Practice Questions

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 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.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(n−7)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.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.If the sum of the first 15 terms of the series ( 43 ) 3 + (1 12 ) 3 + (2 14 ) 3 + 33 + (3 34 ) 3 + … 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

Q64.The sum of all natural numbers 𝑛 such that 100 < 𝑛< 200 and 𝐻. 𝐶. 𝐹. 91, 𝑛> 1 is (1) 3203 (2) 3221 (3) 3121 (4) 3303

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 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.The positive value of λ for which the co-efficient of x2 in the expansion x2(√x + x2λ ) 10 (1) √5 (2) 3 (3) 4 (4) 2√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.If 𝑎, 𝑏 and 𝑐 be three distinct real numbers in G.P. and 𝑎+ 𝑏+ 𝑐= 𝑥𝑏, then 𝑥 cannot be: (1) -3 (2) 2 (3) 4 (4) -2

Q65.If ∑20i=1( 20Ci+20Ci−120Ci−1 ) 3 (1) 200 (2) 100 (3) 50 (4) 400

Q65.The sum of the co-efficient of all even degree terms in 𝑥 in the expansion of 6 6 𝑥+ √𝑥3 - 1 +𝑥- √𝑥3 - 1 , 𝑥> 1 is equal to (1) 26 (2) 32 (3) 24 (4) 29

Q65.Let 𝑎1, 𝑎2, 𝑎3 . . . be an 𝐴. 𝑃. with 𝑎6 = 2 . Then, the common difference of this 𝐴. 𝑃. , which maximise the product 𝑎1 · 𝑎4 · 𝑎5, is : 2 3 (1) (2) 3 2 6 8 (3) (4) 5 5

Q65.The value of cos π ⋅cos π ⋅… ⋅cos π ⋅sin π is: 22 23 210 210 (1) 1 (2) 1 1024 512 (3) 1 (4) 1 2 256

Q65.Let Sk = 1 + 2 + k3+…+k . If S 12 + S 22 + … + S 102 = 125 A, then A is equal to : (1) 301 (2) 303 (3) 156 (4) 283

Q65.If the fourth term in the Binomial expansion of ( x2 + xlog8x)6, (x > 0) is 20 × 87, then a value of (1) 8–2 (2) 8 (3) 83 (4) 82

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.The sum 3×13 + 5×(13+23) + 7×(13+23+33) +. . . . . upto 10th term is 12 12+22 12+22+32 (1) 660 (2) 600 (3) 620 (4) 680

Q65.Let (x + 10)50 + (x −10)50 = a0 + a1x + a2x2 + … . +a50x50 , for all x ∈R; then a2 is equal to : a0 (1) 12.5 (2) 12 (3) 12.25 (4) 12.75

Q66.If the third term in the binomial expansion of (1 + xlog2 x)5 equals 2560, then a possible value of x is (1) 4√2 (2) 18 (3) 2 √2 (4) 14

Q66.If the coefficients of x2 and x3 , are both zero, in the expansion of the expression (1 + ax + bx2)(1 −3x)15 , in powers of x , then the ordered pair (a, b) is equal to (1) (28, 315) (2) (−21, 714) (3) (28, 861) (4) (−54, 315)

Q66.The value of r for which 20Cr20C0 + 20Cr−120C1 + 20Cr−220C2 + … + 20C020Cr is maximum, is: (1) 15 (2) 20 (3) 11 (4) 10