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Day 434: JEE Mains Maths

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

  1. Question 1

    Three Dimensional Geometry. Asked once: 27 Jun 2022 (same type asked 33 times) See all 33 of this type

    164
    The shortest distance between the lines x−32=y−23=z−1−1\frac{x - 3}{2}=\frac{y - 2}{3}=\frac{z - 1}{- 1} and x+32=y−61=z−53\frac{x + 3}{2}=\frac{y - 6}{1}=\frac{z - 5}{3} is :

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  2. Question 2

    Differential Equations. Asked once: 26 Jun 2022 (same type asked 41 times) See all 41 of this type

    160
    If the solution of the differential equation
    dydx+ex(x2−2)y=(x2−2x)(x2−2)e2x  satisfies \frac{dy}{dx}+e^{x}\left( x^{2}- 2 \right)y =\left( x^{2}- 2x \right)\left( x^{2}- 2 \right)e^{2x}\ \text{~satisfies~}
    y(0)=0y(0) = 0, then the value of y(2)y(2) is

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  3. Question 3

    Three Dimensional Geometry. Asked once: 25 Jun 2022 (same type asked 33 times) See all 33 of this type

    180
    Let l1l_{1} be the line in xyxy-plane with xx and yy intercepts 18\frac{1}{8} and 142\frac{1}{4\sqrt{2}} respectively, and l2l_{2} be the line in zx-plane with xx and zz intercepts −18-\frac{1}{8} and −163-\frac{1}{6\sqrt{3}} respectively. If dd is the shortest distance between the line l1l_{1} and l2l_{2}, then d−2d^{- 2} is equal to

  4. Question 4

    Differential Equations. Asked once: 26 Jun 2022 (same type asked 41 times) See all 41 of this type

    85
    Let the solution curve y=y(x)y=y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0\left( 4 +x^{2} \right)dy- 2x\left( x^{2}+ 3y+ 4 \right)dx= 0 pass through the origin. Then y(2)y(2) is equal to

  5. Question 5

    Three Dimensional Geometry. Asked once: 14 Jun 2022 (same type asked 33 times) See all 33 of this type

    164
    If the shortest distance between the lines x−12=y−23=z−3λ\frac{x - 1}{2}=\frac{y - 2}{3}=\frac{z - 3}{\lambda} and x−21=y−44=z−55\frac{x - 2}{1}=\frac{y - 4}{4}=\frac{z - 5}{5} is 13\frac{1}{\sqrt{3}}, then the sum of all possible values of λ\lambda is:

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