Question

The maximum value of the expression $\frac{1}{x^2-2x+3}$ is

• A. \N
• B. 1
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Given expression $\frac{1}{x^2-2x+3}=\frac{1}{{\left(x-1\right)}^2+2}\mathrm{We}\mathrm{know}\mathrm{that}{\left(x-1\right)}^2\ge 0\forall x\in R.\mathrm{Thus},{\left(x-1\right)}^2+2\ge 2\mathrm{or},2\le {\left(x-1\right)}^2+2<\infty \mathrm{or},\frac{1}{2}\ge \frac{1}{{\left(x-1\right)}^2+2}>0\mathrm{or},\frac{1}{x^2-2x+3}\in (0,\frac{1}{2}]$
So, the maximum value of the given expression is $\frac{1}{2}$