Practice Questions

Q59.The angle of elevation of the summit of a mountain from a point on the ground is 45° . After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60° . Then the height (in km) of the summit from the ground is : (1) √3−1 (2) √3+1 √3+1 √3−1 (3) 1 (4) 1 √3−1 √3+1 π

Q59.The negation of the Boolean expression p ∨(~p ∧q) is equivalent to : (1) p ∧~q (2) ~p ∧~q (3) ~p ∨~q (4) ~p ∨q n n

Q59.If p →(p ∧~q) is false, then the truth values of p and q are respectively (1) F, F (2) T, F (3) T, T (4) F, T JEE Main 2020 (09 Jan Shift 2) JEE Main Previous Year Paper

Q59.The proposition p →~(p ∧~q) is equivalent to : (1) q (2) (~p) ∨q (3) (~p) ∧q (4) (~p) ∨(~q)

Q59.Let p, q, r be three statements such that the truth value of (p ∧q) →(~q ∨r) is F . Then the truth values of p, q, r are respectively : (1) T, T, F (2) T, T, T (3) T, F, T (4) F, T, F

Q59.Let the observation xi(1 ≤i ≤10) satisfy the equations ∑10i=1(xi −5) = 10 , ∑10i=1 (xi −5)2 = 40 . If μ and λ are the mean and the variance of the observations, x1 −3, x2 −3, . . . . , x10 −3, then the ordered pair (μ, λ) is equal to: (1) (3,3) (2) (6,3) (3) (6,6) (4) (3,6) Q60. ⎡1 1 2⎤ |adjB| If A = 1 3 4 , B = adjA and C = 3A, then is equal to ⎣1 −1 3⎦ |C| (1) 8 (2) 16 (3) 72 (4) 2

Q59. x(e(√1+x2+x4−1)/x−1) lim x→0 √1+x2+x4−1 (1) is equal to √e (2) is equal to 1 (3) is equal to 0 (4) does not exist

Q59.For some θ ∈(0, π2 ), if the eccentricity of the hyperbola, x2 −y2 sec2 θ = 10 is √5 times the eccentricity of the ellipse, x2 sec2 θ + y2 = 5, then the length of the latus rectum of the ellipse, is (1) 2√6 (2) √30 (3) 2√5 (4) 4√5 3 3

Q59.If A = (29 24 ) and I = (10 01 ), then 10 A−1 , is equal to. (1) A −4I (2) 6I −A (3) A −6I (4) 4I −A

Q59.If R = {(x, y) : x, y ∈Z, x2 + 3y2 ≤8} is a relation on the set of integers Z , then the domain of R−1 is (1) {−2, −1, 1, 2} (2) {0, 1} (3) {−2, −1, 0, 1, 2} (4) {−1, 0, 1}

Q59.The negation of the Boolean expression x ↔~y is equivalent to: (1) (~x ∧y) ∨(~x ∧~y) (2) (x ∧y) ∨(~x ∧~y) (3) (x ∧~y) ∨(~x ∧y) (4) (x ∧y) ∧(~x ∨~y)

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.Let A, B, C and D be four non-empty sets. The contrapositive statement of “If A ⊆B and B ⊆D , then A ⊆C ” is (1) If A ⊈C , then A ⊆B and B ⊆D (2) If A ⊆C , then B ⊂A and D ⊂B (3) If A ⊈C , then A ⊈B and B ⊆D (4) If A ⊈C , then A ⊈B or B ⊈D

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.If Σ −a) = n and Σ −a)2 = na, (n, a > 1), then the standard deviation of n observations i=1(xi i=1(xi x1, x2, … , xn is JEE Main 2020 (06 Sep Shift 1) JEE Main Previous Year Paper (1) a −1 (2) n√(a −1) (3) √n(a −1) (4) √(a −1)

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 xi(1 ≤i ≤10) be ten observation of a random variable X . If ∑10i=1(xi −p) = 3 and ∑10i=1 (xi −p)2 = 9 where 0 ≠p ∈R, then the standard deviation of these observations is: (1) 4 (2) 5 √35 (3) 9 (4) 7 10 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 system of linear equations λx + 2y + 2z = 5 2λx + 3y + 5z = 8 4x + λy + 6z = 10 has (1) no solution when λ = 8 (2) a unique solution when λ = −8 (3) no solution when λ = 2 (4) infinitely many solutions when λ = 2

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. 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

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 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.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