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Day 79: JEE Mains Maths

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

  1. Question 1

    Conic Sections. Asked 2 times: 2021 Paper 4*, 23 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    55
    Let the line y−x=1y - x = 1 intersect the ellipse x22+y21=1\frac{x^{2}}{2}+\frac{y^{2}}{1}= 1 at the points A and B . Then the angle made by the line segment AB at the center of the ellipse is :

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  2. Question 2

    Inverse Trigonometric Functions. Asked 2 times: 27 Jan 2024*, 22 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    68
    The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{- 1}4x +\tan^{- 1}6x =\frac{\pi}{6}, where −126<x<126-\frac{1}{2\sqrt{6}}< x <\frac{1}{2\sqrt{6}} is equal to

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  3. Question 3

    Matrices. Asked 2 times: 24 Jan 2025*, 22 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    72
    Let A be a 3×33 \times 3 matrix such that A+AT=OA + A^{T} = O. If A[1−10]=[332],A2[1−10]=[−319−24]A\begin{bmatrix} 1 \\ - 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix},A^{2}\begin{bmatrix} 1 \\ - 1 \\ 0 \end{bmatrix} = \begin{bmatrix} - 3 \\ 19 \\ - 24 \end{bmatrix} and det(adj(2adj(A+I)))=(2)α⋅(3)β⋅(11)γ,α,β,γdet(adj(2adj(A + I))) = (2)^{\alpha} \cdot (3)^{\beta} \cdot (11)^{\gamma},\alpha,\beta,\gamma are non-negative integers, then α+β+γ\alpha + \beta + \gamma is equal to ____\_\_\_\_

  4. Question 4

    Straight Lines. Asked 2 times: 22 Jan 2025*, 21 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    54
    Let a point A lie between the parallel lines L1L_{1} and L2L_{2} such that its distances from L1L_{1} and L2L_{2} are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC , where the points B and C lie on the lines L1L_{1} and L2L_{2} respectively, is :

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  5. Question 5

    Trigonometric Identities. Asked 2 times: 2021 Paper 10*, 4 Apr 2025 (* = numbers or wording changed that year) See all 2 times it was asked

    70
    If 10sin⁡4θ+15cos⁡4θ=610\sin^{4}\theta+ 15\cos^{4}\theta= 6, then the value of 27csc⁡6θ+8sec⁡6θ16sec⁡8θ\frac{27\csc^{6}\theta+ 8\sec^{6}\theta}{16\sec^{8}\theta} is:

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