Question

Suppose g is a differentiable function such that $g\left(x\right)+g^{\prime }(x{)}^{\prime }1$ for all $x$ and $g\left(0\right)=0$, then the largest possible value of $g\left(1\right)$ is

• A. $e^{-1}$
• B. $e^{-2}$
• C. $1-e^{-2}$
• D. $1-e^{-1}$

Answer 1

The correct option is D $1-e^{-1}$

$g^{\prime }\left(x\right)+g\left(x\right)\le 1$

multiply by $e^x,$

$\Rightarrow g^{\prime }\left(x\right)e^x+g\left(x\right)e^x\le e^x$

$\Rightarrow \frac{d}{dx}\left(g\right(x\left)e^x\right)\le e^x$

On integrating from 0 to 1,

${\left|e^{x}g\right(x\left)\right|}_0^1\le {\left|e^x\right|}_0^1$

$\Rightarrow eg\left(1\right)-e^{0}g\left(0\right)\le e-1$

$\Rightarrow g\left(1\right)\le \frac{e1}{e}$