alleryApproximation by differentials: Δy ≈ f'(x) Δx. Using this, sqrt(25.5) is approximately:
A5.5
B5.05
C5.10
D5.15
Answer & Solution
Correct answer: B. 5.05
Let f(x) = sqrt(x), x = 25, Δx = 0.5. f'(x) = 1/(2 sqrt(x)) = 1/10 at x = 25. So sqrt(25.5) ≈ 5 + 0.1 × 0.5 = 5.05. (Actual value ≈ 5.0498.)
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