Practice Questions

Q75.In a triangle ABC , coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4 . Then area of ΔABC (in sq. units) is : (1) 12 (2) 4 (3) 9 (4) 5

Q76.Consider the following two binary relations on the set A = {a, b, c} : R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}, then : (1) R2 is symmetric but it is not transitive (2) both R1 and R2 are not symmetric (3) both R1 and R2 are transitive (4) R1 is not symmetric but it is transitive

Q76.Consider the following two binary relations on the set A = {a, b, c} : R1 = {(c, a)(b, b), (a, c), (c, c), (b, c), (a, a)} and R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c). Then (1) R2 is symmetric but it is not transitive (2) Both R1 and R2 are transitive (3) Both R1 and R2 are not symmetric (4) R1 is not symmetric but it is transitive is a scalar matrix and |3A| = 108 . Then A2 equals

Q76. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively, 45°, 30° and 30°, then the height of the tower (in m ) is: JEE Main 2018 (08 Apr) JEE Main Previous Year Paper (1) 50√2 (2) 100 (3) 50 (4) 100√3

Q76.Let N denote the set of all natural numbers. Define two binary relations on N as R1 = {(x, y) ∈N × N : 2x + y = 10} and R2 = {(x, y) ∈N × N : x + 2y = 10}. Then (1) both R1 and R2 are transitive relations (2) range of R2 is {1, 2, 3, 4} (3) range of R1 is {2, 4, 8} (4) both R1 and R2 are symmetric relations Q77. ⎡ 1 0 0⎤ Let A = 1 1 0 and B = A20 . Then the sum of the elements of the first column of B is ⎣ 1 1 1⎦ (1) 210 (2) 211 (3) 251 (4) 231 JEE Main 2018 (16 Apr Online) JEE Main Previous Year Paper

Q76.Suppose A is any 3 × 3 non-singular matrix and (A −3I)(A −5I) = O, where I = I3 and O = O3 . If αA+ βA−1 = 4I , then α + β is equal to (1) 8 (2) 12 (3) 13 (4) 7

Q77.Let A be a matrix such that A . [10 23 ] (1) [40 −3236 ] (2) [−324 360 ] (3) [−3236 04] (4) [360 −324 ]

Q77.Let A be a matrix such that A ⋅[10 23 ] is a scalar matrix and |3A| = 108 . Then, A2 equals : (1) [−324 360 ] (2) [360 −324 ] (3) [−3236 04 ] (4) [40 −3236 ] JEE Main 2018 (15 Apr) JEE Main Previous Year Paper

Q77.Let the orthocentre and centroid of a triangle be A(−3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is: (1) 3√5 (2) √10 2 (3) 2√10 (4) 3√52

Q77.If the system of linear equations x + ay + z = 3 x + 2y + 2z = 6 x + 5y + 3z = b has no solution, then (1) a = 1, b ≠9 (2) a ≠−1, b = 9 (3) a = −1, b = 9 (4) a = −1, b ≠9

Q78.The number of values of k for which the system of linear equations (k + 2)x + 10y = k & kx + (k + 3)y = k −1 has no solution is (1) 1 (2) 2 (3) 3 (4) 4

Q78.If the system of linear equations x + ky + 3z = 0 3x + ky −2z = 0 2x + 4y −3z = 0 has a non-zero solution (x, y, z), then xz is equal to: y2 (1) 30 (2) −10 (3) 10 (4) −30

Q78. cos x x 1 f ′(x) If f(x) = 2 sin x x2 2x , then lim x x→0 tan x x 1 (1) does not exist (2) exists and is equal to −2 (3) exists and is equal to 0 (4) exists and is equal to 2

Q78. cos x x 1 f ′(x) If f(x) = 2 sin x x2 2x , then limx→0 x tan x x 1 (1) Exists and is equal to −2 (2) Does not exist (3) Exist and is equal to 0 (4) Exists and is equal to 2

Q78.Let f : A → B be a function defined as f(x) = x−1x−2 , where A = R −{2} and B = R −{1}. Then f is (1) invertible and f −1(y) = 2y+1y−1 (2) invertible and f −1(y) = 3y−1y−1 (3) no invertible (4) invertible and f −1(y) = 2y−1y−1 1 −1) 2−x , x > 1, x ≠2

Q79.Let f(x) = {(x k, x = 2 The value of k for which f is continuous at x = 2 is (1) e−2 (2) e (3) e−1 (4) 1

Q79.If the function f defined as f(x) = x1 − e2x−1k−1 , x ≠0 is continuous at x = 0, then ordered pair (k, f(0)) is equal to (1) (2, 1) (2) (3, 1) (3) (3, 2) (4) ( 13 , 2)

Q79.Let S be the set of all real values of k for which the system of linear equations x + y + z = 2 2x + y −z = 3 3x + 2y + kz = 4 has a unique solution. Then S is (1) an empty set (2) equal to R −{0} (3) equal to {0} (4) equal to R

Q79. x −4 2x 2x If 2x x −4 2x = (A + Bx) (x −A)2, then the ordered pair (A, B) is equal to 2x 2x x −4 (1) (4, 5) (2) (−4, −5) (3) (−4, 3) (4) (−4, 5)

Q79.Let S be the set of all real values of k for which the system of linear equations x + y + z = 2 2x + y −z = 3 3x + 2y + kz = 4 has a unique solution. Then, S is : (1) equal to R –{0} (2) an empty set (3) equal to R (4) equal to {0}

Q80.Let S = {t ∈R : f(x) = |x −π| ⋅(e|x| −1) sin|x| is not differentiable at t}. Then, the set S is equal to: (1) {0, π} (2) ϕ (an empty set) (3) {0} (4) {π}

Q80.Let S = {(λ, μ) ∈R × R : f(t) = (|λ |e|t| −μ) sin(2|t|), t ∈R is a differential function}. Then, S is a subset of : (1) (−∞, 0) × R (2) R × [0 , ∞) (3) [0 , ∞) × R (4) R × (−∞, 0)

Q80.If x = √2cosec−1 t and y = √2sec−1 t, (|t| ≥1), then dxdy is equal to (1) x y (2) −yx (3) −xy (4) xy

Q80.Let S = {(λ, μ) ∈R × R : f(t) = (|λ|et −μ) ⋅sin(2|t|), t ∈R, is a differentiable function } . Then S is a subest of? (1) R × [0, ∞) (2) (−∞, 0) × R (3) [0, ∞) × R (4) R × (−∞, 0)

Q80.If f(x) = sin−1 ( 2×3x1+9x ), then f ′ (−12 ) equals. (1) √3 loge √3 (2) −√3 loge √3 (3) −√3 loge 3 (4) √3 loge 3 JEE Main 2018 (15 Apr Shift 2 Online) JEE Main Previous Year Paper