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Day 500: JEE Mains Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Complex Numbers. Asked once: 22 Jan 2025

    54
    Let z1,z2z_{1},z_{2} and z3z_{3} be three complex numbers on the circle ∣z∣=1|z| = 1 with arg(z1)=−π4,arg(z2)=0arg\left( z_{1} \right)=\frac{- \pi}{4},arg\left( z_{2} \right)= 0 and arg(z3)=π4arg\left( z_{3} \right)=\frac{\pi}{4}. If ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2,α,β∈Z\left| z_{1}{\bar{z}}_{2}+z_{2}{\bar{z}}_{3}+z_{3}{\bar{z}}_{1} \right|^{2}= \alpha + \beta\sqrt{2},\alpha,\beta \in Z, then the value of α2+β2\alpha^{2}+\beta^{2} is:

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  2. Question 2

    Conic Sections. Asked once: 2 Apr 2026 S2

    61
    Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12xy = 12 such that the mid point of the line segment PQ is (12,−12)\left( \frac{1}{2}, -\frac{1}{2} \right). Then the area of the triangle OPQ equals :

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  3. Question 3

    Definite Integration. Asked once: 22 Jan 2026 S2

    72
    Let [•] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha = \int_{0}^{64} \left( x^{1/3} - \left\lbrack x^{1/3} \right\rbrack \right)dx, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\frac{1}{\pi}\int_{0}^{\alpha\pi} \left( \frac{\sin^{2}\theta}{\sin^{6}\theta + \cos^{6}\theta} \right)d\theta is equal to ____\_\_\_\_ .

  4. Question 4

    Determinants. Asked once: 29 Jul 2022

    87
    Let pp and p+2p + 2 be prime numbers and let
    Δ=∣p!(p+1)!(p+2)!(p+1)!(p+2)!(p+3)!(p+2)!(p+3)!(p+4)!∣\Delta =\left| \begin{matrix} p! & (p + 1)! & (p + 2)! \\ (p + 1)! & (p + 2)! & (p + 3)! \\ (p + 2)! & (p + 3)! & (p + 4)! \end{matrix} \right|
    Then the sum of the maximum values of α\alpha and β\beta, such that pαp^{\alpha} and (p+2)β(p + 2)^{\beta} divide Δ\Delta, is

  5. Question 5

    Differential Equations. Asked once: 26 Jun 2022

    90
    Let S=(0,2π)−{π2,3π4,3π2,7π4}S = (0,2\pi) -\left\{ \frac{\pi}{2},\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4} \right\}. Let y=y = y(x),x∈Sy(x),x\in S, be the solution curve of the differential equation dydx=11+sin⁡2x,y(π4)=12\frac{dy}{dx}=\frac{1}{1 + \sin2x},y\left( \frac{\pi}{4} \right)=\frac{1}{2}. If the sum of abscissas of all the points of intersection of the curve y=y(x)y=y(x) with the curve y=2sin⁡xy=\sqrt{2}\sin x is kπ12\frac{k\pi}{12}, then kk is equal to