Practice Questions

Q62.If a complex number z statisfies the equation x + √2|z + 1| + i = 0 , then |z| is equal to : (1) 2 (2) √3 (3) √5 (4) 1

Q62.Let z satisfy |z| = 1 and z = 1 −¯z. Statement 1 : z is a real number. Statement 2 : Principal argument of z is π3 (1) Statement 1 is true Statement 2 is true; Statement (2) Statement 1 is false; Statement 2 is true 2 is a correct explanation for Statement 1 . (3) Statement 1 is true, Statement 2 is false. (4) Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 . JEE Main 2013 (25 Apr Online) JEE Main Previous Year Paper Q63.5 - digit numbers are to be formed using 2, 3, 5, 7 , 9 without repeating the digits. If p be the number of such numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000 , then p : q is: (1) 6 : 5 (2) 3 : 2 (3) 4 : 3 (4) 5 : 3

Q63.The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is : (1) 30C7 (2) 21C8 (3) 21C7 (4) 30C8

Q63.The sum of the series : (2)2 + 2(4)2 + 3(6)2 + … upto 10 terms is : (1) 11300 (2) 11200 (3) 12100 (4) 12300

Q63.If z is a complex number of unit modulus and argument θ, then arg ( 1+1+z−z ) can be equal to (1) θ (2) π −θ (3) −θ (4) π2 −θ

Q63.A committee of 4 persons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes at least 1 lady, at least 1 old man and at most 2 young men. Then the total number of ways in which this committee can be formed is : (1) 40 (2) 41 (3) 16 (4) 32 a1+a2+…+ap p3 a6 is equal to: = ; p ≠q . Then

Q64.Let a1, a2, a3, … be an A.P, such that q3 a1+a2+a3+…+aq a21 (1) 41 (2) 31 11 121 (3) 11 (4) 121 41 1861

Q64.Given sum of the first n terms of an A.P. is 2n+ 3n2 . Another A.P. is formed with the same first term and double of the common difference, the sum of n terms of the new A.P. is : JEE Main 2013 (22 Apr Online) JEE Main Previous Year Paper (1) n + 4n2 (2) 6n2 −n (3) n2 + 4n (4) 3n + 2n2

Q64.If a1, a2, a3, … , an, … . are in A.P. such that a4 −a7 +a10 = m, then the sum of first 13 terms of this A.P., is : (1) 10 m (2) 12 m (3) 13 m (4) 15 m

Q64.Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 −Tn = 10 , then the value of n is : (1) 10 (2) 8 (3) 5 (4) 7

Q64.Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is : (1) 16 (2) 8 (3) 4 (4) 2

Q65. (1) 2925 (2) 1469 (3) 1728 (4) 1456

Q65.If x, y, z are positive numbers in A. P. and tan−1 x, tan−1 y and tan−1 z are also in A. P., then which of the following is correct. (1) 6x = 3y = 2z (2) 6x = 4y = 3z (3) x = y = z (4) 2x = 3y = 6z

Q65.The sum of the rational terms in the binomial expansion of 1 1 10 is : 2 + 3 5 ) (2 (1) 25 (2) 32 (3) 9 (4) 41

Q65.The sum 3 + 5 + 7 + … . upto 11-terms is: 12 12+22 12+22+32 (1) 7 (2) 11 2 4 (3) 11 (4) 60 2 11

Q65.The sum of the series: 1 + 1+21 + 1+2+31 + … …. upto 10 terms, is: (1) 18 (2) 22 11 13 (3) 20 (4) 16 11 9 2 15

Q66.The number of solutions of the equation sin 2x −2 cos x + 4 sin x = 4 in the interval [0, 5π] is : (1) 3 (2) 5 (3) 4 (4) 6

Q66.The ratio of the coefficient of x15 to the term independent of x in the expansion of (x2 + x ) is: (1) 7 : 16 (2) 7 : 64 (3) 1 : 4 (4) 1 : 32

Q66.If the 7th term in the binomial expansion of 9 , x > 0 , is equal to 729 , then x can be: + √3 ln x) ( 3√843 (1) e2 (2) e (3) e (4) 2e 2

Q66.If for positive integers r > 1, n > 2, the coefficients of the (3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n are equal, then n is equal to: (1) 2r + 1 (2) 2r −1 (3) 3r (4) r + 1

Q66.The sum of first 20 terms of the sequence 0. 7, 0. 77, 0. 777, . . . . . . , is : (1) 81 7 (179 + 10−20) (2) 97 (99 + 10−20) (3) 81 7 (179 −10−20) (4) 97 (99 −10−20)

Q67.If two lines L1 and L2 in space, are defined by L1 = {x = √λy + (√λ −1), z = (√λ −1)y + √λ} and L2 = {x = √μy + (1 −√μ), z = (1 −√μ)y + √μ} then L1 is perpendicular to L2 , for all nonnegative reals λ and μ, such that : (1) √λ + √μ = 1 (2) λ ≠μ (3) λ + μ = 0 (4) λ = μ

Q67.The term independent of x in the expansion of 10 ( x2/3−x1/3+1x+1 − x−x1/2x−1 ) is (1) 210 (2) 310 (3) 4 (4) 120

Q67.A value of x for which sin (cot−1(1 + x)) = cos (tan−1 x), is : (1) −12 (2) 1 (3) 0 (4) 1 2

Q67.The number of solutions of the equation, sin−1 x = 2 tan−1 x (in principal values) is : (1) 1 (2) 4 (3) 2 (4) 3