Practice Questions

Q66.If θ1 and θ2 be respectively the smallest and the largest values of θ in (0, 2π) −{π} which satisfy the equation, θ2 2cot2θ − sin5 θ + 4 = 0 , then ∫ cos23θdθ is equal to: θ1 (1) π (2) 2π 3 3 (3) π 3 + 61 (4) π9

Q66.If the value of the integral ∫ 01 3 dx is k6 , then k is equal to: (1−x2) 2 JEE Main 2020 (03 Sep Shift 2) JEE Main Previous Year Paper (1) 2√3 + π (2) 2√3 −π (3) 3√2 + π (4) 3√2 −π

Q67.Let y = y(x) be a solution of the differential equation, √1 −x2 dxdy + √1 −y2 = 0, |x| < 1. If y( 12 ) = √32 , then y( √2−1 ) is equal to (1) √3 (2) −1 2 √2 (3) 1 (4) −√32 √2 →

Q67.Let f(x) = ∫ √x dx (x ≥0). Then f(3) −f(1) is equal to : (1+x)2 (1) −π12 + 12 + √34 (2) π6 + 21 −√34 (3) −π6 + 21 + √34 (4) 12π + 12 −√34 dx is equal to

Q67.Area (in sq. units) of the region outside |x|2 + |y|3 = 1 and inside the ellipse x24 + y29 = 1 is (1) 6(π −2) (2) 3(π −2) (3) 3(4 −π) (4) 6(4 −π) x, y > 0, y(0) = 1. If y(π) = a

Q67.The area (in sq. units) of the region enclosed by the curves y = x2 −1 and y = 1 −x2 is equal to: (1) 4 (2) 8 3 3 (3) 7 (4) 16 2 3 x cosec x is the solution of the differential equation, dxdy + p(x)y = −2π cosec x, 0 < x < 2π ,

Q67.The value of ∫ −ππ 2 1+esin (1) π4 (2) π (3) π2 (4) 3π2

Q67.If x3dy + xy ⋅dx = x2dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to : (1) √e (2) 1 + √e 2 2 (3) 3 2 √e (4) 23 + √e

Q67.Consider a region R = {(x, y) ∈R2 : x2 ≤y ≤2x}. If a line y = α divides the area of region R into two equal parts, then which of the following is true ? (1) α3 −6α2 + 16 = 0 (2) 3α2 −8α3/2 + 8 = 0 (3) 3α2 −8α + 8 = 0 (4) α3 −6α3/2 −16 = 0

Q67.The differential equation of the family of curves, x2 = 4b(y + b), b ∈R, is. (1) x(y') 2 = x + 2yy' (2) x(y') 2 = 2yy' −x (3) xy'' = y' (4) x(y')2 = x −2yy' → → →

Q67.The area (in sq. units) of the region {(x, y) ∈R2 4x2 ≤y ≤8x + 12} is (1) 125 (2) 128 3 3 (3) 124 (4) 127 3 3

Q67.The solution curve of the differential equation, (1 + e−x)(1 + y2) dxdy = y2 which passes through the point (0, 1), is + 2 ) + 2) (1) y2 + 1 = y(loge( 1+e−x2 ) 2) (2) y2 + 1 = y(loge( 1+ex (3) y2 = 1 + y loge( 1+ex2 ) (4) y2 = 1 + y loge( 1+e−x2 )

Q67.The value of ∫2π0 sin8xx+cos8sin8 x x (1) 2π (2) 2π2 (3) π2 (4) 4π

Q67. x, 0 ≤x < 12 ⎧ Given: f(x) = 2 1 , x = 12 ⎨ 1 ⎩1 −x, 2 < x ≤1 x ∈R. Then, the area (in sq. units) of the region bounded by the curves, y = f(x) and and g(x) = (x −12 ) 2, y = g(x) between the lines 2x = 1 and 2x = √3, is: (1) 3 1 + √34 (2) √34 −13 (3) 2 1 −√34 (4) 21 + √34

Q67.The integral ∫ π3 tan3 x ⋅sin2 3x(2 sec2 x ⋅sin2 3x + 3 tan x ⋅sin 6x)dx is equal to: 6 (1) 18 7 (2) −19 (3) −118 (4) 29 dy y+3x

Q67.If ∫ 5+7 sincosθ−2θ cos2 θ dθ =Aloge B(θ) (1) 2 sin θ+1 (2) 2 sin θ+1 sin θ+3 5(sin θ+3) (3) 5(sin θ+3) (4) 5(2 sin θ+1) 2 sin θ+1 sin θ+3

Q67.If y = y(x) is the solution of the differential equation , ey( dxdy −1) equal to (1) 1 + loge2 (2) 2 + loge2 (3) 2e (4) loge2 →

Q67.The area (in sq. units) of the region A = {(x, y) : x + y ≤1, 2y2 ≥x } (1) 1 (2) 7 3 6 (3) 1 (4) 5 6 6

Q68.Let the volume of a parallelepiped whose coterminous edges are given by u = ˆi + ˆj + λˆk,→v = ˆi + ˆj + 3ˆk and → → → w = 2ˆi + ˆj + ˆk be 1 cu. unit. If θ be the angle between the edges u and w, then the value of cos θ can be (1) 7 (2) 7 6√6 6√3 (3) 5 (4) 5 7 3√3 y−8

Q68.The area (in sq. units) of the region A = {(x, y) : (x −1)[x] ≤y ≤2√x, 0 ≤x ≤2}, where [t] denotes the greatest integer function, is : (1) 3 8 √2 −12 (2) 34 √2 + 1 (3) 8 3 √2 −1 (4) 43 √2 −12

Q68.If dy = xy ; y(1) = 1; then a value of x satisfying y(x) = e is: dx x2+y2 e (1) 1 √3e (2) 2 √2 (3) √2e (4) √3e

Q68.Let y = y(x) be the solution of the differential equation, 2+siny+1 x . dxdy = −cos and dy at x = π is b, then the ordered pair (a, b) is equal to dx (1) (2, 32 ) (2) (1, −1) (3) (1, 1) (4) (2, 1)

Q68.The solution of the differential equation − + 3 = 0 is dx loge(y+3x) (where C is a constant of integration) (1) x −12 (loge(y + 3x))2 = C (2) x −loge(y + 3x) = C (3) y + 3x −12 (loge x)2 = C (4) x −2 loge(y + 3x) = C

Q68.A vector →a = αˆi + 2ˆj + βˆk(α, β ∈R) lies in the plane of the vectors, b = ˆi + ˆj and →c= ˆi −ˆj + 4ˆk. If →a → bisects the angle between b and →c, then (1) →a⋅ˆi + 3 = 0 (2) →a⋅ˆi + 1 = 0 (3) →a⋅ˆk + 2 = 0 (4) →a⋅ˆk + 4 = 0

Q68.Let f(x) = |x −2| and g(x) = f(f(x)), x ∈[0, 4]. Then ∫30 (g(x) −f(x)) (1) 1 (2) 0 (3) 1 (4) 3 2 2