Practice Questions

Q61.If the solution of the equation 1, 𝑥∈0, 𝜋 is sin-1𝛼+ √𝛽 , where 𝛼, 𝛽 are integers, logcos𝑥cot𝑥+ 4logsin𝑥tan𝑥= 2 2 then 𝛼+ 𝛽 is equal to: (1) 3 (2) 5 (3) 6 (4) 4 -2

Q61.Let w = zz + k1z + k2iz + λ(1 + i), k1, k2 ∈R. . Let Re(w) = 0 be the circle C of radius 1 in the first quadrant touching the line y = 1 and the y−axis. If the curve Im(w) = 0 intersects C at A and B, then 30(AB)2 is equal to _______. JEE Main 2023 (13 Apr Shift 1) JEE Main Previous Year Paper

Q61.The number of real roots of the equation √𝑥2 - 4𝑥+ 3 + √𝑥2 - 9 = √4𝑥2 - 14𝑥+ 6, is: (1) 0 (2) 1 (3) 3 (4) 2

Q61.Let S = {α : log2(92α−4 + 13) −log2( 25 ⋅32α−4 + 1) = 2}. Then the maximum value of β for which the equation x2 −2(∑α∈s α) 2x + ∑a∈s (α + 1)2β = 0 has real roots, is _____ .

Q61.Let m and n be the numbers of real roots of the quadratic equations x2 −12x + [x] + 31 = 0 and x2 −5 x + 2 −4 = 0 respectively, where [x] denotes the greatest integer ≤x. Then m2 + mn + n2 is equal to

Q61.Let α1, α2, … , α7α1, α2, … , α7 be the roots of the equation x7 + 3x5 −13x3 −15x = 0 and |α1| ≥|α2| ≥… ≥|α7|. Then, α1α2 −α3α4 + α5α6 is equal to _______ ¯

Q61.Let α, β be the roots of the quadratic equation x2 + √6x + 3 = 0. Then α15+β15+α10+β10α23+β23+α14+β14 (1) 81 (2) 9 (3) 72 (4) 729

Q62.The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is _______.

Q62.Let a, b be two real numbers such that ab < 0 . If the complex number 1+aib+i is of unit modulus and a + ib lies on the circle |z −1| = |2z| , then a possible value of 1+[a]4b , where [t] is greatest integer function, is : (1) 0 (2) −1 (3) 1 (4) 21

Q62.Let 𝑆= 𝑧∈ℂ: ¯𝑧= 𝑖𝑧2 + Re ( ¯𝑧) . Then ∑𝑧∈𝑆| 𝑧| 2 is equal to (1) 5 (2) 4 2 (3) 7 (4) 3 2

Q62.Let α = 8 −14i, A = {z ∈C : z2−(¯z)2−112iαz−α¯z = 1} and B = {z ∈C : |z + 3i| = 4} Then, ∑z∈A∩B(Re z −Imz) is equal to ________

Q63.The sum to 10 terms of the series 1 2 3 + + + … is :- 1 + 12 + 14 1 + 22 + 24 1 + 32 + 34 59 55 (1) (2) 111 111 (3) 56 (4) 58 111 111

Q64.Let a circle 𝐶1 be obtained on rolling the circle 𝑥2 + 𝑦2 - 4𝑥- 6𝑦+ 11 = 0 upwards 4 units on the tangent T to it at the point 3, 2. Let 𝐶2 be the image of 𝐶1 in 𝑇. Let 𝐴 and 𝐵 be the centers of circles 𝐶1 and 𝐶2 respectively, and 𝑀 and 𝑁 be respectively the feet of perpendiculars drawn from 𝐴 and 𝐵 on the 𝑥-axis. Then the area of the trapezium AMNB is: (1) 22 + √2 (2) 41 + √2 (3) 3 + 2√2 (4) 21 + √2

Q64.The sum 12 −2. 32 + 3. 52 −4. 72 + 5. 92 −… . . +15. 292 is _____ . , is

Q64.Let a, b, c > 1, a3, b3 and c3 be in A. P. and loga b, logc a and logb c be in G. P. If the sum of first 20 terms of an A. P., whose first term is a+4b+c3 and the common difference is a−8b+c10 is −444, then abc is equal to (1) 343 (2) 216 (3) 343 (4) 125 8 8

Q64.Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? = p1 p2 p3 . . . pm , where

Q64.Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____ .

Q64.Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 orange, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is _____ .

Q64.Let 𝑥1, 𝑥2, … , 𝑥100 be in an arithmetic progression, with 𝑥1 = 2 and their mean equal to 200 . If 𝑦𝑖= 𝑖𝑥𝑖- 𝑖, 1 ≤𝑖≤100, then the mean of 𝑦1, 𝑦2, … , 𝑦100 is (1) 10100 (2) 10101 . 50 (3) 10049 . 50 (4) 10051 . 50

Q64.The total number of 4 -digit numbers whose greatest common divisor with 54 is 2 , is

Q65.If the maximum distance of normal to the ellipse 𝑥2 + 𝑦2 = 1, 𝑏< 2, from the origin is 1 , then the eccentricity 4 𝑏2 of the ellipse is: (1) 1 (2) √3 √2 2 (3) 1 (4) √3 2 4

Q65.The sum ∑∞n=1 2n2+3n+4(2n)! is equal to : (1) 11e 2 + 2e7 (2) 13e4 + 4e5 −4 (3) 11e 2 + 2e7 −4 (4) 13e4 + 4e5

Q65.If (30C1)2 + 2(30C2)2 + 3(30C3)2. . . . . . . . . . 30(30C30)2 = (30!)2α60! , then (1) 30 (2) 60 (3) 15 (4) 10

Q65.Let a, b, c and d be positive real numbers such that a + b + c + d = 11 . If the maximum value of a5b3c2d is 3750β, then the value of β is (1) 90 (2) 110 (3) 55 (4) 108

Q65.Let a1 = b1 = 1 and an = an−1 + (n −1), bn = bn−1 + an−1, ∀n ≥2. If S = ∑10n=1( 2nbn ) and T = ∑8n=1 2n−1n then 27(2S −T) is equal to