Practice Questions

Q53.If the number of five digit numbers with distinct digits and 2 at the 10th place is 336k , then k is equal to: (1) 4 (2) 6 (3) 7 (4) 8

Q53.Total number of 6− digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appears, is (1) 1 2 (6!) (2) 6! (3) 56 (4) 25 (6!)

Q53.The common difference of the A. P. b1, b2, . . . . , bm is 2 more than common difference of A. P. a1, a2, . . . . . , an . If a40 = −159, a100 = −399 and b100 = a70 , then b1 is equal to : (1) 81 (2) −127 (3) −81 (4) 127 is 405 , then |k| equals :

Q53.If the first term of an A. P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A. P.is (1) 1 (2) 1 6 5 (3) 1 (4) 1 4 7

Q53.Let u = z−ki2z+i , z = x + iy and k > 0. If the curve represented by Re (u)+ Im (u) = 1 intersects the y-axis at points P and Q where PQ = 5 then the value of k is (1) 3 (2) 1 2 2 (3) 4 (4) 2

Q54.The product 2 41 ∙4 161 ∙8 481 ∙16 1 128 ∙. . . . to ∞ is equal to: (1) 2 21 (2) 2 41 (3) 1 (4) 2

Q54.If the constant term in the binomial expansion of (√x − x2k ) 10 (1) 9 (2) 1 (3) 3 (4) 2

Q54.Let two points be A(1, −1) and B(0, 2). If a point P(x', y') be such that the area of ΔPAB = 5 sq. units and it lies on the line 3x + y −4λ = 0, then a value of λ is (1) 4 (2) 3 (3) 1 (4) −3

Q54.If α and β, be the coefficients of x4 and x2 , respectively in the expansion of 6 6 + √x2 + −√x2 (x −1) (x −1) , then (1) α + β = 60 (2) α + β = −30 (3) α −β = 60 (4) α −β = −132

Q54.If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n+5 are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is : (1) 462 (2) 330 (3) 792 (4) 252

Q54.Let an be the nth term of a G.P. of positive terms. If ∑100n=1 a2n+1 = 200 and ∑100n=1 a2n = 100, then ∑200n=1 an is equal to: (1) 300 (2) 225 (3) 175 (4) 150

Q54.If 1 + (1 −22 ⋅1) + (1 −42 ⋅3) + (1 −62 ⋅5) + … … + (1 −202 ⋅19) = α −220 β , then an ordered pair (α, β) is equal to: (1) (10, 97) (2) (11, 103) (3) (10, 103) (4) (11, 97)

Q54.Let a, b, c, d and p be non-zero distinct real numbers such that (a2 + b2 + c2)p2 −2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then (1) a, b, c are in A.P. (2) a, c, p are in G.P. (3) a, b, c, d are in G.P. (4) a, b, c, d are in A.P. is equal to

Q54.If the sum of first 11 terms of an A.P. , a1, a2, a3 … … is 0(a1 ≠0) then the sum of the A.P a1, a3, a5, … . . a23 is ka1 where k is equal to (1) −12110 (2) 12110 (3) 725 (4) −725

Q54.If the term independent of x in the expansion of ( 23 x2 − 3x1 )9 is (1) 11 (2) 5 (3) 9 (4) 7

Q54.If |x| < 1, |y| < 1 and x ≠1 , then the sum to infinity of the following series (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3)+. . . . . is (1) x+y−xy (2) x+y+xy (1+x)(1+y) (1+x)(1+y) (3) x+y−xy (4) x+y+xy (1−x)(1−y) (1−x)(1−y)

Q54.If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the first 50 terms of this G.P. is : (1) 26 1 (349 −1) (2) 261 (350 −1) (3) 13 2 (350 −1) (4) 131 (350 −1)

Q54.Five numbers are in A. P. , whose sum is 25 and product is 2520. If one of these five numbers is −12 , then the greatest number amongst them is (1) 27 (2) 7 (3) 21 (4) 16 2

Q54.If 32 sin 2α−1, 14 and 34−2 sin 2α are the first three terms of an A.P. for some α , then the sixth term of this A.P. is (1) 66 (2) 81 (3) 65 (4) 78

Q54.The value of ( 2 ⋅1 P0 −3 ⋅2 P1 + 4 ⋅3 P2−. . . . . . . . up to 51th term) +( 1! −2! + 3!−. . . . . . . up to 51th term) is equal to (1) 1 −51(51)! (2) 1 + (51)! (3) 1 + (52)! (4) 1 1 1 n 2 + 5 8 is exactly 33, then the least value of n is

Q54.If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19+. . . . is (102)m, then m is equal to (1) 20 (2) 25 (3) 5 (4) 10

Q55.The value of cos3( π8 ). cos( 3π8 ) + sin3( π8 ). sin( 3π8 ) is: (1) 1 (2) 1 √2 2√2 (3) 1 (4) 1 2 4

Q55.Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (−1, −4) in the line is : (1) ( 115 , 285 ) (2) ( 295 , 85 ) (3) ( 85 , 295 ) (4) ( 295 , 115 )

Q55.The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + … + 492 + 49 + 1, is (1) 32 (2) 63 (3) 60 (4) 35

Q55.If a line y = mx + c, is a tangent to the circle (x −3)2 + y2 = 1 , and it is perpendicular to a line L1, where , 1 ), then L1 is the tangent to the circle x2 + y2 = 1 , at the point ( √21 √2 (1) c2 −7c + 6 = 0 (2) c2 + 7c + 6 = 0 (3) c2 + 6c + 7 = 0 (4) c2 −6c + 7 = 0