Practice Questions
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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 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.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 {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.The remainder when (11)1011 + (1011)11 is divided by 9 is _____ . (1) 1 (2) 8 (3) 6 (4) 4
Q63.The remainder when (2021)2022 + (2022)2021 is divided by 7 is (1) 0 (2) 1 (3) 2 (4) 6
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
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 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 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 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.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
Q64.The remainder when 72022 + 32022 is divided by 5 is (1) 0 (2) 2 (3) 3 (4) 4
Q64.Let the area of the triangle with vertices A(1, Ξ±), B(Ξ±, 0) and C(0, Ξ±) be 4 sq. units. If the points (Ξ±, βΞ±), (βΞ±, Ξ±) and (Ξ±2, Ξ²) are collinear, then Ξ² is equal to (1) 64 (2) β8 (3) β64 (4) 512
Q64.The locus of the mid-point of the line segment joining the point (4, 3) and the points on the ellipse x2 + 2y2 = 4 is an ellipse with eccentricity (1) β3 (2) 1 2 2β2 (3) 1 (4) 1 β2 2
Q64.Let n β₯5 be an integer. If 9n β8n β1 = 64Ξ± and 6n β5n β1 = 25 Ξ², then Ξ± βΞ² is equal to: (1) 1 + nC2(8 β5) + nC3(82 β52) + β¦ + nCn(8nβ1(2)β5nβ2)1 + nC3(8 β5) + nC4(82 β52) + β¦ + nCn(8nβ2 β5nβ2 (3) nC3(8 β5) + nC4(82 β52) + β¦ + nCn(8nβ2 β5nβ2)(4) nC4(8 β5) + nC5(82 β52) + β¦ + nCn(8nβ3 β5nβ3)
Q64.If the tangents drawn at the point O(0, 0) and P(1 + β5, 2) on the circle x2 the point Q, then the area of the triangle OPQ is equal to (1) 3+β5 (2) 4+2β5 2 2 (3) 5+3β5 (4) 7+3β5 2 2
Q64.Let C be a circle passing through the points A(2, β1) and B(3, 4). The line segment AB is not a diameter of C . If r is the radius of C and its centre lies on the circle (x β5)2 + (y β1)2 = 132 , then r2 is equal to (1) 32 (2) 652 (3) 61 (4) 30 2
Q64.The value of 2 sin 22Ο sin 3Ο22 sin 5Ο22 sin 7Ο22 sin 9Ο22 is equal to: (1) 1 (2) 5 16 16 (3) 7 (4) 3 16 16 JEE Main 2022 (25 Jul Shift 2) JEE Main Previous Year Paper
Q64.Let A(1, 1), B(β4, 3), C(β2, β5) be vertices of a triangle ABC, P be a point on side BC , and Ξ1 and Ξ2 be the areas of triangle APB and ABC . Respectively. If Ξ1 : Ξ2 = 4 : 7 , then the area enclosed by the lines AP, AC and the x -axis is (1) 1 (2) 3 4 4 (3) 1 (4) 1 2
Q64.In an isosceles triangle ABC , the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4 . Let the point B lie on the line x + 3y = 7. If (Ξ±, Ξ²) is the centroid ΞABC , then 15(Ξ± + Ξ²) is equal to (1) 51 (2) 39 (3) 41 (4) 49 y2
Q64.A line, with the slope greater than one, passes through the point π΄4, 3 and intersects the line π₯- π¦- 2 = 0 at the point π΅. If the length of the line segment π΄π΅ is β29 , then π΅ also lies on the line 3 (1) 2π₯+ π¦= 9 (2) 3π₯- 2π¦= 7 (3) π₯+ 2π¦= 6 (4) 2π₯- 3π¦= 3
Q64.The distance between the two points A and Aβ² which lie on y = 2 such that both the line segments AB and Aβ²B (where B is the point (2, 3)) subtend angle Ο4 at the origin, is equal to (1) 10 (2) 485 (3) 52 (4) 3 5
Q64.The term independent of x in the expression of (1 βx2 + 11 5x2 1 ) 3x3)( 25 x3 (1) 7 (2) 33 40 200 (3) 39 (4) 11 200 50