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Practice Questions

7,135 questions across 23 years of JEE Main β€” find and practise any topic!

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Q63.For x β©Ύ0, the least value of K, for which 41+x + 41βˆ’x, K2 , 16x + 16βˆ’x are three consecutive terms of an A.P., is equal to : (1) 8 (2) 4 (3) 10 (4) 16

202405 Apr Shift 2Sequences & Series
MathsMedium

Q63.If the set R = {(a, b) : a + 5b = 42, a, b ∈N} has m elements and βˆ‘mn=1 (1 βˆ’in!) = x + iy, where i = βˆšβˆ’1 , then the value of m + x + y is (1) 12 (2) 4 (3) 8 (4) 5

202408 Apr Shift 1Sets Relations Functions
MathsMedium

Q63.Suppose 28 - 𝑝, 𝑝, 70 - 𝛼, 𝛼 are the coefficient of four consecutive terms in the expansion of ( 1 + π‘₯) 𝑛. Then the value of 2𝛼- 3𝑝 equals (1) 7 (2) 10 (3) 4 (4) 6 πœ‹

202430 Jan Shift 2Binomial Theorem
MathsMedium

Q63.The sum of the series + + + . ... up to 10 terms is 1 βˆ’3 β‹…12 + 14 1 βˆ’3 β‹…22 + 24 1 βˆ’3 β‹…32 + 34 (1) 45 (2) - 45 109 109 55 55 (3) (4) - 109 109

202431 Jan Shift 1Sequences & Series
MathsMedium

Q63.Let three real numbers a, b, c be in arithmetic progression and a + 1, b, c + 3 be in geometric progression. If a > 10 and the arithmetic mean of a, b and c is 8, then the cube of the geometric mean of a, b and c is (1) 128 (2) 316 (3) 120 (4) 312

202404 Apr Shift 2Sequences & Series
MathsMedium

Q63.Suppose ΞΈΟ΅ [0, Ο€4 ] is a solution of 4 cos ΞΈ βˆ’3 sin ΞΈ = 1. Then cos ΞΈ is equal to : (1) 4 (2) 6+√6 (3√6+2) (3√6+2) (3) 4 (4) 6βˆ’βˆš6 (3√6βˆ’2) (3√6βˆ’2)

202405 Apr Shift 1Trigonometric Functions & Equations
MathsMedium

Q63.Let A = {n ∈[100, 700] ∩N : n is neither a multiple of 3 nor a multiple of 4 }. Then the number of elements in A is (1) 290 (2) 280 (3) 300 (4) 310

202406 Apr Shift 1Sets Relations Functions
MathsMedium

Q63.If loge a, loge b, loge c are in an A. P. and loge a βˆ’loge 2b, loge 2b βˆ’loge 3c, loge 3c βˆ’loge a are also in an A. P., then a : b : c is equal to (1) 9 : 6 : 4 (2) 16 : 4 : 1 (3) 25 : 10 : 4 (4) 6 : 3 : 2

202429 Jan Shift 2Sequences & Series
MathsMedium

Q63.Let 𝑆𝑛 denote the sum of the first n terms of an arithmetic progression. If 𝑆10 = 390 and the ratio of the tenth and the fifth terms is 15 : 7, then 𝑆15 βˆ’π‘†5 is equal to: (1) 800 (2) 890 (3) 790 (4) 690 1 18 1 1

202401 Feb Shift 2Sequences & Series
MathsMedium

Q63.If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to (1) 7 (2) 4 (3) 5 (4) 6

202429 Jan Shift 1Sequences & Series
MathsMedium

Q63.If 𝑛 is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then 𝑛 is equal to: (1) 47 (2) 53 (3) 51 (4) 43

202401 Feb Shift 1Permutation & Combination
MathsMedium

Q63.In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 70 and the 3 product of the third and fifth terms is 49 . Then the sum of the 4th , 6th and 8th terms is equal to : (1) 96 (2) 91 (3) 84 (4) 78

202408 Apr Shift 2Sequences & Series
MathsMedium

Q63.Let a, ar, ar2 , be an infinite G.P. If βˆ‘βˆžn=0 arn = 57 and βˆ‘βˆžn=0 a3r3n = 9747, then a + 18r is equal to (1) 46 (2) 38 (3) 31 (4) 27 is

202409 Apr Shift 2Sequences & Series
MathsMedium

Q63.If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315th position in this arrangement is : (1) NRAGUP (2) NRAPUG (3) NRAPGU (4) NRAGPU

202406 Apr Shift 2Permutation & Combination
MathsMedium

Q63.There are 5 points P1, P2, P3, P4, P5 on the side AB, excluding A and B, of a triangle ABC . Similarly there are 6 points P6, P7, … , P11 on the side BC and 7 points P12, P13, … , P18 on the side CA of the triangle. The number of triangles, that can be formed using the points P1, P2, … , P18 as vertices, is : (1) 776 (2) 796 (3) 751 (4) 771

202404 Apr Shift 1Permutation & Combination
MathsMedium

Q64.The sum of the coefficient of x2/3 and xβˆ’2/5 in the binomial expansion of (x2/3 + 12 xβˆ’2/5) 9 (1) 21/4 (2) 63/16 (3) 19/4 (4) 69/16

202409 Apr Shift 2Binomial Theorem
MathsMedium

Q64.Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then : (1) P2 = 6√3Q (2) P2 = 36√3Q (3) P = 36√3Q2 (4) P2 = 72√3Q

202406 Apr Shift 2Sequences & Series
MathsMedium

Q64.If the term independent of x in the expansion of (√ax2 + 2x31 )10 is 105 , then a2 is equal to : (1) 2 (2) 4 (3) 6 (4) 9 JEE Main 2024 (08 Apr Shift 2) JEE Main Previous Year Paper cos 36∘+5 sin 18∘

202408 Apr Shift 2Binomial Theorem
MathsMedium

Q64.Let |cos ΞΈ cos(60 βˆ’ΞΈ) cos(60 + ΞΈ)| ≀18 , ΞΈΟ΅[0, 2Ο€]. Then, the sum of all ΞΈΟ΅[0, 2Ο€], where cos 3ΞΈ attains its maximum value, is : (1) 15Ο€ (2) 18Ο€ (3) 6Ο€ (4) 9Ο€

202409 Apr Shift 1Trigonometric Functions & Equations
MathsMedium

Q64.If the constant term in the expansion of 12 + , x β‰ 0, is Ξ± Γ— 28 Γ— 5√3, then 25Ξ± is equal to : ( 5√3x 2x ) 3√5 (1) 724 (2) 742 (3) 639 (4) 693

202405 Apr Shift 2Binomial Theorem
MathsMedium

Q64. nβˆ’1Cr = (k2 βˆ’8)nCr+1 if and only if : (1) 2√2 < k ≀3 (2) 2√3 < k ≀3√2 (3) 2√3 < k < 3√3 (4) 2√2 < k < 2√3 JEE Main 2024 (27 Jan Shift 1) JEE Main Previous Year Paper

202427 Jan Shift 1Permutation & Combination
MathsMedium

Q64.Let the first three terms 2, p and q , with q β‰ 2, of a G.P. be respectively the 7th , 8th and 13th terms of an A.P. If the 5th term of the G.P. is the nth term of the A.P., then n is equal to: (1) 163 (2) 151 (3) 177 (4) 169

202404 Apr Shift 1Sequences & Series
MathsMedium

Q64.If the coefficients of x4, x5 and x6 in the expansion of (1 + x)n are in the arithmetic progression, then the maximum value of n is: (1) 7 (2) 21 (3) 28 (4) 14

202404 Apr Shift 2Binomial Theorem
MathsMedium

Q64.If each term of a geometric progression a1, a2, a3, … with a1 = 18 and a2 β‰ a1 , is the arithmetic mean of the next two terms and Sn = a1 + a2 + … + an , then S20 βˆ’S18 is equal to (1) 215 (2) βˆ’218 (3) 218 (4) βˆ’215

202429 Jan Shift 2Sequences & Series
MathsMedium

Q64.If Ξ±, βˆ’Ο€2 < Ξ± < Ο€2 is the solution of 4 cos ΞΈ + 5 sin ΞΈ = 1, then the value of tan Ξ± is (1) 10βˆ’βˆš10 (2) 10βˆ’βˆš10 6 12 (3) √10βˆ’10 (4) √10βˆ’10 12 6

202429 Jan Shift 1Trigonometric Functions & Equations
MathsMedium

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