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Jee Main 2025 — Medium JEE math MCQ

Let \(f(x) = \frac{2x^2 + 16}{2x^3 + 2x^2 + 4 + 32}\). Then the value of \(8 \left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right)\) is equal to
  1. A. 92
  2. B. 118
  3. C. 102
  4. D. 108

Solution

The correct option is **B**. (B. 118)

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let f(x)=2x2+162x3+2x2+4+32f(x) = \frac{2x^2 + 16}{2x^3 + 2x^2 + 4 + 32}. Then the value of 8(f(115)+f(215)+…+f(5915))8 \left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right) is equal to
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep