Let S = sum_{k=1}^{n} (log_2 k) and T = int_{1}^{n} (log_2 x) dx. Which of the following statements is true? A. S > T B. S = T C. S < T and 2S > T D. 2S <= T

GATE 2000 · Engineering Mathematics · Integration · medium

Answer: S > T

1. Interpret S as a Riemann sum: Each term log_2(k) equals the value of log_2 at the right endpoint of [k-1, k]. Since the subintervals have width 1, S is the right-endpoint Riemann sum for log_2(x) over [1, n].
2. Right Riemann sum overestimates for increasing functions: Since log_2(x) is strictly increasing, on each sub-interval [k-1, k] the value at the right endpoint log_2(k) exceeds the average value (the integral). Summing over all intervals: S > T.
3. Conclusion: Statement A (S > T) is true. This can also be verified numerically: for n=4, S approx 4.585, T = int_1^4 log_2(x) dx = [x log_2 x - x/ln2]_1^4 = 4*2 - 4/ln2 - (0 - 1/ln2) = 8 - 3/ln2 approx 8 - 4.328 = 3.672. Indeed S > T.