Question

The maximum of value of ${\left(\sqrt{-3+4x-x^2}+4\right)}^2+{\left(x-5\right)}^2,\left(where1\le x\le 3\right)$ is

• A. 50
• B. 36
• C. 32
• D. 20

Answer 1

The correct option is A 50

$\begin{array}{c}f\left(x\right)=\left[\sqrt{-3+4x-x^2}\right]+{\left(x-5\right)}^2\\ maximumvalueis\\ \Rightarrow {\left[\sqrt{-\left(x^2-4x+3\right)}+4\right]}^2+{\left(x-5\right)}^2\\ \Rightarrow {\left[\sqrt{-\left(x^2-4x+4-1\right)}+4\right]}^2+{\left(x-5\right)}^2\\ \Rightarrow {\left[\sqrt{1-{\left(x-2\right)}^2}+4\right]}^2+{\left(x-5\right)}^2\\ Maximumvalue={\left[1+4\right]}^2+25\\ =25+25=50\end{array}$