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Day 6: JEE Mains Maths

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

  1. Question 1

    Sequences and Series. This type asked 10 times: 2021 Paper 25, 26 Jun 2022, 28 Jun 2022, 29 Jan 2023, 13 Apr 2023, 8 Apr 2024, 22 Jan 2025, 3 Apr 2025, 7 Apr 2025, 21 Jan 2026 S1. This one: 21 Jan 2026 S1 See all 10 of this type

    56
    Let a1,a2,a3,…a_{1},a_{2},a_{3},\ldots. be a G.P. of increasing positive terms such that a2⋅a3⋅a4=64a_{2}\cdot a_{3}\cdot a_{4}= 64 and a1+a3+a5=8137a_{1}+a_{3}+a_{5}=\frac{813}{7}. Then a3+a5+a7a_{3}+a_{5}+a_{7} is equal to :

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  2. Question 2

    Matrices. This type asked 9 times: 2021 Paper 17, 29 Jun 2022, 25 Jul 2022, 12 Apr 2023, 4 Apr 2024, 29 Jan 2025, 21 Jan 2026 S2, 28 Jan 2026 S2, 6 Apr 2026 S2. This one: 6 Apr 2026 S2 See all 9 of this type

    55
    Let A=[100310931]A =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix} and B=[bij],1≤i,j≤3B =\left\lbrack b_{ij} \right\rbrack,1 \leq i,j \leq 3. If B=A99−IB=A^{99}-I, then the value of b31−b21b32\frac{b_{31}-b_{21}}{b_{32}} is :

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  3. Question 3

    Application of Integrals. This type asked 9 times: 2021 Paper 2, 25 Jun 2022, 26 Jun 2022, 27 Jul 2022, 8 Apr 2023, 31 Jan 2024, 5 Apr 2024, 22 Jan 2025, 4 Apr 2026 S2. This one: 4 Apr 2026 S2 See all 9 of this type

    68
    The area of the region bounded by the curves x+3y2=0x+ 3y^{2}= 0 and x+4y2=1x+ 4y^{2}= 1 is equal to :

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  4. Question 4

    Three Dimensional Geometry. This type asked 9 times: 27 Jan 2024, 9 Apr 2024, 23 Jan 2025, 28 Jan 2025, 3 Apr 2025, 22 Jan 2026 S2, 28 Jan 2026 S1, 28 Jan 2026 S2, 4 Apr 2026 S1. This one: 4 Apr 2026 S1 See all 9 of this type

    65
    The square of the distance of the point ( −2,−8,6- 2, - 8,6 ) from the line x−11=y−12=z−1\frac{x- 1}{1}=\frac{y- 1}{2}=\frac{z}{- 1} along the line x+51=y+5−1=z2\frac{x+ 5}{1}=\frac{y+ 5}{- 1}=\frac{z}{2} is equal to:

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  5. Question 5

    Three Dimensional Geometry. This type asked 9 times: 27 Jun 2022, 25 Jan 2023, 13 Apr 2023, 30 Jan 2024, 6 Apr 2024, 22 Jan 2025, 29 Jan 2025, 21 Jan 2026 S1, 2 Apr 2026 S2. This one: 2 Apr 2026 S2 See all 9 of this type

    65
    Let the point A be the foot of perpendicular drawn from the point P(a,b,0)P(a,b,0) on the line x−12=y−21=z−α3\frac{x- 1}{2}=\frac{y- 2}{1}=\frac{z-\alpha}{3}. If the midpoint of the line segment PA is (0,34,−14)\left( 0,\frac{3}{4},\frac{- 1}{4} \right), then the value of a2+b2+α2a^{2}+b^{2}+\alpha^{2} is equal to :

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