Practice Questions

Q65.If sin4α + 4cos4β + 2 = 4√2sinαcosβ, α, β ∈[0, π] , then cos(α + β) −cos(α −β) is equal to (1) −1 (2) −√2 (3) √2 (4) 0 JEE Main 2019 (12 Jan Shift 2) JEE Main Previous Year Paper

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.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) −25 (2) −35 (3) 25 (4) −36

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.The sum of the following series 1 + 6 + 9(12+22+32)7 + 12(12+22+32+42)9 + 15(12+22+…+52)11 +. . . . is: (1) 7520 (2) 7510 (3) 7830 (4) 7820

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

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 ∑ 𝑘 is equal to 𝑘= 1 2𝑘 11 21 (1) 1 – (2) 2 – 220 220 3 11 (3) 2 – (4) 2 – 217 219 6 Q66. 1 1 If the fourth term in the binomial expansion of √𝑥 1 + log10𝑥+ 𝑥 12 is equal to 200, and 𝑥> 1, then the value of 𝑥 is (1) 100 (2) 104 (3) 103 (4) 10

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

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(2β), then the ordered pair (A, β) 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) − 3 ) 6 is equal to x2

Q65.Let 𝑆𝑛 denote the sum of the first 𝑛 terms of an 𝐴. 𝑃. . If 𝑆4 = 16 and 𝑆6 = - 48 , then 𝑆10 is equal to: (1) -320 (2) -380 (3) -260 (4) -410

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

Q66.The product of three consecutive terms of a G. P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A. P., then the sum of the original three terms of the given G. P. is : (1) 28 (2) 24 (3) 32 (4) 36

Q66.Let 𝑎, 𝑏 and 𝑐 be in 𝐺. 𝑃. with common ratio 𝑟, where 𝑎≠0 and 0 < 𝑟≤1 . If 3𝑎, 7𝑏 and 15𝑐 are the 2 first three terms of an 𝐴. 𝑃. , then the 4𝑡ℎ term of this 𝐴. 𝑃. is : 7 (1) 𝑎 (2) 3𝑎 2 (3) 5𝑎 (4) 3𝑎 1 𝑛

Q66.If some three consecutive coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is: (1) 227 (2) 964 (3) 625 (4) 232

Q66.Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant? (1) Fourth (2) Second (3) Third (4) First

Q66.If the fractional part of the number 2403 is 𝑘 then 𝑘 is equal to 15 15, (1) 4 (2) 14 (3) 8 (4) 6 𝜋 𝜋

Q66.The value of cos2 10°– cos 10° cos 50°+cos250° is (1) 3 (2) 3 4 4 + cos 20° (3) 3 (4) 3 2 2 (1 + cos 20°)

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.The sum of the series 2 . 20𝐶0 + 5 . 20𝐶1 + 8 . 20𝐶2 + 11 . 20𝐶3 + . . . . . . . + 62 . 20𝐶20 is equal to (1) 226 (2) 225 (3) 224 (4) 223

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

Q66.The coefficient of 𝑥18 in the product 1 + 𝑥1 - 𝑥101 + 𝑥+ 𝑥29 is (1) 84 (2) -84 (3) -126 (4) 126

Q66.If nC4, nC5 and nC6 are in A.P., then n can be (1) 9 (2) 14 (3) 12 (4) 11