Question

The approximate value of $(1.0002{)}^{3000}$ is

• A. 1.2
• B. 1.4
• C. 1.6
• D. 1.8

Answer 1

Answer: (C)

Let $y=x^{3000},x=1$ and $x+\Delta x=1.0002$
Then, $\Delta x=1.0002-1=0.0002$
Also, $y=1$, when $x=1$
Now, $y=x^{3000}$
$\Rightarrow \frac{dy}{dx}=3000x^{2999}$
$\Rightarrow {\left[\frac{dy}{dx}\right]}_{x=1}^{dx}=3000$
$\because \Delta y=\frac{dy}{dx}\Delta x$
$\therefore \Delta y=3000\times 0.0002=\frac{6}{10}=0.6$
Hence, $(1.0002{)}^{3000}=y+\Delta y=1+0.6=1.6$