Practice Questions

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

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.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

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 →

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 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.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.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.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.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.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

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.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