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Day 290: JEE Mains Maths

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

  1. Question 1

    Differential Equations. Asked once: 4 Apr 2026 S2 (same type asked 41 times) See all 41 of this type

    69
    Let y=y(x)y=y(x) be the solution of the differential equation : dydx+(6x2+(3x2+2x3+4)e−2x(x3+2)(2+e−2x))y=2+e−2x\frac{dy}{dx}+\left( \frac{6x^{2}+\left( 3x^{2}+ 2x^{3}+ 4 \right)e^{- 2x}}{\left( x^{3}+ 2 \right)\left( 2 +e^{- 2x} \right)} \right)y= 2 +e^{- 2x},x∈(−1,2)x \in ( - 1,2), satisfying y(0)=32y(0) =\frac{3}{2}. If y(1)=α(2+e−2)y(1) =\alpha\left( 2 +e^{- 2} \right), then α\alpha is equal to :

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  2. Question 2

    Three Dimensional Geometry. Asked once: 22 Jan 2026 S1 (same type asked 33 times) See all 33 of this type

    56
    Let P(α,β,γ)P(\alpha,\beta,\gamma) be the point on the line x−12=y+1−3=z\frac{x - 1}{2}=\frac{y + 1}{- 3}= z at a distance 4144\sqrt{14} from the point (1,−1,0)(1, - 1,0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3\frac{x - \alpha}{1}=\frac{y - \beta}{2}=\frac{z - \gamma}{3} and x+52=y−101=z−31\frac{x + 5}{2}=\frac{y - 10}{1}=\frac{z - 3}{1} , is equal to

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  3. Question 3

    Determinants. Asked once: 23 Jan 2026 S2 (same type asked 26 times) See all 26 of this type

    56
    The system of linear equations x+y+z=6x + y + z = 6, 2x+5y+az=362x + 5y + az = 36, x+2y+3z=bx + 2y + 3z = b has

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  4. Question 4

    Three Dimensional Geometry. Asked once: 23 Jan 2026 S2 (same type asked 19 times) See all 19 of this type

    72
    If the image of the point P(a,2,a)P(a,2,a) in the line x2=y+a1=z1\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1} is Q and the image of Q in the line x−2b2=y−a1=z+2b−5\frac{x - 2b}{2} = \frac{y - a}{1} = \frac{z + 2b}{- 5} is PP, then a+ba + b is equal to ____\_\_\_\_

  5. Question 5

    Differential Equations. Asked once: 31 Jan 2024 (same type asked 19 times) See all 19 of this type

    69
    The solution curve of the differential equation ydxdy=x(log⁡ex−log⁡ey+1),x>0,y>0y\frac{dx}{dy}=x\left( \log_{e}x-\log_{e}y+ 1 \right),x> 0,y> 0 passing through the point (e,1)(e,1) is

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