Practice Questions

Q74.The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is : (1) 3 (2) 5 (3) 7 (4) 1

Q74.Consider the statement: " P(n) : n2 −n + 41 is prime". Then which one of the following is true? (1) P(3) is false but P(5) is true (2) Both P(3) and P(5) are false (3) Both P(3) and P(5) are true (4) P(5) is false but P(3) is true

Q74.If the Boolean expression 𝑝⊕𝑞∧~𝑝⊙𝑞 is equivalent to 𝑝∧𝑞, where ⊕, ⊙∈∧, ∨, then the ordered pair ⊕, ⊙ is (1) ∨, ∧ (2) ∧, ∧ (3) ∨, ∨ (4) ∧, ∨ Q75.5 students of a class have an average height 150 𝑐𝑚 and variance 18 𝑐𝑚2 . A new student, whose height is 156 𝑐𝑚, joined them. The variance in 𝑐𝑚2 of the height of these six students is: (1) 22 (2) 16 (3) 18 (4) 20

Q74.If [10 11 ][10 21 ][10 31 ] [10 n −11 ] [10 781 ], [10 n1 ] (1) [10 –121 ] (2) [121 10 ] (3) [131 10 ] (4) [10 –131 ]

Q74.If for some x ∈R, the frequency distribution of the marks obtained by 20 students in a test is: Marks 2 3 5 7 Frequency distribution (x + 1)2 (2x −5) x2 −3x x JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper Then the mean of the marks is : (1) 3.0 (2) 2.5 (3) 3.2 (4) 2.8

Q74. lim cot3x−tanxπ is x→π4 cos(x+ 4 ) (1) 4√2 (2) 8√2 (3) 4 (4) 8

Q74. lim x+2sinx is x→0 √x2+2 sin x+1 − √sin2x−x+1 (1) 3 (2) 1 (3) 2 (4) 6

Q74.A student scores the following marks in five tests: 45,54,41,57,43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is: 10 100 (1) (2) 3 3 (3) 10 (4) 100 √3 √3

Q74.The negation of the Boolean expression ~𝑠∨~𝑟∧𝑠 is equivalent to (1) 𝑟 (2) 𝑠∧𝑟 (3) 𝑠∨𝑟 (4) ~𝑠∧~𝑟

Q74.Let [x] denote the greatest integer less than or equal to X . Then : limx→0 tan(π sin2 x)+(|x|−sin(x[x]))2x2 (1) does not exist (2) equals π (3) equals π + 1 (4) equals 0

Q74.The mean and variance for seven observations are 8 and 16 respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is (1) 48 (2) 45 (3) 49 (4) 40

Q75.The Boolean expression ((p ∧q) ∨(p ∨~q)) ∧(~p ∧~q) is equivalent to (1) p ∧(~q) (2) (~p) ∧(~q) (3) p ∨(~q) (4) p ∧q

Q75.Let A and B be two invertible matrices of order 3 × 3. If det (ABAT) (BA−1 BT) is equal to (1) 1 (2) 1 4 (3) 1 (4) 16 16

Q75.If the angle of elevation of a cloud from a point P which is 25m above a lake be 30o and the angle of depression of reflection of the could in the lake from P be 60o , then the height of the cloud (in meters) from JEE Main 2019 (12 Jan Shift 2) JEE Main Previous Year Paper the surface of the lake is : (1) 50 (2) 60 (3) 45 (4) 42 and B = {x ∈Z : −3 < 2x −1 < 9},

Q75.The number of values of θ ∈(0, π) for which the system of linear equations x + 3y + 7z = 0 −x + 4y + 7z = 0 (sin 3θ)x + (cos 2θ)y + 2z = 0 has a non-trivial solution, is: (1) Two (2) Three (3) Four (4) One

Q75.Two vertical poles of height, 20 𝑚 and 80 𝑚 stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is: (1) 16 (2) 12 (3) 18 (4) 15

Q75.Let 𝐴= cos𝛼-sin𝛼 𝑎∈𝑅 such that 𝐴32 = 0 -1 . Then, a value of 𝛼 is: sin𝛼 cos𝛼, 1 0 (1) 0 (2) 𝜋 (3) 𝜋 (4) 𝜋 16 64 32 JEE Main 2019 (08 Apr Shift 1) JEE Main Previous Year Paper

Q75.The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is (1) 10 : 3 (2) 4 : 9 (3) 6 : 7 (4) 5 : 8

Q75.If 𝐴 is a symmetric matrix and 𝐵 is skew- symmetric matrix such that 𝐴+ 𝐵= 2 3 , then 𝐴𝐵 is equal to: 5 -1 (1) -4 2 (2) 4 -2 1 4 1 -4 (3) 4 -2 (4) -4 -2 -1 -4 -1 4

Q75. y + 1 α β Let α and β be the roots of the equation x2 + x + 1 = 0. Then for y ≠0 in R, α y + β 1 is equal β 1 y + α to (1) y3 (2) y(y2–1) (3) y3–1 (4) y(y2–3)

Q75.If q is false and p ∧q ↔r is true, then which one of the following statements is a tautology? (1) (p ∨r) →(p ∧r) (2) (p ∧r) →(p ∨r) (3) p ∧r (4) p ∨r

Q75.The Boolean expression ∼(p ⇒(∼q)) is equivalent to (1) (∼p) ⇒q (2) q ⇒∼p (3) p ∨q (4) p ∧q

Q75. ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at, A and B are cot−1(3√2) and cosec−1(2√2) respectively, then the height of the tower (in metres) is (1) 100 (2) 20 3√3 (3) 25 (4) 10√5

Q75.If both the mean and the standard deviation of 50 observations 𝑥1, 𝑥2, … , 𝑥50 are equal to 16, then the mean of 𝑥1 - 42, 𝑥2 - 42, … , 𝑥50 - 42 is (1) 525 (2) 480 (3) 400 (4) 380

Q75.The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then xy is equal to: (1) 9 (2) 7 4 3 (3) 7 (4) 8 2 3