Practice Questions

Q68.Consider the following system of equations: x + 2y −3z = a 2x + 6y −11z = b x −2y + 7z = c where a, b and c are real constants. Then the system of equations : (1) has a unique solution when 5a = 2b + c (2) has no solution for all a, b and c (3) has infinite number of solutions when (4) has a unique solution for all a, b and c 5a = 2b + c

Q68.Let the equation of the pair of lines, y = px and y = qx, can be written as (y −px)(y −qx) = 0. Then the equation of the pair of the angle bisectors of the lines x2 −4xy −5y2 = 0 is: (1) x2 −3xy + y2 = 0 (2) x2 + 4xy −y2 = 0 (3) x2 + 3xy −y2 = 0 (4) x2 −3xy −y2 = 0

Q68.If in a triangle ABC, AB = 5 units, ∠B = cos−1( 53 ) and radius of circumcircle of ΔABC is 5 units, then the area (in sq. units) of ΔABC is: (1) 10 + 6√2 (2) 8 + 2√2 (3) 6 + 8√3 (4) 4 + 2√3 a ∈R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix.

Q68.A ray of light through (2, 1) is reflected at a point P on the y− axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity 1 and the distance of the nearer focus from this 3 directrix is 8 , then the equation of the other directrix can be: √53 JEE Main 2021 (27 Jul Shift 1) JEE Main Previous Year Paper (1) 11x + 7y + 8 = 0 or 11x + 7y −15 = 0 (2) 11x −7y −8 = 0 or 11x + 7y + 15 = 0 (3) 2x −7y + 29 = 0 or 2x −7y −7 = 0 (4) 2x −7y −39 = 0 or 2x −7y −7 = 0 x2f(2)−4f(x) is equal to:

Q68.The value of lim [r]+[2r]+...+[nr] , where r is non-zero real number and [r] denotes the greatest integer less than n→∞ n2 or equal to r, is equal to : (1) r (2) r 2 (3) 2r (4) 0

Q69.Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is: (1) 12 (2) 4 (3) 1 (4) 6

Q69.Consider the system of linear equations -𝑥+ 𝑦+ 2𝑧= 0 3𝑥- 𝑎𝑦+ 5𝑧= 1 2𝑥- 2𝑦- 𝑎𝑧= 7 Let 𝑆1 be the set of all 𝑎∈𝑅 for which the system is inconsistent and 𝑆2 be the set of all 𝑎∈𝑅 for which the system has infinitely many solutions. If nS1 and nS2 denote the number of elements in S1 and S2 respectively, then (1) nS1 = 2, nS2 = 0 (2) nS1 = 2, nS2 = 2 (3) nS1 = 0, nS2 = 2 (4) nS1 = 1, nS2 = 0

Q69.Let in a series of 2n observations, half of them are equal to a and remaining half are equal to −a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of a2 + b2 is equal to : (1) 425 (2) 650 (3) 250 (4) 925

Q69.A possible value of tan( 41 sin−1 √638 ) (1) 2√2 −1 (2) 1 2√2 (3) √7 −1 (4) 1 √7

Q69.A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is : JEE Main 2021 (31 Aug Shift 1) JEE Main Previous Year Paper (1) 8√10 (2) 6√10 (3) 12√10 (4) 12√15

Q69.The system of linear equations 3𝑥- 2𝑦- 𝑘𝑧= 10 2𝑥- 4𝑦- 2𝑧= 6 𝑥+ 2𝑦- 𝑧= 5 𝑚 is inconsistent if : 4 4 (1) 𝑘= 3, 𝑚≠ (2) 𝑘= 3, 𝑚= 5 5 (3) 𝑘≠3, 𝑚∈𝑅 (4) 𝑘≠3, 𝑚≠4 5 1 2 Then the composition

Q69.The value of k ∈R, for which the following system of linear equations 3x −y + 4z = 3 x + 2y −3z = −2 JEE Main 2021 (20 Jul Shift 2) JEE Main Previous Year Paper 6x + 5y + kz = −3 has infinitely many solutions, is: (1) 3 (2) −5 (3) 5 (4) −3

Q69.Consider three observations a, b and c such that b = a + c . If the standard deviation of a + 2, c + 2 is d , then which of the following is true? (1) b2 = 3(a2 + c2) + 9d2 (2) b2 = a2 + c2 + 3d2 (3) b2 = 3(a2 + c2 + d2) (4) b2 = 3(a2 + c2) −9d2 has : i = √−1. Then, the system of linear equations = A8[ xy ]

Q69.If the Boolean expression (p ⇒q) ⇔(q*(~p)) is a tautology, then the Boolean expression p*(~q) is equivalent to: (1) q ⇒p (2) ~q ⇒p (3) p ⇒~q (4) p ⇒q

Q69.The values of λ and μ such that the system of equations x + y + z = 6, 3x + 5y + 5z = 26 and x + 2y + λz = μ has no solution, are: (1) λ = 3, μ = 5 (2) λ = 3, μ ≠10 (3) λ ≠2, μ = 10 (4) λ = 2, μ ≠10

Q69.If the truth value of the Boolean expression ((p ∨q) ∧(q →r) ∧(~r)) →(p ∧q) is false, then the truth values of the statements p, q, r respectively can be: (1) FTF (2) TFF (3) TFT (4) FFT

Q69.Let [λ] be the greatest integer less than or equal to λ. The set of all values of λ for which the system of linear equations x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [λ])z = [λ] has a solution is: (1) R (2) (−∞, −9) ∪[−8, ∞) (3) (−∞, −9) ∪(−9, ∞) (4) [−9, −8) Q70. ⎡[x + 1] [x + 2] [x + 3]⎤ Let A = [x] [x + 3] [x + 3] , where [x] denotes the greatest integer less than or equal to x. If ⎣ [x] [x + 2] [x + 4] ⎦ det (A)= 192 , then the set of values of x is in the interval: (1) [62, 63) (2) [65, 66) (3) [60, 61) (4) [68, 69) = x ∈( π2 , π), then dxdy at x = 5π6 is:

Q69.Let A be a 3 × 3 matrix with det (A) = 4. Let Ri denote the ith row of A . If a matrix B is obtained by performing the operation R2 →2R2 + 5R3 on 2 A , then det (B) is equal to : (1) 64 (2) 16 (3) 128 (4) 80

Q69.Which of the following is the negation of the statement "for all M > 0, there exists x ∈S such that x ≥M ′′? (1) there exists M > 0, such that x < M for all (2) there exists M > 0, there exists x ∈S such that x ∈S x ≥M (3) there exists M > 0, there exists x ∈S such that (4) there exists M > 0 such that x ≥M for all x < M x ∈S

Q69.The statement (p ∧(p →q) ∧(q →r)) →r is (1) a tautology (2) equivalent to q →~r (3) a fallacy (4) equivalent to p →~r JEE Main 2021 (27 Aug Shift 1) JEE Main Previous Year Paper

Q69.The equation of one of the straight lines which passes through the point (1, 3) and makes an angles with the straight line, y + 1 = 3√2x is tan−1(√2) + + = 0 (1) 4√2x + 5y −(15 4√2) = 0 (2) 5√2x + 4y −(15 4√2) + = 0 (3) 4√2x + 5y −4√2 = 0 (4) 4√2x −5y −(5 4√2)

Q69.Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin−1( 3x5 ) + sin−1( 4x5 ) = sin−1 x is equal to: (1) 2 (2) 1 (3) 3 (4) 0

Q69.Let A = {1, 2, 3, … , 10} and f : A →A be defined as + 1 if k is odd f(k) = {k k if k is even JEE Main 2021 (26 Feb Shift 2) JEE Main Previous Year Paper Then the number of possible functions g : A →A such that gof = f is: (1) 10C5 (2) 55 (3) 5! (4) 105

Q69.The value of the limit lim tan(π cos2 θ) is equal to : θ→0 sin(2π sin2 θ) (1) −12 (2) −14 (3) 0 (4) 14

Q69.The mean and variance of 7 observations are 8 and 16 respetively. If two observations are 6 and 8, then the variance of the remaining 5 observations is : (1) 92 (2) 134 5 5 112 536 (3) (4) 5 25