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Day 303: JEE Mains Maths

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

  1. Question 1

    Three Dimensional Geometry. Asked once: 29 Jul 2022 (same type asked 7 times) See all 7 of this type

    69
    If the foot of the perpendicular from the point A(−1,4,3)A( - 1,4,3) on the plane P:2x+my+nz=4P:2x + my + nz = 4, is (−2,72,32)\left( - 2,\frac{7}{2},\frac{3}{2} \right), then the distance of the point A from the plane PP, measured parallel to a line with direction ratios 3,−1,−43, - 1, - 4, is equal to:

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  2. Question 2

    Probability. Asked once: 14 Jun 2022 (same type asked 7 times) See all 7 of this type

    77
    If a random variable XX follows the Binomial distribution B(33,p)B(33,p) such that 3P(X=0)=P(X=1)3P(X = 0) = P(X = 1), then the value of P(X=15)P(X=18)−P(X=16)P(X=17)\frac{P(X = 15)}{P(X = 18)}-\frac{P(X = 16)}{P(X = 17)} is equal to

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  3. Question 3

    Binomial Theorem. Asked once: 22 Jan 2025 (same type asked 6 times) See all 6 of this type

    72
    If ∑r=05 11C2r+12r+2=mn,gcd(m,n)=1\sum_{r = 0}^{5} \frac{\ ^{11}C_{2r + 1}}{2r + 2}=\frac{m}{n},gcd(m,n) = 1, then m−nm - n is equal to

  4. Question 4

    Matrices. Asked once: 27 Jan 2024 (same type asked 6 times) See all 6 of this type

    89
    Let A=[201110101],B=[B1, B2, B3]A =\begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix},B =\left\lbrack B_{1},{\text{ }B}_{2},{\text{ }B}_{3} \right\rbrack, where B1B_{1}, B2, B3B_{2},{\text{ }B}_{3} are column matrices, and AB1=[100]AB_{1}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
    AB2=[230],AB3=[321]AB_{2}=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix},AB_{3}=\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}
    If α=∣B∣\alpha = |B| and β\beta is the sum of all the diagonal elements of BB, then α3+β3\alpha^{3}+\beta^{3} is equal to .

  5. Question 5

    Three Dimensional Geometry. Asked once: 29 Jan 2024 (same type asked 6 times) See all 6 of this type

    76
    Let PQRPQR be a triangle with R(−1,4,2)R( - 1,4,2). Suppose M(2,1,2)M(2,1,2) is the mid point of PQPQ. The distance of the centroid of △PQR\triangle PQR from the point of intersection of the lines x−20=y2=z+3−1\frac{x - 2}{0}=\frac{y}{2}=\frac{z + 3}{- 1} and x−11=y+3−3=z+11\frac{x - 1}{1}=\frac{y + 3}{- 3}=\frac{z + 1}{1} is

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