Practice Questions

Q90.Let Q be the foot of the perpendicular from the point P(7, −2, 13) on the plane containing the lines y−1 x+1 6 = 7 = z−38 and x−13 = y−25 = z−37 Then (PQ)2, is equal to ______. JEE Main 2021 (26 Aug Shift 2) JEE Main Previous Year Paper

Q90.Let there be three independent events E1, E2 and E3. The probability that only E1 occurs is α only E2 occurs is β and only E3 occurs is γ. Let ′p′ denote the probability of none of events occurs that satisfies the equations (α −2β)p = αβ and (β −3γ)p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1). Then, Probability of occurrence of E1 is equal to ________. Probability of occurrence of E3 JEE Main 2021 (17 Mar Shift 1) JEE Main Previous Year Paper

Q90.If Im,n = ∫10 xm−1(1 −x)n−1dx, for m, n ⩾1, and ∫10 xm−1+xn−1(1+x)m+n ________. JEE Main 2021 (26 Feb Shift 2) JEE Main Previous Year Paper

Q90.Let →a = ˆi + 5ˆj + αˆk, b = ˆi + 3ˆj + βˆk and →c= −ˆi + 2ˆj −3ˆk be three vectors such that, b ×→c = 5√3 and →a → 2 is ________. is perpendicular to b. Then the greatest amongst the values of →a JEE Main 2021 (27 Aug Shift 1) JEE Main Previous Year Paper

Q52.If z1, z2 are complex numbers such that Re (z1) = |z1 −1| and Re (z2) = |z2 −1| and arg(z1 −z2) = π6 , then Im(z1 + z2) is equal to : (1) 2√3 (2) √3 2 (3) 1 (4) 2 √3 √3

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.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.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 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

Q55.Let S be the sum of the first 9 term of the series : {x + ka} + {x2 + (k + 2)a} + {x3 + (k + 4)a} + {x4 + (k + 6)a} + … where a ≠0 and x ≠ 1 . If x10−x+45a(x−1) S = x−1 , then k is equal to (1) −5 (2) 1 (3) −3 (4) 3

Q55.Let α > 0, β > 0 be such that α3 + β2 = 4 . If the maximum value of the term independent of x in the 1 10 10k, then k is equal to binomial expansion of (αx 9 + βx−16 ) is (1) 336 (2) 352 (3) 84 (4) 176

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

Q56.For a > 0, let the curves C1 : y2 = ax and C2 : x2 = ay intersect at origin O and a point P. Let the line x = b(0 < b < a) intersect the chord OP and the x -axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, C1 and C2, and the area of ΔOQR = 21 , then ‘ a ’ satisfies the equation: (1) x6 −6x3 + 4 = 0 (2) x6 −12x3 + 4 = 0 (3) x6 + 6x3 −4 = 0 (4) x6 −12x3 −4 = 0

Q56.Let P be a point on the parabola, y2 = 12x and N be the foot of the perpendicular drawn from P , on the axis of the parabola. A line is now drawn through the mid-point M of PN , parallel to its axis which meets the parabola at Q . If the y−intercept of the line NQ is 43 , then : JEE Main 2020 (03 Sep Shift 1) JEE Main Previous Year Paper (1) PN = 4 (2) MQ = 13 (3) MQ = 14 (4) PN = 3

Q56.In the expansion of ( cosx θ + x sin1 θ )16, if l1 is the least value of the term independent of 8 ≤θ ≤π4 and l2 is the least value of the term independent of x when 16π ≤θ ≤π8 , then the ratio l2 : l1 is equal to: (1) 1 : 8 (2) 16 : 1 (3) 8 : 1 (4) 1 : 16

Q56.A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If ∠BAC = 90o,and ar (Δ ABC) = 5√5 sq. units, then the abscissa of the vertex C is : JEE Main 2020 (04 Sep Shift 1) JEE Main Previous Year Paper (1) 1 + √5 (2) 1 + 2√5 (3) 2 + √5 (4) 2√5 −1 y2

Q56.The centre of the circle passing through the point (0, 1) and touching the parabola y = x2 at the point (2, 4) is : JEE Main 2020 (06 Sep Shift 2) JEE Main Previous Year Paper (1) ( −5310 , 165 ) (2) ( 65 , 5310 ) (3) ( 103 , 165 ) (4) ( −165 , 5310 )

Q57.If the point P on the curve, 4x2 + 5y2 = 20 is farthest from the point Q(0, −4), then PQ2 is equal to (1) 36 (2) 48 (3) 21 (4) 29

Q57.Let the line y = mx and the ellipse 2x2 + y2 = 1 intersect at a point P in the first quadrant. If the normal to , (0, β), then β is equal to 0) and this ellipse at P meets the co-ordinate axes at (− 3√21 (1) 2√2 (2) 2 3 √3 (3) 2 (4) √2 3 3 JEE Main 2020 (08 Jan Shift 1) JEE Main Previous Year Paper Q58. 3x2+2 x21 lim is equal to x→0 ( 7x2+2 ) (1) 1 (2) 1 e e2 (3) e2 (4) e

Q58.If the line y = m x + c is a common tangent to the hyperbola 100x2 −y264 = 1 and the circle x2 + y2 = 36, then which one of the following is true? (1) c2 = 369 (2) 5m = 4 (3) 4c2 = 369 (4) 8m + 5 = 0

Q58.If α is the positive root of the equation, p(x) = x2 −x −2 = 0, then lim √1−cosx+α−4p(x) is equal to x→α+ (1) 23 (2) √23 (3) 1 (4) 12 √2

Q58.Let P(3, 3) be a point on the hyperbola, x2 −y2 = 1. If the normal to it at P intersects the x-axis at (9, 0) a2 b2 and e is its eccentricity, then the ordered pair (a2, e2) is equal to: (1) ( 29 , 3) (2) ( 32 , 2) (3) ( 29 , 2) (4) (9, 3)

Q58.The area (in sq. units) of an equilateral triangle inscribed in the parabola y2 = 8x, with one of its vertices on the vertex of this parabola is (1) 64√3 (2) 256√3 (3) 192√3 (4) 128√3 JEE Main 2020 (02 Sep Shift 2) JEE Main Previous Year Paper

Q60.Let A be a 2 × 2 real matrix with entries from {0, 1} and |A| ≠0 . Consider the following two statements; (P) If A ≠l2 , then |A| = −1 (Q) If |A| = 1 , then tr(A) = 2 Where l2 denotes 2 × 2 identity matrix and tr(A) denotes the sum of the diagonal entries of A . Then (1) (P) is false and (Q) is true (2) Both (P) and (Q) are false (3) (P) is true and (Q) is false (4) Both (P) and (Q) are true

Q60.If the system of linear equations 2x + 2ay + az = 0 2x + 3by + bz = 0 2x + 4cy + cz = 0, where a, b, c ∈R are non-zero and distinct; has a non-zero solution, then (1) a 1 , 1b , 1c are in A. P. (2) a, b, c are in G. P. (3) a + b + c = 0 (4) a, b, c are in A. P.