Practice Questions

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.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 {(x, y) ∈R2 4x2 ≤y ≤8x + 12} is (1) 125 (2) 128 3 3 (3) 124 (4) 127 3 3

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.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

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

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.Let y = y(x) be the solution curve of the differential equation, (y2 −x) dxdy = 1 , satisfying y(0) = 1 . This curve intersects the X−axis at a point whose abscissa is (1) 2 −e (2) −e (3) 2 (4) 2 + e → → → → →

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.If y = y(x) is the solution of the differential equation 5+ex2+y ⋅dydx + ex = 0 satisfying y(0) = 1 then value of y(loge 13) is (1) 1 (2) −1 (3) 0 (4) 2

Q68.If a curve y = f(x) , passing through the point (1, 2), is the solution of the differential equation 2x2dy = (2xy + y2)dx, then f( 21 ) is equal to JEE Main 2020 (02 Sep Shift 2) JEE Main Previous Year Paper (1) 1 (2) 1 1+loge 2 1−loge 2 (3) 1 + loge 2 (4) 1+loge−1 2

Q68.The general solution of the differential equation √1 + x2 + y2 + x2y2 + xy dxdy = 0 (where C is a constant of integration) + C (1) √1 + y2 + √1 + x2 = 12 loge( √1+x2+1√1+x2−1 ) + C (2) √1 + y2 −√1 + x2 = 12 loge( √1+x2+1√1+x2−1 ) + C (3) √1 + y2 + √1 + x2 = 12 loge( √1+x2−1√1+x2+1 ) (4) 1 √1+x2+1 + C √1 + y2 −√1 + x2 = 2 loge( √1+x2−1 )

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.If f '(x) = tan−1(sec x + tan x), −π2 < x < π2 and f(0) = 0 , then f(1) is equal to: (1) π+1 (2) 1 4 4 (3) π−1 (4) π+2 4 4

Q68.The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, −2, 3) and (1, 1, 0) lies on the plane (1) 2 x + y −z = 1 (2) x −y −2 z = 1 (3) x −2 y + z = 1 (4) x + 2 y −z = 1 + + +

Q68.If y = ( 2π −1) then the function p(x) is equal to : (1) cot x (2) cosec x (3) sec x (4) tan x

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.Let →a = ˆi −2ˆj + ˆk and b = ˆi −ˆj + ˆk, be two vectors. If →c, is a vector such that b ×→c= b ×→a and →c⋅→a = 0, → then →c⋅ b, is equal to. (1) −32 (2) 21 (3) −12 (4) −1

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.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 a, b, c ∈R be such that a2 + b2 + c2 = 1. If a cos θ = b cos(θ + 2π3 ) = c cos(θ + 4π3 ),where θ = π9 , then the angle between the vectors aˆi + bˆj + cˆk and bˆi + cˆj + aˆk is: (1) 0 (2) 2π3 (3) π (4) π 2 9

Q69.Let y = y(x) be the solution of the differential equation, xy′ −y = x2(x cos x + sin x), x > 0. If y(π) = π, then y′′( π2 ) + y( π2 ) is equal to : (1) 2 + π2 (2) 1 + π2 + π24 (3) 2 + π2 + π24 (4) 1 + π2 b where→a = xˆi −2ˆj + 3ˆk, →b = −2ˆi + xˆj −ˆk and

Q69.The shortest distance between the lines x−3 3 = −1 = z−31 and x+3−3 = y+72 = z−64 is (1) 2√30 (2) 72 √30 (3) 3√30 (4) 3

Q69.Let y = y(x) be the solution of the differential equation cos x dxdy + 2y sin x = sin 2x, x ∈(0, π2 ) If y(π/3) = 0, then y(π/4) is equal to : (1) 2 −√2 (2) 2 + √2 (3) √2 −2 (4) 1 −1 √2

Q69.If the volume of a parallelopiped, whose coterminous edges are given by the vectors →a = ˆi + ˆj + nˆk , → b = 2ˆi + 4ˆj − nˆk and,→c= ˆi + nˆj + 3ˆk (n ≥0) is 158 cubic units, then : → (1) →a⋅→c= 17 (2) b ⋅→c= 10 (3) n = 7 (4) n = 9