Practice Questions

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 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.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

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

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

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 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.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 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.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 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.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

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.Let A = {θ : sin(θ) = tan(θ)} and B = (θ : cos(θ) = 1\} be two sets. Then: (1) A = B (2) A ⊄ B (3) B ⊄ A (4) A ⊂B and B −A ≠ϕ

Q68.The expression 1−cotA tanA + 1−tanAcotA can be written as : (1) tanA + cotA (2) secA + cosecA (3) sinAcosA + 1 (4) secAcosecA + 1

Q68.If the image of point P(2, 3) in a line L is Q(4, 5), then the image of point R(0, 0) in the same line is: (1) (2, 2) (2) (4, 5) (3) (3, 4) (4) (7, 7)

Q68.A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is: (1) 1 (2) 3 (3) 7 (4) 5 2 2

Q69.If the x-intercept of some line L is double as that of the line, 3x + 4y = 12 and the y-intercept of L is half as that of the same line, then the slope of L is : (1) −3 (2) −3/8 (3) −3/2 (4) −3/16

Q69.If the three lines x −3y = p, ax + 2y = q and ax + y = r form a right-angled triangle then : (1) a2 −9a + 18 = 0 (2) a2 −6a −12 = 0 (3) a2 −6a −18 = 0 (4) a2 −9a + 12 = 0

Q69.A ray of light along x + √3y = √3 gets reflected upon reaching X−axis, the equation of the reflected ray is (1) y = √3x −√3 (2) √3y = x −1 (3) y = x + √3 (4) √3y = x −√3