Question

If $\frac{6x^2-5x-3}{x^2-2x+6}\le 4,$ the least and the highest values of $4x^2$ are

• A. $0,81$
• B. $9,81$
• C. $36,81$
• D. none of these

Answer 1

The correct option is A $0,81$

$\Rightarrow \frac{6x^2-5x-3}{x^2-2x+6}\le 4$

$\Rightarrow 6x^2-5x-3\le 4x^2-8x+24$

$\Rightarrow 2x^2+3x-27\le 0$

$\Rightarrow (2x+9)(x-3)\le 0$

$\Rightarrow xϵ(-9/2,3)$

$\therefore$the greatest and least values of $4x^2$ occurs when $x=\frac{-9}{2}$ and when $x=0$ respectively

$\therefore$ greatest value $=81$ and least value $=0$