Practice Questions

Q60.Let θ = and A = . If B = A + A4 , then det (B) : 5 [−sinθcosθ cosθsinθ ] (1) is one (2) lies in (2, 3) (3) is zero (4) lies in (1, 2)

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.The statement (p →(q →p)) →(p →(p ∨q)) is : (1) equivalent to (p ∧q) ∨(~q) (2) a contradiction (3) equivalent to (p ∨q) ∧(~p) (4) a tautology

Q60.The mean and variance of 8 observations are 10 and 13. 5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is : (1) 9 (2) 5 (3) 3 (4) 7

Q60.The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p ≠0 and q ≠0. If the new mean and new s.d. become half of their original values, then q is equal to (1) −5 (2) 10 (3) −20 (4) −10

Q60.Let 50∪ = ∪n = T , where each Xi contains 10 elements and each Yi contains 5 elements. If each element i=1Xi i=1Yi of the set T is an element of exactly 20 of sets Xi 's and exactly 6 of sets Yi 's then n is equal to : (1) 15 (2) 50 (3) 45 (4) 30

Q60.The following system of linear equations 7x + 6y −2z = 0 3x + 4y + 2z = 0 x −2y −6z = 0, has (1) infinitely many solutions, (x, y, z) satisfying (2) no solution y = 2z (3) infinitely many solutions, (x, y, z) satisfying (4) only the trivial solution x = 2z

Q60.The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14 then the absolute difference of the remaining two observations is : (1) 1 (2) 4 (3) 2 (4) 3 JEE Main 2020 (05 Sep Shift 1) JEE Main Previous Year Paper

Q60. lim (tan( π4 + x))1/x is equal to x→0 (1) e (2) 2 (3) 1 (4) e2

Q60.For the frequency distribution: Variate (x) : x1, x2, x3, … , x15 Frequency (f) : f1, f2, f3, … , f15 where 0 < x1 < x2 < x3 < … < x15 = 10 and ∑15i=1 fi > 0, the standard deviation cannot be (1) 4 (2) 1 (3) 6 (4) 2

Q61.For a suitably chosen real constant a, let a function, f : R −{−a} →R be defined by f(x) = a+xa−x . Further supposed that for any real number x ≠−a,and f(x) ≠−a, (fof)(x) = x. Then f(−12 ) is equal to : (1) 3 1 (2) −13 (3) −3 (4) 3

Q61.Let A = [aij] and B = [bij] be two 3 × 3 real matrices such that bij = (3)(i+j−2)aij , where i, j = 1,2, 3 . If the determinant of B is 81 , then determinant of A is (1) 1 (2) 3 3 (3) 1 (4) 1 81 9

Q61.Let R1 and R2 be two relations defined as follows : R1 = {(a, b) ∈R2 : a2 + b2 ∈Q} and R2 = {(a, b) ∈R2 : a2 + b2 ∉Q} , where Q is the set of all rational numbers, then (1) R1 is transitive but R2 is not transitive. (2) R2 is transitive but R1 is not transitive. (3) Neither R1 nor R2 is transitive. (4) R1 and R2 are both transitive. Q62. ⎡ 2 −1 1 ⎤ Let A be a 3 × 3 matrix such that adj A = −1 0 2 and B =adj (adjA). If |A| = λ and ⎣ 1 −2 −1 ⎦ (B−1) ⊤= μ, then the ordered pair (|λ|, μ) is equal to (1) (3, 811 ) (2) (9, 91 ) (3) (3, 81) (4) (9, 811 )

Q61.If g(x) = x2 + x −1 and (gof)(x) = 4x2 −10x + 5, then f( 54 ) is equal to (1) 3 2 (2) −12 (3) 2 1 (4) −32 tanα+cotα 1 3π dy 5π + sin2α , α ∈( 4 , π), then dα at α = 6 is 1+tan2α )

Q61.If the mean and the standard deviation of the data 3, 5, 7, a, b are 5and 2 respectively, then a and b are the roots of the equation: (1) x2 −10x + 18 = 0 (2) 2x2 −20x + 19 = 0 (3) x2 −10x + 19 = 0 (4) x2 −20x + 18 = 0

Q61. cos2 x 1 + sin2 x sin 2x Let m and M be respectively the minimum and maximum value values of 1 + cos2 x sin2 x sin 2x cos2 x sin2 x 1 + sin 2x Then the ordered pair (m, M) is equal to: (1) (3, 3) (2) (−3, −1) (3) (4, 1) (4) (1, 3)

Q61.A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be: (1) 63 (2) 36 (3) 54 (4) 38

Q61.Let f : (1, 3) →R, be a function defined by f(x) = x[x] , where [x], denotes the greatest integer ≤x. Then 1+x2 the range of f , is (1) ( 25 , 35 ] ∪( 34 , 45 ) (2) ( 25 , 12 ) ∪( 35 , 45 ] (3) ( 25 , 45 ] (4) ( 35 , 45 )

Q61.Which of the following is a tautology? (1) (~p) ∧(p ∨q) →q (2) (q →p) ∨~(p →q) (3) (~q) ∨(p ∧q) →q (4) (p →q) ∧(q →p) Q62. ⎡ 1 2 1 ⎤ Let A = where P = −2 3 −4 then the set A {X = (x, y, z)T : PX = 0 and x2 + y2 + z2 = 1} ⎣ 1 9 −1 ⎦ (1) Is a singleton. (2) Is an empty set. (3) Contains more than two elements (4) Contains exactly two elements Q63. ⎡a b c ⎤ Let a, b, c ∈R be all non-zero and satisfies a3 + b3 + c3 = 2. If the matrix A = b c a satisfies ⎣ c a b ⎦ ATA = I, then a value of abc can be (1) −13 (2) 13 (3) 3 (4) 23

Q61.Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m ) above the line AC is : (1) 20/3 (2) 5 (3) 10/3 (4) 6

Q61.For which of the following ordered pairs (μ, δ), the system of linear equations x + 2y + 3z = 1 3x + 4y + 5z = μ 4x + 4y + 4z = δ is inconsistent? (1) (4, 3) (2) (4, 6) (3) (1, 0) (4) (3, 4)

Q61.Let a −2b + c = 1. x + a x + 2 x + 1 If f(x) = x + b x + 3 x + 2 , then: x + c x + 4 x + 3 (1) f(−50) = 501 (2) f(−50) = −1 (3) f(50) = −501 (4) f(50) = 1 4 ] = A. Then the function, f(x) = [x2] sin(πx) is x

Q61.Let S be the set of all λ ∈R for which the system of linear equations 2x −y + 2z = 2 x −2y + λz = −4 x + λy + z = 4 has no solution. Then the set S (1) Contains more than two elements (2) Is an empty set (3) Is a singleton (4) Contains exactly two elements

Q61. x −2 2x −3 3x −4 If Δ = 2x −3 3x −4 4x −5 = Ax3 + Bx2 + Cx + D , then B + C is equal to : 3x −5 5x −8 10x −17 (1) −1 (2) 1 (3) −3 (4) 9 Q62. 2π −(sin−1 45 + sin−1 135 + sin−1 1665 ) is equal to : (1) π (2) 5π 2 4 (3) 3π (4) 7π 2 4

Q61.If the system of equations x + y + z = 2 2 x + 4 y −z = 6 3x + 2y + λz = μ has infinitely many solutions, then : (1) λ + 2μ = 14 (2) 2λ −μ = 5 (3) λ −2μ = −5 (4) 2λ + μ = 14