Question

For the real values of x, maximum value of $\frac{x}{x^2-5x+9}$ is

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

Answer: (A)

1

Given, $\frac{x}{x^2-5x+9}$

let, $y=\frac{x}{x^2-5x+9}$

or, $x^{2}y-5xy+ay=x$

or, $x^{2}y+x(-5y-1)+ay=0$

Therefore the above equation is a quadratic equation of x , nd for all real value of x we get root of it. i.e. , the discriminant will be greater than zero.

$\therefore D\ge 0\Rightarrow b^2-4ac\ge 0$

$\therefore (-5y-1{)}^2-4y(ay)\ge 0\Rightarrow -11y^2+10y+1\ge 0\Rightarrow 11y^2-10y-1\le 0\Rightarrow (11y+1\left)\right(y-1)\le 0$

either $y=\frac{-1}{11}$ or $y=1$

$\therefore y\in [-1/11,1]$

$\therefore$ The maximum value of the given expression is 1

$\therefore$ option A is correct.