CTET Paper II (Mathematics and Science) Feb 2014 Solved Paper

Concept:For very large exponents, terms like $\frac{16}{3^{500}}$ and $\frac{2}{3^{500}}$ become extremely small and can be ignored to simplify the expression.Explanation:Start with the given fraction: $\frac{3^{502} - 3^{500} + 16}{3^{500} + 2}$.Factor $3^{500}$ from the first two terms in the numerator: $3^{500}(3^2 - 1) + 16 = 3^{500}(9 - 1) + 16 = 3^{500} \cdot 8 + 16$.Divide both numerator and denominator by $3^{500}$:$\frac{8 + \frac{16}{3^{500}}}{1 + \frac{2}{3^{500}}}$.Since $3^{500}$ is astronomically large, $\frac{16}{3^{500}} \approx 0$ and $\frac{2}{3^{500}} \approx 0$.Thus the expression simplifies to $\frac{8 + 0}{1 + 0} = 8$.Answer:The value is 8, option C.