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Hint:– Using the theory of approximation let us simplify the problem. Then we proceed by applying the concepts of derivatives, partial derivatives and some basic rules of calculus to arrive at the answer.

Data Given\[{\left( {1.0002} \right)^{3000}}\], which can be expressed as ${\left( {{\text{1 + 0}}{\text{.0002}}} \right)^{3000}}$

Now let us consider a variable y =${{\text{x}}^{3000}}$, which can also be expressed as

y + ∆y = ${\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}}$ (where x is the whole number part and ∆x is the very small decimal part)

\[ \Rightarrow \Delta {\text{y = }}{\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}} - {{\text{x}}^{3000}}\] --- Equation 1

Now let’s consider the original function, y =${{\text{x}}^{3000}}$=f(x)

On differentiating we get, $\Delta {\text{y = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)\Delta {\text{x}}$.

Substituting this in Equation 1, we get

$

\Rightarrow {\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}} - {{\text{x}}^{3000}} = 3000{{\text{x}}^{2900}}\Delta {\text{x}} \\

{\text{Now put x = 1, }}\Delta {\text{x = 0}}{\text{.0002}} \\

\Rightarrow {\left( {1.0002} \right)^{3000}} - 1 = 3000\left( {0.0002} \right) \\

\Rightarrow {\left( {1.0002} \right)^{3000}} = 1.6 \\

$

Hence Option C is the correct answer.

Note:- In these types of questions, start off by making sure the number with the decimal units and without the decimal units are computed individually, an easy way to go about the answer. Then proceed by using the concepts we know from derivatives and partial derivatives to further simplify the answer. Making our subject, i.e. in this case \[{\left( {1.0002} \right)^{3000}}\]as a different variable makes the job easy.

__Complete step-by-step solution -__Data Given\[{\left( {1.0002} \right)^{3000}}\], which can be expressed as ${\left( {{\text{1 + 0}}{\text{.0002}}} \right)^{3000}}$

Now let us consider a variable y =${{\text{x}}^{3000}}$, which can also be expressed as

y + ∆y = ${\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}}$ (where x is the whole number part and ∆x is the very small decimal part)

\[ \Rightarrow \Delta {\text{y = }}{\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}} - {{\text{x}}^{3000}}\] --- Equation 1

Now let’s consider the original function, y =${{\text{x}}^{3000}}$=f(x)

On differentiating we get, $\Delta {\text{y = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)\Delta {\text{x}}$.

Substituting this in Equation 1, we get

$

\Rightarrow {\left( {{\text{x + }}\Delta {\text{x}}} \right)^{3000}} - {{\text{x}}^{3000}} = 3000{{\text{x}}^{2900}}\Delta {\text{x}} \\

{\text{Now put x = 1, }}\Delta {\text{x = 0}}{\text{.0002}} \\

\Rightarrow {\left( {1.0002} \right)^{3000}} - 1 = 3000\left( {0.0002} \right) \\

\Rightarrow {\left( {1.0002} \right)^{3000}} = 1.6 \\

$

Hence Option C is the correct answer.

Note:- In these types of questions, start off by making sure the number with the decimal units and without the decimal units are computed individually, an easy way to go about the answer. Then proceed by using the concepts we know from derivatives and partial derivatives to further simplify the answer. Making our subject, i.e. in this case \[{\left( {1.0002} \right)^{3000}}\]as a different variable makes the job easy.

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