Practice Questions
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Q80.The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to: (1) 5 (2) 9 16 16 (3) 11 (4) 13 16 16
Q80.Let the plane ax + by + cz = d pass through (2, 3, β5) and is perpendicular to the planes 2x + y β5z = 10 and 3x + 5y β7z = 12 If a, b, c, d are integers d > 0 and gcd(|a|, |b|, |c|, d) = 1 then the value of a + 7b + c + 20d is equal to JEE Main 2022 (28 Jun Shift 2) JEE Main Previous Year Paper (1) 18 (2) 20 (3) 24 (4) 22 Β―
Q80.If A and B are two events such that P(A) = 31 , P(B) = 15 and P(A βͺB) = 12 , then P(A Bβ²) + P(B Aβ²) is equal to (1) 3 (2) 5 4 8 (3) 5 (4) 7 4 8
Q80.Let S = {1, 2, 3, β¦ , 2022}. Then the probability, that a randomly chosen number n from the set S such that HCF(n, 2022) = 1, is (1) 128 (2) 166 1011 1011 (3) 127 (4) 112 337 337
Q80.Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is 4 5 (1) (2) 9 18 (3) 1 (4) 3 6 10
Q80.If the numbers appeared on the two throws of a fair six faced die are πΌ and π½, then the probability that π₯2 + πΌπ₯+ π½> 0, for all π₯βπ , is 17 4 (1) (2) 36 9 (3) 1 (4) 19 2 36
Q80.If the mirror image of the point (2, 4, 7) in the plane 3x βy + 4z = 2 is (a, b, c), the 2a + b + 2c is equal to (1) 54 (2) β6 (3) 50 (4) β42 Β―
Q80.A random variable X has the following probability distribution: X 0 1 2 3 4 P(X) k 2k 4k 6k 8k The value of P( 1<x<4xβ€2 )is equal to (1) 4 (2) 2 7 3 (3) 3 (4) 4 7 5 Β―
Q80.If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x β6y = 30, then the probability that y < 1 is (1) 16 (2) 56 (3) 2 (4) 6 3 7
Q80.Let E1, E2, E3 be three mutually exclusive events such that P(E1) = 2+3p6 , P(E2) = 2βp8 and P(E3) = 1βp2 . If the maximum and minimum values of p are p1 and p2 then (p1 + p2) is equal to: (1) 2 (2) 5 3 3 (3) 5 (4) 1 4
Q80.Let π be a binomially distributed random variable with mean 4 and variance 3. Then 54 ππβ€2 is equal to (1) 73 (2) 146 27 27 146 126 (3) (4) 81 81
Q80.Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is (1) 46 (2) 275 64 65 (3) 41 (4) 36 55 54
Q80.If a random variable X follows the Binomial distribution B(33, p) such that 3P(X = 0) = P(X = 1), then the value of P(X=15) βP(X=16) is equal to P(X=18) P(X=17) (1) 1320 (2) 1088 (3) 1088 (4) 120 1089 1331
Q80.Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is (1) 105 (2) 77 16 36 (3) 3631 (4) 3536
Q80.If the lines βr= (Λi βΛj + Λk) Ξ»(3Λj βΛk) and βr (Ξ±Λi βΛj) ΞΌ(2Λi β3Λk) are co-planar, the the distance of the plane containing these two lines from the point (Ξ±, 0, 0) is (1) 2 (2) 2 9 11 (3) 4 (4) 2 11
Q80.Let S be the sample space of all five digit numbers. If p is the probability that a randomly selected number from S , is a multiple of 7 but not divisible by 5 , then 9p is equal to (1) 1. 0146 (2) 1. 2085 (3) 1. 0285 (4) 1. 1521 Β―
Q80.Let A and B be two events such that P(B β£A) = 25 , P(A β£B) = 71 and P(A β©B) = 19 . Consider (S1)P(Aβ² βͺB) = 65 , (S2)P(Aβ² β©Bβ²) = 181 . Then (1) Both (S1) and (S2) are true (2) Both (S1) and (S2) are false (3) Only (S1) is true (4) Only (S2) is true
Q80.Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is 2 11 (1) (2) 3 16 23 13 (3) (4) 32 16
Q81.Let Ξ±, Ξ² be the roots of the equation x2 β4Ξ»x + 5 = 0 and Ξ±, Ξ³ be the roots of the equation + + 7 + 3Ξ»β3 = 0. If Ξ² + Ξ³ = 3β2, then (Ξ± + 2Ξ² + Ξ³)2 is equal to x2 β(3β2 2β3)x is 939,
Q87.Let π΄= 1 -1 and π΅= π½1 , πΌ, π½βπ . Let πΌ1 be the value of πΌ which satisfies π΄+ π΅2 = π΄2 + 2 2 and 2 πΌ 1 0 2 2 πΌ2 be the value of πΌ which satisfies π΄+ π΅2 = π΅2. Then πΌ1 - πΌ2 is equal to
Q88.Let π: 0, 1 βπ be a twice differentiable function in 0, 1 such that π0 = 3 and π1 = 5. If the line π¦= 2π₯+ 3 intersects the graph of π at only two distinct points in 0, 1, then the least number of points π₯β0, 1, at which π''π₯= 0, is β3 15π₯3
Q61.The equation arg( z+1zβ1 ) = Ο4 represents a circle with: (1) centre at (0, 0) and radius β2 (2) centre at (0, 1) and radius 2 (3) centre at (0, β1) and radius β2 (4) centre at (0, 1) and radius β2 22
Q61.The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is: (1) 77 (2) 82 (3) 42 (4) 35
Q61.The sum of 10 terms of the series 3 + 5 + 7 + β¦ is : 12Γ22 22Γ32 32Γ42 (1) 143 (2) 99 144 100 (3) 1 (4) 120121
Q61.Let n denote the number of solutions of the equation z2 + 3z = 0, where z is a complex number. Then the value of ββk=0 nk1 is equal to (1) 1 (2) 34 (3) 32 (4) 2