Practice Questions
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Q62.The sum β21n=1 (4nβ1)(4n+3)3 is equal to (1) 7 (2) 7 87 29 (3) 14 (4) 21 87 29
Q62.If π§= π₯+ ππ¦ satisfies π§- 2 = 0 and π§- π- π§+ 5π= 0, then (1) π₯+ 2π¦- 4 = 0 (2) π₯2 + π¦- 4 = 0 (3) π₯+ 2π¦+ 4 = 0 (4) π₯2 - π¦+ 3 = 0 Q63. βπ,π π= 0 ππΆπ ππΆπ is equal to πβ π (1) 22π- 2ππΆπ (2) 22π- 1 - 2π- 1πΆπ- 1 1 1 2π- (3) 22π- 2 2ππΆπ (4) 2π- + 1πΆπ
Q62.Let Ξ±, Ξ² be the roots of the equation x2 ββ2x + β6 = 0 and 1 + 1, 1 + 1 be the roots of the equation Ξ±2 Ξ²2 x2 + ax + b = 0 . Then the roots of the equation x2 β(a + b β2)x + (a + b + 2) = 0 are : (1) non-real complex numbers (2) real and both negative (3) real and both positive (4) real and exactly one of them is positive
Q63.If the constant term in the expansion of (3x3 β2x2 + x5 ) is 2k. l, where l is an odd integer, then the value of k is equal to (1) 6 (2) 7 (3) 8 (4) 9
Q63.If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is (1) 21 (2) 22 (3) 23 (4) 24 β , x β 0 is
Q63.Let S = {ΞΈ β[0, 2Ο] : 82 sin2 ΞΈ + 82 cos2 ΞΈ = 16} . Then to: (1) 0 (2) β2 (3) β4 (4) 12
Q63.Let πππ=β 0 be a sequence such that π0 = π1 = 0 and ππ+ 2 = 3ππ+ 1 - 2ππ+ 1, βπβ₯0. Then π25π23 - 2π25π22 - 2π23π24 + 4π22π24 is equal to (1) 483 (2) 528 (3) 575 (4) 624 Q64. βπ=20 1 π2 + 1π! is equal to (1) 22! - 21! (2) 22! - 221! (3) 21! - 220! (4) 21! - 20!
Q63.Let π§1 and π§2 be two complex numbers such that Β―π§1 = πΒ―π§2 and arg = π, then the argument of π§1 is Β―π§2 (1) arg π§2 = Ο (2) arg π§2 = - 3Ο 4 4 Ο 3Ο (3) arg π§1 = 4 (4) arg π§1 = - 4
Q63.Let the sum of an infinite G. P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 98 . Then the sum of the first 21 terms of an AP, whose first term is 10ar, nth term is an and the 25 common difference is 10 ar2 , is equal to (1) 21a11 (2) 22a11 (3) 15a16 (4) 14a16
Q63.Let S = 2 + 76 + 1272 + 2073 + 3074 + β¦ . . then 4S is equal to JEE Main 2022 (27 Jun Shift 2) JEE Main Previous Year Paper (1) ( 27 ) 2 (2) ( 73 ) 3 (3) 3 7 (4) ( 37 ) 4
Q63.Consider the sequence π1, π2, π3, β¦ β¦ such that π1 = 1, π2 = 2 and ππ+ 2 = + ππ for π= 1, 2, 3, β¦ ππ+ 1 1 1 1 1 π1 + π2 π2 + π3 π3 + π4 π30 + π31 If Β· Β· β¦ = 2πΌ61πΆ31 then πΌ is equal to π3 π4 π5 π32 (1) -30 (2) -31 (3) -60 (4) -61
Q63.The number of solutions of the equation cos(x + Ο3 ) cos( Ο3 βx) = 14 cos2 2x, x β[β3Ο, 3Ο] is: (1) 8 (2) 5 (3) 6 (4) 7
Q63.Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1, k2), then k1 + k2 is equal to (1) 2 (2) 47 (3) 2 (4) 4 7
Q63.The sum of the infinite series 1 + 65 + 1262 + 2263 + 3564 + 5165 + 7066 + β¦ is equal to: (1) 425 (2) 429 216 216 (3) 288 (4) 280 125 125
Q63.The remainder when (11)1011 + (1011)11 is divided by 9 is _____ . (1) 1 (2) 8 (3) 6 (4) 4
Q63.Let the tangents at two points A and B on the circle x2 + y2 β4x + 3 = 0 meet at origin O(0, 0). Then the area of the triangle of OAB is (1) 3β3 (2) 3β3 2 4 (3) 3 (4) 3 2β3 4β3
Q63.If m is the slope of a common tangent to the curves x2 16 + 9 = 1 and x2 + y2 = 12 , then 12m2 is equal to JEE Main 2022 (26 Jun Shift 2) JEE Main Previous Year Paper (1) 6 (2) 9 (3) 10 (4) 12
Q63.If {ai}ni=1 , where n is an even integer, is an arithmetic progression with common difference 1 , and n βni=1 ai = 192, β i=12 a2i = 120 , then n is equal to (1) 18 (2) 36 (3) 96 (4) 48 JEE Main 2022 (24 Jun Shift 1) JEE Main Previous Year Paper
Q63.Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of ΞPQR is (1) 25 (2) 25β3 4β3 2 (3) 25 (4) 25 β3 2β3
Q63.If β31k=1(31Ck)(31Ckβ1) ββ30k=1(30Ck)(30Ckβ1) = (30!)(31!)Ξ±(60!) , where (1) 1411 (2) 1320 (3) 1615 (4) 1855 + y2 β2x β4y = 0 intersect at
Q63.The value of cos( 2Ο7 ) + cos( 4Ο7 ) + cos( 6Ο7 ) is equal to (1) β1 (2) β12 (3) β13 (4) β14
Q63.The remainder when (2021)2022 + (2022)2021 is divided by 7 is (1) 0 (2) 1 (3) 2 (4) 6
Q63.The number of solutions of cosπ₯= sinπ₯, such that -4πβ€π₯β€4π is (1) 4 (2) 6 (3) 8 (4) 12
Q63.Let a circle πΆ touch the lines πΏ1: 4π₯- 3π¦+ πΎ1 = 0 and πΏ2: 4π₯- 3π¦+ πΎ2 = 0, πΎ1, πΎ2 βπ . If a line passing through the centre of the circle πΆ intersects πΏ1 at -1, 2 and πΏ2 at 3, - 6, then the equation of the circle πΆ is (1) π₯- 12 + π¦- 22 = 4 (2) π₯- 12 + π¦+ 22 = 16 (3) π₯+ 12 + π¦- 22 = 4 (4) π₯- 12 + π¦- 22 = 16
Q64.Let S = {ΞΈ β(0, Ο2 ) : β9m=1 sec(ΞΈ + (m β1) Ο6 ) sec(ΞΈ + mΟ6 ) = β8β3 }. Then (1) S = { 12Ο } (2) S = { 2Ο3 } (3) βΞΈβS ΞΈ = Ο2 (4) βΞΈβS ΞΈ = 3Ο4