Practice Questions
10,208 questions across 23 years of JEE Main β find and practise any topic!
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Q61.Let πΌ and π½ be the roots of the equation ππ₯2 + ππ₯βπ= 0, where πβ 0. If π, π and π be the consecutive terms of a non-constant G.P and 1 1 3 then the value of πΌβπ½2 is: πΌ+ π½= 4, (1) 80 (2) 9 9 20 (3) (4) 8 3
Q61.Let Ξ±, Ξ² be the roots of the equation x2 + 2β2x β1 = 0. The quadratic equation, whose roots are Ξ±4 + Ξ²4 and 1 (Ξ±6 + Ξ²6), is : 10 (1) x2 β190x + 9466 = 0 (2) x2 β180x + 9506 = 0 (3) x2 β195x + 9506 = 0 (4) x2 β195x + 9466 = 0
Q62.Let π and π be two distinct positive real numbers. Let 11th term of a GP, whose first term is π and third term is π, is equal to πth term of another GP, whose first term is π and fifth term is π. Then π is equal to (1) 20 (2) 25 (3) 21 (4) 24
Q62.In an A.P., the sixth term a6 = 2. If the a1a4a5 is the greatest, then the common difference of the A.P., is equal to (1) 3 (2) 8 2 5 (3) 2 (4) 5 3 8
Q62.If the sum of the series 1 + 1 + β¦ + 1 is equal to 5 , then 50 d is equal to : 1β (1+d) (1+d)(1+2 d) (1+9 d)(1+10 d) (1) 10 (2) 5 (3) 15 (4) 20
Q62.If π§ is a complex number such that π§β€1, then the minimum value of π§+ 1 + 4π is: 23 5 (1) 2 (2) 2 3 (3) (4) 3 2
Q62.If 1 + 1 + β¦ + 1 = m and 1β 21 + 2β 31 + β¦ + 99β 1001 = n , then the point (m, n) lies on the β1+β2 β2+β3 β99+β100 line (1) 11(x β1) β100(y β2) = 0 (2) 11x β100y = 0 (3) 11(x β2) β100(y β1) = 0 (4) 11(x β1) β100y = 0
Q62.Let z be a complex number such that the real part of zβ2i is zero. Then, the maximum value of |z β(6 + 8i)| z+2i is equal to (1) 12 (2) 10 (3) 8 (4) β
Q62.The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is (1) 48 (2) 56 (3) 24 (4) 16
Q62.The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to : (1) 179 (2) 177 (3) 181 (4) 175
Q62.Let π= π§βπΆ: π§β1 = 1 and β2 β1π§+ Β―π§- ππ§- Β―π§= 2β2. Let π§1, π§2 βπ be such that π§1 = maxπ§βπ π§ and 2 π§2 = minπ§βπ π§. Then β2π§1 βπ§2 equals: (1) 1 (2) 4 (3) 3 (4) 2
Q62.Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to (1) 18 (2) 16 (3) 12 (4) 15
Q62.The number of common terms in the progressions 4, 9, 14, 19, β¦ β¦, up to 25th term and 3, 6, 9, 12,.... up to 37th term is : (1) 9 (2) 5 (3) 7 (4) 8 n
Q62.The value of 1Γ22+2Γ32+β¦+100Γ(101)2 is 12Γ2+22Γ3+β¦.+1002Γ101 (1) 32 (2) 31 31 30 (3) 306 (4) 305 305 301 JEE Main 2024 (04 Apr Shift 2) JEE Main Previous Year Paper
Q62.For 0 < π< π< π, let ( π+ πβ 2π) π₯2 + ( π+ πβ 2π) π₯+ ( π+ πβ 2π) = 0 and πΌβ 1 be one of its root. Then, among the two statements (I) If πΌβ-1, 0, then π cannot be the geometric mean of π and π. (II) If πΌβ0, 1, then π may be the geometric mean of π and π. (1) Both (I) and (II) are true (2) Neither (I) nor (II) is true (3) Only (II) is true (4) Only (I) is true 1 2 3
Q62.Let 0 β€r β€n. If n+1Cr+1 : nCr : nβ1Crβ1 = 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 z be a complex number such that |z + 2| = 1 and Im ( z+2 ) = 5 . Then the value of |Re(z + 2)| is (1) 2β6 (2) 24 5 5 (3) 1+β6 (4) β6 5 5
Q62.Let π§1 and π§2 be two complex number such that π§1 + π§2 = 5 and π§13 + π§23 = 20 + 15π. Then π§14 + π§24 equals- (1) 30β3 (2) 75 (3) 15β15 (4) 25β3
Q62.Let Ξ± and Ξ² be the sum and the product of all the non-zero solutions of the equation (Β―z)2 + |z| = 0, z β C. Then 4 (Ξ±2 + Ξ²2) is equal to : (1) 6 (2) 8 (3) 2 (4) 4
Q63.Let A = {n β[100, 700] β©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.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.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, β¦ , P11 on the side BC and 7 points P12, P13, β¦ , P18 on the side CA of the triangle. The number of triangles, that can be formed using the points P1, P2, β¦ , P18 as vertices, is : (1) 776 (2) 796 (3) 751 (4) 771
Q63.If A denotes the sum of all the coefficients in the expansion of (1 β3x + 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 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