

If the expression (mx1+1x)\left( mx - 1 + \frac{1}{x} \right) is non-negative for all positive real x, then the minimum value of m must be

A

–1/2

B

0

C

¼

D

½

Answer

¼

Explanation

Solution

We know that ax2 + bx + c ≥ 0 if a > 0 and b2 – 4ac ≤ 0.

So, mx – 1 + 1x\frac{1}{x} ≥ 0 ⇒ mx2x+1x0\frac{mx^{2} - x + 1}{x} \geq 0

⇒ mx2 – x + 1 ≥ 0 as x > 0.

Now, mx2 – x + 1 ≥ 0 if m > 0 and 1 – 4m ≤ 0

or if m > 0 and m ≥ 1/4.

Thus, the minimum value of m is ¼