Practice Questions

Q61.Let a complex number be w = 1 −√3i . Let another complex number z be such that |zw| = 1 and arg(z) −arg(w) = π2 . Then the area of the triangle (in sq. units) with vertices origin, z and w is equal to (1) 4 (2) 12 (3) 1 (4) 2 4

Q61.The integer k, for which the inequality x2 −2(3k −1)x + 8k2 −7 > 0 is valid for every x in R is: (1) 4 (2) 2 (3) 3 (4) 0 JEE Main 2021 (25 Feb Shift 1) JEE Main Previous Year Paper ¯¯

Q61.Let 𝑆𝑛 be the sum of the first 𝑛 terms of an arithmetic progression. If 𝑆3𝑛= 3𝑆2𝑛, then the value of 𝑆4𝑛 is : 𝑆2𝑛 JEE Main 2021 (25 Jul Shift 1) JEE Main Previous Year Paper (1) 6 (2) 4 (3) 2 (4) 8

Q61.Let S1, S2 and S3 be three sets defined as : z −1 S1 = ≤√2}, {z ∈C S2 = {z ∈C : Re((1 −i)z) ≥1} and S3 = {z ∈C : Im(z) ≤1}. Then, the set S1 ∩S2 ∩S3 (1) is a singleton (2) has exactly two elements (3) has infinitely many elements (4) has exactly three elements

Q61.The value of 4 + 1 1 is: 5+ 1 4+ 1 5+ 4+……∞ (1) 2 + 52 √30 (2) 2 + √54 √30 (3) 4 + 4 √30 (4) 5 + 25 √30 √5

Q61.If for x ∈(0, π2 ), log10 sin x + log10 cos x = −1 and log10(sin x + cos x) = 12 (log10 n −1), n > 0 , then the value of n is equal to : (1) 20 (2) 12 (3) 9 (4) 16

Q61.The number of real solutions of the equation, x2 −|x| −12 = 0 is: (1) 2 (2) 3 (3) 1 (4) 4

Q61.Let n denote the number of solutions of the equation z2 + 3z = 0, where z is a complex number. Then the value of ∑∞k=0 nk1 is equal to (1) 1 (2) 34 (3) 32 (4) 2

Q61.Let α and β be the roots of x2 −6x −2 = 0. If an = αn −βn for n ⩾1, then the value of a10−2a83a9 is: (1) 1 (2) 3 (3) 2 (4) 4

Q61.The number of pairs 𝑎, 𝑏 of real numbers, such that whenever 𝛼 is a root of the equation 𝑥2 + 𝑎𝑥+ 𝑏= 0, 𝛼2 - 2 is also a root of this equation, is : (1) 6 (2) 8 (3) 4 (4) 2

Q62.The sum of the series ∑∞n=1 n2+6n+10(2n+1)! is equal to (1) 41 8 e + 198 e−1 + 10 (2) 418 e + 198 e−1 −10 (3) −418 e + 198 e−1 −10 (4) 418 e −198 e−1 −10 + + …

Q62.Let C be the set of all complex numbers. Let S1 = {z ∈C |z–3–2i|2 = 8}, S2 = z ∈C| Re(z) ≥5 and ¯S3 = {z ∈C| |z–z| ≥8}. Then the number of elements in S1 ∩S2 ∩S3 is equal to (1) 1 (2) 0 (3) 2 (4) Infinite b ≠0, are equal, then the value of b is equal

Q62.Consider a rectangle ABCD having 5, 6, 7, 9 points in the interior of the line segments AB, BC, CD, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β −α) is equal to (1) 795 (2) 1173 (3) 1890 (4) 717

Q62.Let 𝑃1, 𝑃2 … , 𝑃15 be 15 points on a circle. The number of distinct triangles formed by points 𝑃𝑖, 𝑃𝑗, 𝑃𝑘 such that 𝑖+ 𝑗+ 𝑘≠15, is : (1) 455 (2) 419 (3) 12 (4) 443

Q62.Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3r2, then r2 −d is equal to : (1) 7 −√3 (2) 7 + 3√3 (3) 7 −7√3 (4) 7 + √3

Q62.Let a complex number z, |z| ≠1, satisfy log 1 |z|+11 ≤2 . Then, the largest value of |z| is equal to √2 ( (|z|−1)2 ) _________. (1) 8 (2) 7 (3) 6 (4) 5

Q62.The area of the triangle with vertices P(z), Q(iz) and R(z + iz) is (1) 1 (2) 12 z 2 (3) 1 (4) 1 z + iz 2 2 2

Q62.If 𝑏 is very small as compared to the value of 𝑎, so that the cube and other higher powers of 𝑏 can be neglected 𝑎 in the identity 1 1 1 1 … . + 𝛼𝑛+ 𝛽𝑛2 + 𝛾𝑛3 𝑎- 𝑏+ 𝑎- 2𝑏+ 𝑎- 3𝑏+ 𝑎- 𝑛𝑏= then the value of 𝛾 is : (1) 𝑎2 + 𝑏 (2) 𝑎+ 𝑏 3𝑎3 3𝑎2 (3) 𝑏2 (4) 𝑎+ 𝑏2 3𝑎3 3𝑎3

Q62.Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 −S6 is equal to: (1) 1862 (2) 1842 (3) 1852 (4) 1872

Q62.The sum of the series 1 + 2 + + … + 2100 when x = 2 is: x+1 x2+1 x4+1 x2100+1 (1) 1 − 2101 (2) 1 + 2101 4101−1 4101−1 (3) 1 + 2100 (4) 1 − 2100 4101−1 4201−1

Q62.The number of solutions of the equation 32tan2𝑥+ 32sec2𝑥= 81, 0 ≤𝑥≤ 𝜋 is : 4 (1) 0 (2) 2 (3) 1 (4) 3 JEE Main 2021 (31 Aug Shift 2) JEE Main Previous Year Paper 𝑧- 𝑖

Q62.A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is: (1) 1050 (2) 1625 (3) 575 (4) 560

Q62.The sum of all those terms which are rational numbers in the expansion of 1 1 12 3 + 3 4 (2 ) is: (1) 89 (2) 27 (3) 35 (4) 43 , then the

Q62.If n ⩾2 is a positive integer, then the sum of the series n+1C2 + 2(2C2 + 3C2 + 4C2 + … + nC2) is (1) n(n−1)(2n+1) (2) n(n+1)(2n+1) 6 6 (3) n(n+1)2(n+2) (4) n(2n+1)(3n+1) 12 6

Q62.If sum of the first 21 terms of the series log91/2 x + log91/3 x + log91/4 x + … . . where x > 0 is 504, then x is equal to (1) 243 (2) 9 (3) 7 (4) 81