Question

Given that $f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$ for all real $x$, and $f\left(0\right)=g\left(0\right)$, then $f\left(x\right)<g\left(x\right)$ for all $x$ belongs to

• A. $(0,\infty )$
• B. $(-\infty ,0)$
• C. $R$
• D. None of the above.

Answer 1

The correct option is B $(-\infty ,0)$

The given equation is:

$f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$. This means that $f\left(x\right)$ is increasing at a faster rate than $g\left(x\right)$.

Now, at one point $x=0$ we have $f\left(0\right)=g\left(0\right)$. This means increasing at a faster rate than $g\left(x\right)$, $f\left(x\right)$ equals $g\left(x\right)$ values. Hence below this region the value of $f\left(x\right)$ must be lower than $g\left(x\right)$ otherwise the two function values have never been equal.

Hence $f\left(x\right)<g\left(x\right)$ in the region $(-\infty ,0)$ only. .....Answer