Practice Questions

Q62.Let 𝑧1 and 𝑧2 be two complex number such that 𝑧1 + 𝑧2 = 5 and 𝑧13 + 𝑧23 = 20 + 15𝑖. Then 𝑧14 + 𝑧24 equals- (1) 30√3 (2) 75 (3) 15√15 (4) 25√3

Q62.Let z be a complex number such that |z + 2| = 1 and Im ( z+2 ) = 5 . Then the value of |Re(z + 2)| is (1) 2√6 (2) 24 5 5 (3) 1+√6 (4) √6 5 5

Q62.Let α and β be the sum and the product of all the non-zero solutions of the equation (¯z)2 + |z| = 0, z ∈ C. Then 4 (α2 + β2) is equal to : (1) 6 (2) 8 (3) 2 (4) 4

Q62.Let 0 ≤r ≤n. If n+1Cr+1 : nCr : n−1Cr−1 = 55 : 35 : 21, then 2n + 5r is equal to: JEE Main 2024 (06 Apr Shift 2) JEE Main Previous Year Paper (1) 50 (2) 62 (3) 55 (4) 60

Q62.Let 𝑎 and 𝑏 be two distinct positive real numbers. Let 11th term of a GP, whose first term is 𝑎 and third term is 𝑏, is equal to 𝑝th term of another GP, whose first term is 𝑎 and fifth term is 𝑏. Then 𝑝 is equal to (1) 20 (2) 25 (3) 21 (4) 24

Q62.If the sum of the series 1 + 1 + … + 1 is equal to 5 , then 50 d is equal to : 1⋅(1+d) (1+d)(1+2 d) (1+9 d)(1+10 d) (1) 10 (2) 5 (3) 15 (4) 20

Q63.If A denotes the sum of all the coefficients in the expansion of (1 −3x + 10x2) and B denotes the sum of all the coefficients in the expansion of (1 + x2)n , then : (1) A = B3 (2) 3 A = B (3) B = A3 (4) A = 3 B

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

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

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

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

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

Q63.For x ⩾0, the least value of K, for which 41+x + 41−x, K2 , 16x + 16−x are three consecutive terms of an A.P., is equal to : (1) 8 (2) 4 (3) 10 (4) 16

Q63.If loge a, loge b, loge c are in an A. P. and loge a −loge 2b, loge 2b −loge 3c, loge 3c −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

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

Q63.The 20th term from the end of the progression 20, 191 181 173 … , - 1291 is :- 4, 2, 4, 4 (1) -118 (2) -110 (3) -115 (4) -100

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

Q63.The coefficient of x70 in x2(1 + x)98 + x3(1 + x)97 + x4(1 + x)96 + … + x54(1 + x)46 is 99Cp −46Cq . Then a possible value of p + q is : (1) 55 (2) 83 (3) 61 (4) 68

Q63.If 2 sin3 x + sin 2x cos x + 4 sin x −4 = 0 has exactly 3 solutions in the interval [0, nπ2 ⌉, n ∈N , then the roots of the equation x2 + nx + (n −3) = 0 belong to : (1) (0, ∞) (2) (−∞, 0) (3) (−√172 , √172 ) (4) Z

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 −𝑆5 is equal to: (1) 800 (2) 890 (3) 790 (4) 690 1 18 1 1

Q63.The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is (1) 406 (2) 130 (3) 142 (4) 136

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

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

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

Q63.Suppose θϵ [0, π4 ] is a solution of 4 cos θ −3 sin θ = 1. Then cos θ is equal to : (1) 4 (2) 6+√6 (3√6+2) (3√6+2) (3) 4 (4) 6−√6 (3√6−2) (3√6−2)