Question

Define a functions $f\left(x\right)=\frac{16x^2-96x+153}{x-3}$ for all real $x\ne 3$. The least positive value of f(x) is?

• A. $16$
• B. $18$
• C. $22$
• D. $24$

Answer 1

The correct option is D $24$

$\frac{16x^2-96x+153}{x-3}$

$=\frac{16x^2-96x+144+9}{x-3}$

$=\frac{16(x^2-2\ast 3\ast x+3^2)}{x-3}+\frac{9}{x-3}$

$=\frac{16(x-3{)}^2}{x-3}+\frac{9}{x-3}$

Using, AM$\ge$GM

$=16(x-3)+\frac{9}{x-3}\ge 2\times \sqrt{16(x-3)\times \frac{9}{x-3}}\ge 24$