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Day 1: JEE Mains Maths

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

  1. Question 1

    Differential Equations. This type asked 41 times: 7 in 2021, 6 in 2022, 7 in 2023, 7 in 2024, 6 in 2025, 8 in 2026. This one: 8 Apr 2026 S2 See all 41 of this type

    67
    Let y=y(x)y=y(x) be the solution of the differential equation x1−x2dy+(y1−x2−xcos⁡−1x)dx=0x\sqrt{1 -x^{2}}dy+\left( y\sqrt{1 -x^{2}}-x\cos^{- 1}x \right)dx= 0, x∈(0,1),lim⁡x→1−y(x)=1x\in (0,1),\lim_{x\rightarrow1^{-}} y(x) = 1. Then y(12)y\left( \frac{1}{2} \right) equals :

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  2. Question 2

    Three Dimensional Geometry. This type asked 33 times: 2 in 2021, 4 in 2022, 10 in 2023, 8 in 2024, 5 in 2025, 4 in 2026. This one: 6 Apr 2026 S2 See all 33 of this type

    64
    The shortest distance between the lines x−41=y−32=z−2−3\frac{x - 4}{1}=\frac{y - 3}{2}=\frac{z - 2}{- 3} and x+22=y−64=z−5−5\frac{x + 2}{2}=\frac{y - 6}{4}=\frac{z - 5}{- 5} is:

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  3. Question 3

    Determinants. This type asked 26 times: 1 in 2021, 5 in 2022, 5 in 2023, 7 in 2024, 4 in 2025, 4 in 2026. This one: 8 Apr 2026 S2 See all 26 of this type

    53
    If the system of linear equations : x+y+z=6x+y+z= 6, x+2y+5z=10x+ 2y+ 5z= 10, 2x+3y+λz=μ2x + 3y +\lambda z =\mu. has infinitely many solutions, then the value of λ+μ\lambda+\mu equals.

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  4. Question 4

    Three Dimensional Geometry. This type asked 19 times: 2 in 2021, 2 in 2022, 7 in 2024, 2 in 2025, 6 in 2026. This one: 6 Apr 2026 S1 See all 19 of this type

    65
    Let the image of the point P(1,6,a)P(1,6,a) in the line L:x1=y−12=z−a+1 b, b>0L:\frac{x}{1}=\frac{y - 1}{2}=\frac{z - a + 1}{\text{ }b},\text{ }b > 0, be (a3,0,a+c)\left( \frac{a}{3},0,a + c \right). If S(α,β,γ),α>0S(\alpha,\beta,\gamma),\alpha> 0, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is 2142\sqrt{14}, then α+β+γ\alpha+\beta+\gamma is equal to :

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  5. Question 5

    Differential Equations. This type asked 19 times: 3 in 2021, 6 in 2022, 4 in 2023, 3 in 2024, 1 in 2025, 2 in 2026. This one: 2 Apr 2026 S2 See all 19 of this type

    70
    Let x=x(y)x = x(y) be the solution of the differential equation 2y2dxdy−2xy+x2=0,y>1,x(e)=e2y^{2}\frac{dx}{dy}- 2xy+x^{2}= 0,y> 1,x(e) =e. Then x(e2)x\left( e^{2} \right) is equal to:

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