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

The approximate value of $(1.0002)^{3000}$ is
  1. $1.2$
  2. $1.4$
  3. $1.6$
  4. $1.8$

Solution

Let $y=f(x)=x^{3000}$ here $ \begin{aligned} x & =1, \Delta x=0.0002 \\ \Delta y & =f^{\prime}(x)=3000 x^{2999} \Delta x \\ \Delta y & =(3000)(0.0002)=0.6 \\ f(x+\Delta x) & =y+\Delta y=1+0.6=1.6 \end{aligned} $

Asked in: AP EAMCET 2002

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