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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 5 April 2026, Shift 2, Maths Q3

Question 3 of 75 in this shift, Maths question 3 of 25, Section A.

If f:N→Zf: \mathbf{N} \to \mathbf{Z} is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣f(n)=\begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, k∈Nk \in \mathbf{N}, and ∑n=1kf(n)=98\sum_{n=1}^{k} f(n)=98, then k is equal to :
  1. (1)33Official answer
  2. (2)44
  3. (3)55
  4. (4)66

Official answer

Option 1

NTA final key.

Chapter
Determinants
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.