Practice Questions

Q69.The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y where x < y are 6 and 49 respectively. Then x4 + y2 is equal to (1) 320 (2) 420 (3) 162 (4) 674

Q69.The function f : R →R defined by f(x) = lim cos(2πx)−x2n sin(x−1) is continuous for all x in n→∞ 1+x2n+1−x2n (1) R −{−1} (2) R −{−1, 1} (3) R −{1} (4) R −{0} Q70. π 1+( dxdy ) 2 π Let x(t) = 2√2 cos t√sin 2t and y(t) = 2√2 sin t√sin 2t, t ∈(0, 2 ). Then d2y at t = 4 is equal to dx2 (1) −2√2 (2) 2 3 3 (3) 1 (4) −2 3 3

Q69.The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 45° . Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is 60° . If ∠BAQ = 30°, AB = d and the area of the trapezium PQRB is α, then the ordered pair (d, α) is (1) (10(√3 −1), 25) (2) (10(√3 −1), 252 ) + + (3) (10(√3 1), 25) (4) (10(√3 1), 252 ) . If A2 + γA + 18I = O, then det (A) is equal to _______.

Q69.Let R be a relation from the set {1, 2, 3 … … … , 60} to itself such that R ={ (a, b) : b = pq , where p, q ≥3 are prime numbers}. Then, the number of elements in R is (1) 600 (2) 660 (3) 540 (4) 720

Q69.Negation of the Boolean expression 𝑝↔𝑞→𝑝 is (1) ~𝑝∧𝑞 (2) 𝑝∧~𝑞 (3) ~𝑝∨~q (4) ~𝑝∧~𝑞 Q70. 1 92 -102 112 Let 𝐴= 1 and 𝐵= 122 132 -142 , then the value of 𝐴'𝐵𝐴 is; 1 -152 162 172 (1) 1224 (2) 1042 (3) 540 (4) 539

Q69. tan(2 tan−1 51 + sec−1 √52 + 2 tan−1 18 ) is equal to: (1) 1 (2) 2 (3) 1 (4) 5 4 4

Q69.Let A be a 3 × 3 invertible matrix. If |adj(24A)| =adj (3 adj (2A))|, then |A|2 is equal to (1) 26 (2) 212 (3) 512 (4) 66

Q69. sin−1(sin 2π3 ) + cos−1(cos 7π6 ) + tan−1(tan 3π4 ) is equal to JEE Main 2022 (27 Jun Shift 1) JEE Main Previous Year Paper (1) 11π (2) 17π 12 12 (3) 31π 12 (4) −3π4

Q69.The number of 𝜃∈0, 4𝜋 for which the system of linear equations 3sin3𝜃𝑥- 𝑦+ 𝑧= 2 3cos2𝜃𝑥+ 4𝑦+ 3𝑧= 3 6𝑥+ 7𝑦+ 7𝑧= 9 has no solution is (1) 6 (2) 7 (3) 8 (4) 9

Q69.Let a vertical tower AB of height 2h stands on a horizontal ground. Let from a point P on the ground a man can see upto height h of the tower with an angle of elevation 2α. When from P , he moves a distance d in the −→ direction of AP , he can see the top B of the tower with an angle of elevation α. If d = √7h , then tan α is equal to (1) √5 −2 (2) √3 −1 (3) √7 −2 (4) √7 −√3

Q70.Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then (1) The system of linear equations AX = 0 has a (2) The system of linear equations AX = 0 has unique solution infinitely many solutions (3) B is an invertible matrix (4) adj(A) is an invertible matrix

Q70.The value of n→∞6lim tan{∑nr=1 tan−1( r2+3r+31 )} is equal to (1) 1 (2) 2 (3) 3 (4) 6

Q70.Let A and B be two 3 × 3 matrices such that AB = I and |A| = 18 then |adj(Badj(2A))| is equal to (1) 128 (2) 32 (3) 64 (4) 102

Q70.The ordered pair (a, b), for which the system of linear equations 3x −2y + z = b 5x −8y + 9z = 3 2x + y + az = −1 has no solution, is (1) (3, 13 ) (2) (−3, 31 ) (3) (−3, −13 ) (4) (3, −13 )

Q70.If the system of equations 𝑥+ 𝑦+ 𝑧= 6 2𝑥+ 5𝑦+ 𝛼𝑧= 𝛽 𝑥+ 2𝑦+ 3𝑧= 14 has infinitely many solutions, then 𝛼+ 𝛽 is equal to (1) 8 (2) 36 (3) 44 (4) 48

Q70.The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to (1) 133 (2) 19 104 103 (3) 18 (4) 271 103 104

Q70.If the inverse trigonometric functions take principal values, then cos−1( 103 cos(tan−1( 43 )) + 25 sin(tan−1( 43 ))) is equal to (1) 0 (2) π4 (3) π (4) π 3 6

Q70.The number of values of α for which the system of equations x + y + z = α αx + 2αy + 3z = −1 x + 3αy + 5z = 4 is inconsistent, is (1) 0 (2) 1 (3) 2 (4) 3

Q70.The negation of the Boolean expression ~𝑞∧𝑝⇒~𝑝∨𝑞 is logically equivalent to (1) 𝑝⇒𝑞 (2) 𝑞⇒𝑝 (3) ~𝑝⇒𝑞 (4) ~𝑞⇒𝑝

Q70.Let A = (α4 −2β ) (1) −18 (2) 18 (3) −50 (4) 50 1 [t] is the greatest

Q70.The total number of functions, 𝑓: 1, 2, 3, 4 →1, 2, 3, 4, 5, 6 such that 𝑓1 + 𝑓2 = 𝑓3, is equal to (1) 60 (2) 90 (3) 108 (4) 126

Q70.The number of values of a ∈N such that the variance of 3, 7, 12, a, 43 −a is a natural number is: (1) 0 (2) 2 (3) 5 (4) infinite

Q70.If cos−1( 2y ) = loge ( x5 ) 5, |y| < 2, then (1) x2y′′ + xy′ −25y = 0 (2) x2y′′ −xy′ −25y = 0 (3) x2y′′ −xy′ + 25y = 0 (4) x2y′′ + xy′ + 25y = 0

Q70.Consider the following statements: P : Ramu is intelligent. Q : Ramu is rich. R : Ramu is not honest. The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as: (1) ((P ∧(~R)) ∧Q) ∧((~Q) ∧((~P) ∨R)) (2) ((P ∧R) ∧Q) ∨((~Q) ∧((~P) ∨(~R))) (3) ((P ∧R) ∧Q) ∧((~Q) ∧((~P) ∨(~R))) (4) ((P ∧(~R)) ∧Q) ∨((~Q) ∧((~P) ∧R))

Q70.Let f : R →R be a continuous function such that f(3x) −f(x) = x. If f(8) = 7 , then f(14) is equal to: (1) 4 (2) 10 (3) 11 (4) 16