Practice Questions

Q61.The sum of the roots of the equation, x2 + |2x −3| −4 = 0, is: (1) 2 (2) −2 (3) √2 (4) −√2

Q61.If a ∈ R and the equation −3(x − [x])2 + 2(x − [x]) + a2 = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval (1) (−2, −1) (2) ( −∞, −2) ∪(2,∞) (3) (−1, 0) ∪(0, 1) (4) (1, 2)

Q61.The equation √3x2 + x + 5 = x −3, where x is real, has (1) no solution (2) exactly four solutions (3) exactly one solution (4) exactly two solutions JEE Main 2014 (19 Apr Online) JEE Main Previous Year Paper

Q62.If z1, z2 and z3, z4 are 2 pairs of complex conjugate numbers, then arg ( z1z4 ) ( z2z3 ) equals: (1) 0 (2) π 2 (3) 3π (4) π 2 JEE Main 2014 (11 Apr Online) JEE Main Previous Year Paper

Q62.Let z ≠−i be any complex number such that z+iz−i is a purely imaginary number. Then z + 1z is: (1) 0 (2) any non-zero real number other than 1 . (3) any non-zero real number. (4) a purely imaginary number. Q63.8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do no occupy odd places, is: (1) 160 (2) 120 (3) 60 (4) 48

Q62.For all complex numbers z of the form 1 + iα, α ∈R, if z2 = x + iy, then (1) y2 −4x + 4 = 0 (2) y2 + 4x −4 = 0 (3) y2 −4x + 2 = 0 (4) y2 + 4x + 2 = 0

Q62.If equations ax2 + bx + c = 0, (a, b, c ∈R, a ≠0) and 2x2 + 3x + 4 = 0 have a common root, then a : b : c equals : (1) 2 : 3 : 4 (2) 4 : 3 : 2 (3) 1 : 2 : 3 (4) 3 : 2 : 1

Q62.Let α and β be the roots of equation px2 + qx + r = 0, p ≠0. If p, q, r are in A.P. and α1 + β1 = 4, then the value of |α −β| is (1) √34 (2) 2√13 9 9 (3) √61 (4) 2√17 9 9

Q63.If z is a complex number such that |z| ≥2, then the minimum value of z + 12 : (1) Is strictly greater than 5 (2) Is strictly greater than 3 but less than 5 2 2 2 (3) Is equal to 5 (4) Lies in the interval (1, 2) 2

Q63.Let w(Im w≠0) be a complex number. Then, the set of all complex numbers z satisfying the equation ¯w −wz = k(1 −z), for some real number k, is (1) {z : z ≠1} (2) {z : |z| = 1, z ≠1} ¯(3) {z : z = z} (4) {z : |z| = 1}

Q63.Two women and some men participated in a chess tournament in which every participant played two games with each of the other participants. If the number of games that the men played between them-selves exceeds the number of games that the men played with the women by 66 , then the number of men who participated in the tournament lies in the interval (1) (11, 13] (2) (14, 17) (3) [10, 12) (4) [8, 9]

Q63.An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is: (1) 72(7!) (2) 18(7!) (3) 40(7!) (4) 36(7!)

Q64.In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49 , and the sum of the first and the third term is 35. Then the first term of this geometric progression is: (1) 7 (2) 21 (3) 28 (4) 42

Q64.If (10)9 + 2(11)1(10)8 + 3(11)2(10)7 + ...... + 10(11)9 = k(10)9, then k is equal to : JEE Main 2014 (06 Apr) JEE Main Previous Year Paper (1) 100 (2) 110 (3) 121 (4) 441 10 100

Q64.Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of 1a and 1b . If 1 : G is 4 : 5, then a : b can be: M JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper (1) 1 : 4 (2) 1 : 2 (3) 2 : 3 (4) 3 : 4

Q64.Let f(n) = [ 13 + 1003n ]n, where [n] denotes the greatest integer less than or equal to n. Then ∑56n=1 f(n) is equal to (1) 56 (2) 1287 (3) 1399 (4) 689

Q64.The sum of the digits in the unit's place of all the 4 - digit numbers formed by using the numbers 3, 4, 5 and 6 , without repetition is : (1) 18 (2) 36 (3) 108 (4) 432

Q65.Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12 , then its 4th term is : (1) 8 (2) 24 (3) 20 (4) 16

Q65.The sum of the first 20 terms common between the series 3 + 7 + 11 + 15+ and 1 + 6 + 11+ 16 + … .. is (1) 4000 (2) 4020 (3) 4200 (4) 4220

Q65.The least positive integer n such that 1 −23 − 322 −… . − 3n−12 < 1001 , is: (1) 4 (2) 5 (3) 6 (4) 7

Q65.The number of terms in an A. P. is even, the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by 10 12 , then the number of terms in the A. P. is (1) 4 (2) 8 (3) 16 (4) 12

Q65.Three positive numbers form an increasing G. P. If the middle term in this G. P. is doubled, the new numbers are in A. P. Then the common ratio of the G. P. is : (1) 2 −√3 (2) 2 + √3 (3) √2 + √3 (4) 3 + √2

Q66.If 1 + x4 + x5 = ∑5i=0 ai (1 + xi), for all x in R, then a2 is: (1) −4 (2) 6 (3) −8 (4) 10 is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive

Q66.If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2)(1 −2x)18 in powers of x are both zero, then (a, b) is equal to (1) (14, 2723 ) (2) (16, 2723 ) (3) (16, 2513 ) (4) (14, 2513 )

Q66.If the sum 3 + 5 + 7 + .... .+ up to 20 terms is equal to 21k , then k is equal to 12 12+22 12+22+32 (1) 240 (2) 120 (3) 60 (4) 180