Approximation Application

Question

Using the first-order binomial approximation for small quantities, the approximate value of (1.002)^{10} computed to a sensible number of decimal places is closest to which figure?

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For small x, the binomial theorem provides the first-order approximation (1 + x)^n ≈ 1 + nx, where higher powers of the tiny quantity x are negligible, a practical JEE Advanced application. Here write 1.002 = 1 + 0.002 with x = 0.002 and n = 10, so (1.002)^{10} = (1 + 0.002)^{10} ≈ 1 + 10(0.002) = 1 + 0.020 = 1.020. The neglected term is C(10, 2)(0.002)^2 = 45 × 0.000004 = 0.00018, which is small enough not to affect the third decimal materially. Hence the approximation is 1.020. Option 1.002 ignores the exponent entirely. Option 1.200 misplaces the decimal by treating x as 0.02. Option 1.010 halves the linear correction incorrectly. The validity rests on x being small so that quadratic and higher terms are negligible. Plausibility check: since the second-order correction 0.00018 is two orders of magnitude smaller than the first-order term 0.020, the linear approximation 1.020 is reliable to three decimal places, confirming the estimate.

Approximation Application

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