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Friday, 9 October

JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 30 January 2024, Shift 2, Maths Q10

Question 10 of 90 in this shift, Maths question 10 of 30, Section A.

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be defined as f(x)=ae2x+bex+cxf(x) = ae^{2x} + be^x + cx. If f(0)=−1f(0) = -1, f′(log⁡e2)=21f'(\log_e 2) = 21 and ∫0log⁡e4(f(x)−cx)dx=392\int_0^{\log_e 4} (f(x) - cx)dx = \frac{39}{2}, then the value of ∣a+b+c∣|a + b + c| equals
  1. (1)88Official answer
  2. (2)1010
  3. (3)1212
  4. (4)1616

Official answer

Option 1

NTA final key.

Chapter
Integrals
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.