Question

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

• A. $1.06$
• B. $1.008$
• C. $1.6$
• D. $1.08$

Answer 1

The correct option is C $1.6$

$\text{Let},f\left(x\right)=x^{3000}$
consider, $x=1,\Delta x=0.0002$
We know,
$f(x+\Delta x)=f\left(x\right)+\Delta x\cdot f^{\prime }\left(x\right)$
$f(1+0.0002)=f\left(1\right)+\Delta x\cdot 3000(1{)}^{2999}$
$\Rightarrow f\left(1.0002\right)=1+(0.0002\times 3000)=1.6$