GATE CS 2000 Q2.3 — Maxima and Minima, Mean Value Theorem, Integration

Let $S = \sum_{i=3}^{100} i \log_{2} i$ , and $T = \int_{2}^{100} x \log_{2}x dx$ .

Which of the following statements is true?

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Answer: A

Let $f(x) = x \log_{2} x$ . For $x \ge 2$ , the derivative of $f(x)$ is:

f'(x) = \log_{2} x + \frac{1}{\ln 2} > 0

Since $f'(x) > 0$ , the function $f(x)$ is strictly increasing on the interval $[2, 100]$ .

For any interval $[i-1, i]$ where $i \ge 3$ , because $f(x)$ is strictly increasing, we have:

f(x) < f(i) \quad \text{for all } x \in [i-1, i)

Integrating both sides over the interval $[i-1, i]$ gives:

\int_{i-1}^{i} f(x) dx < \int_{i-1}^{i} f(i) dx = f(i)

Summing this inequality from $i = 3$ to $100$ :

\sum_{i=3}^{100} \int_{i-1}^{i} f(x) dx < \sum_{i=3}^{100} f(i)

\int_{2}^{100} x \log_{2} x dx < \sum_{i=3}^{100} i \log_{2} i

This simplifies to $T < S$ , or equivalently, $S > T$ .

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Published 10 Oct 2026

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