Practice Questions
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Q63.Let z0 be a root of quadratic equation, x2 + x + 1 = 0. If z = 3 + 6iz810 β3iz930 , then arg (z) is equal to: (1) 0 (2) Ο4 (3) Ο (4) Ο 6 3
Q64.Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square, whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is (1) 262 (2) 190 (3) 225 (4) 157
Q64.The number of natural numbers less than 7000 which can be formed by using the digits 0, 1, 3, 7, 9 (repetition of digits allowed) is equal to: (1) 375 (2) 250 (3) 374 (4) 372 up to 15 terms,
Q64.Consider three boxes, each containing 10 balls labelled 1, 2, β¦ . , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni , the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is : (1) 240 (2) 82 (3) 120 (4) 164
Q64.The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27 . 19 Then the common ratio of this series is: JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper (1) 1 (2) 2 3 3 (3) 2 (4) 4 9 9
Q64.Let the sum of the first n terms of a non-constant A. P. , a1, a2, a3, . . . . , an be 50n + n(nβ7)2 A, where a constant. If d is the common difference of this A. P., then the ordered pair (d, a50) is equal to (1) (50, 50 + 46A) (2) (A, 50 + 45A) (3) (50, 50 + 45A) (4) (A, 50 + 46A) x is
Q64.If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is: 1 3 (1) (2) 10 10 (3) 1 (4) 3 5 20
Q64.The sum of all natural numbers π such that 100 < π< 200 and π». πΆ. πΉ. 91, π> 1 is (1) 3203 (2) 3221 (3) 3121 (4) 3303
Q64.If π, π and π be three distinct real numbers in G.P. and π+ π+ π= π₯π, then π₯ cannot be: (1) -3 (2) 2 (3) 4 (4) -2
Q64.The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is (1) 1356 (2) 1365 (3) 1256 (4) 1465 = 21k , then k equals
Q64.The number of four-digit numbers strictly greater than 4321 that can be formed using the digit 0,1, 2,3, 4,5 (repetition of digits is allowed) is: (1) 360 (2) 288 (3) 306 (4) 310 JEE Main 2019 (08 Apr Shift 2) JEE Main Previous Year Paper 20 1
Q64.If a1, a2, a3, . . . . are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P is: (1) 280 (2) 120 (3) 150 (4) 200
Q64.The sum 1 + 13 + 23 + 13 + 23 + 33 + . . . + 13 + 23 + 33 + . . . + 153 - 1 + 2 + 3 + . . . + 15 is equal to 1 + 2 1 + 2 + 3 1 + 2 + 3 + . . . + 15 21 (1) 620 (2) 1240 (3) 1860 (4) 660
Q64.Let Sn = 1 + q + q2 + β¦ . +qn and Tn = 1 + ( q+12 ) ( q+12 ) ( q+12 ) and q β 1. If 101C1 + 101C2 β S1 + β¦ . +101C101 β S100 = Ξ±T100, then Ξ± is equal to : (1) 299 (2) 202 (3) 200 (4) 2100
Q64.The positive value of Ξ» for which the co-efficient of x2 in the expansion x2(βx + x2Ξ» ) 10 (1) β5 (2) 3 (3) 4 (4) 2β2
Q64.If the sum of the first 15 terms of the series ( 43 ) 3 + (1 12 ) 3 + (2 14 ) 3 + 33 + (3 34 ) 3 + β¦ then K is equal to : (1) 9 (2) 27 (3) 54 (4) 108
Q64.If a1, a2, a3. . . . . . . . . , an are in A. P. and a1 + a4 + a7. . . . . . . . . +a16 = 114 , then a1 + a6 + a11 + a16 is equal to : (1) 64 (2) 98 (3) 38 (4) 76 JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper
Q65.If the sum and product of the first three terms in an A. P. are 33 and 1155, respectively, then a value of its 11th term is: (1) β25 (2) β35 (3) 25 (4) β36
Q65.Let ππ denote the sum of the first π terms of an π΄. π. . If π4 = 16 and π6 = - 48 , then π10 is equal to: (1) -320 (2) -380 (3) -260 (4) -410
Q65.The sum of the following series 1 + 6 + 9(12+22+32)7 + 12(12+22+32+42)9 + 15(12+22+β¦+52)11 +. . . . is: (1) 7520 (2) 7510 (3) 7830 (4) 7820
Q65.The sum of the real values of x for which the middle term in the binomial expansion of 8 + x3 ) equals ( x33 5670 is : (1) 0 (2) 6 (3) 4 (4) 8
Q65.If 20C1 + (22) 20C2 + (32) 20C3+. . . . . +(202) 20C20 = A(2Ξ²), then the ordered pair (A, Ξ²) is equal to JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper (1) (380, 19) (2) (420, 18) (3) (420, 19) (4) (380, 18) β 3 ) 6 is equal to x2
Q65.Let Sk = 1 + 2 + k3+β¦+k . If S 12 + S 22 + β¦ + S 102 = 125 A, then A is equal to : (1) 301 (2) 303 (3) 156 (4) 283
Q65.Let (x + 10)50 + (x β10)50 = a0 + a1x + a2x2 + β¦ . +a50x50 , for all x βR; then a2 is equal to : a0 (1) 12.5 (2) 12 (3) 12.25 (4) 12.75
Q65.The value of cos Ο β cos Ο β β¦ β cos Ο β sin Ο is: 22 23 210 210 (1) 1 (2) 1 1024 512 (3) 1 (4) 1 2 256