Question

Let $g:[1,6]\to [0,\infty ]$ be a real valued differentiable function satisfying $g^{\prime }\left(x\right)=\frac{2}{x+g\left(x\right)}$ and $g\left(1\right)=0$, the maximum value of $g$ cannot exceed.

• A. $ln2$
• B. $ln6$
• C. $6ln2$
• D. $2ln6$

Answer 1

The correct option is A $ln2$

$f\left(x\right)=\left(x\right)$ for $x\ne 0$

$=\perp$ $x=0$

$g\left(x\right)=\frac{2}{x+g\left(x\right)}$

$g\left(x\right)+x=\frac{2}{g\left(x\right)}\Rightarrow g\left(x\right)=\frac{2}{g\left(x\right)}-x$

$g\left(1\right)=\frac{2}{1}=2\int g\left(x\right)={\int }_0^1\frac{2}{g\left(x\right)}-x$

$=2\ln \left[g\right(x\left)\right]-\frac{x^2}{2}$

$=2\ln \left(2\right)-\frac{1}{2}$

Value of $g\left(x\right)$ will never be greater than $\ln 2$