Question

Let $f\left(x\right)$ and $g\left(x\right)$ be defined and differentaible for $x\ge x_0$ and $f\left(x_0\right)=g\left(x_0\right),f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$ for $4x>x_0,$ then

• A. $f\left(x\right)<g\left(x\right),x>x_0$
• B. $f\left(x\right)=g\left(x\right),x>x_0$
• C. $f\left(x\right)>g\left(x\right),x>x_0$
• D. none of these

Answer 1

The correct option is C $f\left(x\right)>g\left(x\right),x>x_0$

Since $f\left(x_0\right)=g\left(x_0\right)$ and $f^{\prime }\left(x\right)>g^{\prime }\left(x\right),\forall x>x_0$

$\therefore f^{\prime }\left(x\right)-g^{\prime }\left(x\right)>0$

$\Rightarrow f\left(x\right)-g\left(x\right)$ is an increasing function $\forall x>x_0.$

$\therefore f\left(x\right)-g\left(x\right)>f\left(x_0\right)-g\left(x_0\right)=0$

$\therefore f\left(x\right)>g\left(x\right)\forall x>x_0$