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Day 49: JEE Mains Maths

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

  1. Question 1

    Binomial Theorem. This type asked 4 times: 2021 Paper 23, 29 Jun 2022, 25 Jan 2023, 10 Apr 2023. This one: 10 Apr 2023 See all 4 of this type

    81
    The coefficient of x7x^{7} in (1−x+2x3)10\left( 1 - x + 2x^{3} \right)^{10} is

  2. Question 2

    Three Dimensional Geometry. This type asked 4 times: 2021 Paper 1, 2021 Paper 17, 2021 Paper 22, 10 Apr 2023. This one: 10 Apr 2023 See all 4 of this type

    73
    Let PP be the point of intersection of the line x+33=y+21=1−z2\frac{x + 3}{3}=\frac{y + 2}{1}=\frac{1 - z}{2} and the plane x+y+z=2x + y + z = 2. If the distance of the point PP from the plane 3x−4y+12z=323x - 4y + 12z = 32 is qq, then qq and 2q2q are the roots of the equation

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  3. Question 3

    Trigonometric Equations. This type asked 4 times: 2021 Paper 5, 2021 Paper 23, 26 Jul 2022, 10 Apr 2023. This one: 10 Apr 2023 See all 4 of this type

    165
    Let S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S =\left\{ x \in\left( -\frac{\pi}{2},\frac{\pi}{2} \right):9^{1 -\tan^{2}x}+9^{\tan^{2}x}= 10 \right\} and β=∑x∈Stan⁡2(x3)\beta =\sum_{x \in S} \tan^{2}\left( \frac{x}{3} \right), then 16(β−14)2\frac{1}{6}(\beta - 14)^{2} is equal to

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  4. Question 4

    Conic Sections. This type asked 4 times: 2021 Paper 3, 2021 Paper 23, 2021 Paper 24, 6 Apr 2023. This one: 6 Apr 2023 See all 4 of this type

    179
    Let the eccentricity of an ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1 is reciprocal to that of the hyperbola 2x2−2y2=12x^{2}- 2y^{2}= 1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is

  5. Question 5

    Vector Algebra. This type asked 4 times: 2021 Paper 25, 29 Jul 2022, 25 Jan 2023, 6 Apr 2023. This one: 6 Apr 2023 See all 4 of this type

    157
    The sum of all values of α\alpha, for which the points whose position vectors i^−2j^+3k^,2i^−3j^+4k^\widehat{i}- 2\widehat{j}+ 3\widehat{k},2\widehat{i}- 3\widehat{j}+ 4\widehat{k}, (α+1)i^+2k^(\alpha + 1)\widehat{i}+ 2\widehat{k} and 9i^+(α−8)j^+6k^9\widehat{i}+ (\alpha - 8)\widehat{j}+ 6\widehat{k} are coplanar, is equal to

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