Practice Questions

Q85.The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is (1) 17 (2) 21 4 2 (3) 15 (4) 15 2 4

Q85.The area (in sq. units) of the region A = {(x, y) : x2 ≤y ≤x + 2} is (1) 136 (2) 316 (3) 9 (4) 10 2 3 dy

Q85.The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y −2 is : (1) 5 (2) 9 4 8 (3) 7 (4) 3 8 4

Q85.The region represented by |x −y| ≤2 and |x + y| ≤2 is bounded by a (1) rhombus of area 8√2 sq. units. (2) rhombus of side length 2 units. (3) square of area 16 sq. units. (4) square of side length 2√2 units. x ∈(−π2 , π2 ) , such that

Q85.If ∫ dθ = 1 − , > , then the value of k is √2k sec θ √2 0 (k 0) (1) 21 (2) 1 (3) 2 (4) 4 JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper

Q85.The solution of the differential equation, dy dx = (x −y)2 , when y(1) = 1, is: (1) loge 2−x2−y = x −y (2) −loge 1+x−y1−x+y = 2(x −1) (3) −loge 1−x+y1+x−y = x + y −2 (4) loge 2−x2−y = 2(y −1)

Q85.A curve amongst the family of curves represented by the differential equation, (x2 −y2) dx + 2xy dy = 0 which passes through (1, 1), is (1) A circle with centre on the x− axis. (2) A circle with centre on the y− axis. (3) A hyperbola with transverse axis along the x− (4) An ellipse with major axis along the y− axis. axis. x f( x1 )

Q85.The area (in sq. units) of the region bounded by the curves 𝑦= 2𝑥 and 𝑦= 𝑥+ 1, in the first quadrant is 3 1 1 (1) - (2) 2 log𝑒⁡2 2 3 3 (3) log𝑒⁡2 + 2 (4) 2

Q85.If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 sq. unit. Then k is (1) √3 (2) 1 √3 (3) √3 (4) 2 2 √3 JEE Main 2019 (10 Jan Shift 1) JEE Main Previous Year Paper 3 1

Q86.Let 𝑆𝛼= 𝑥, 𝑦: 𝑦2 ≤𝑥, 0 ≤𝑥≤𝛼 and A𝛼 is area of the region 𝑆𝛼. If for a 𝜆, 0 < 𝜆< 4, A𝜆: A4 = 2: 5, then 𝜆 equals: (1) 2 13 (2) 4 13 4 2 5 25 (3) 4 13 (4) 2 13 4 2 25 5

Q86.Let y = y(x) be the solution of the differential equation, x dxdy + y = x loge x, (x > 1). If 2y(2) = loge 4 −1, then y(e) is equal to (1) −e2 (2) 4e (3) −e22 (4) e24

Q86.If y = y(x) is the solution of the differential equation dxdy = (tanx −y)sec2x , y(0) = 0, then y(−π4 ) is equal to: (1) 1 e −2 (2) 2 + 1e (3) e −2 (4) 12 −e

Q86.The solution of the differential equation x y(1) = 1, is dx + 2y = x2, (x ≠0) with (1) y = x35 + 5x21 (2) y = 34 x2 + 4x21 (3) y = x24 + 4x23 (4) y = 45 x3 + 5x21 → → → → → → → → →−−−−

Q86.If cosx dxdy −ysinx = 6x, (0 < x < π2 ) and y( π3 ) = 0, then y( π6 ) is equal to (1) −π2 (2) π2 4√3 2√3 (3) −π22 (4) −π22√3 π

Q86.Let 𝑦= 𝑦𝑥 be the solution of the differential equation, 𝑑𝑦 𝑦tan𝑥= 2𝑥+ 𝑥2tan𝑥, 𝑥∈- π π such that 𝑑𝑥+ 2, 2, 𝑦0 = 1 . Then JEE Main 2019 (10 Apr Shift 2) JEE Main Previous Year Paper π π π (1) 𝑦'π - 𝑦'- 4 4 = π - √2 (2) y' 4 + y'- 4 = - √2 π2 (3) 𝑦π - 𝑦-π = √2 (4) y'π + y'- π = + 2 4 4 4 4 2

Q86.Consider the differential equation, 𝑦2𝑑𝑥+ 𝑥- 𝑦𝑑𝑦= 0. If value of 𝑦 is 1 when 𝑥= 1, then the value of 𝑥 for which 𝑦= 2, is (1) 3 - 1 (2) 3 - 2 2 √𝑒 √𝑒 1 1 5 1 (3) + (4) + 2 √𝑒 2 √𝑒

Q86.If y(x) is the solution of the differential equation dxdy + ( 2x+1x )y = e−2x, x > 0, where y(1) = 21 e−2, then: (1) y (loge 2) = loge 4 (2) y (loge 2) = loge4 2 (3) y(x) is decreasing in ( 12 , 1) (4) y(x) is decreasing in (0,1)

Q86.The area of the region A = {(x, y) : 0 ≤y ≤x|x| + 1 and −1 ≤x ≤1} in sq. units, is (1) 4 (2) 2 3 (3) 1 (4) 2 3 3 →

Q86.If dy + y = dx , x ∈(−π3 , π3 ), and y( π4 ) = 34 , then y(−π4 ) equals x x cos2 cos2 (1) 1 (2) 1 3 3 + e3 (3) 3 1 + e6 (4) −43 →

Q86.If 𝑦= 𝑦( 𝑥) is the solution of the differential equation, 𝑥 𝑑𝑦 2𝑦= 𝑥2 satisfying 𝑦1 = 1, then 𝑦 1 is equal to 𝑑𝑥+ 2 (1) 7 (2) 1 64 4 13 49 (3) (4) 16 16 2

Q86.Let α ∈R and the three vectors →a = αˆi + ˆj + 3ˆk, b = 2ˆi + ˆj −αˆk and →c= αˆi −2ˆj + 3ˆk. Then the set S = { → α :→a, b and →care coplanar} (1) is singleton (2) contains exactly two positive numbers (3) is empty (4) contains exactly two numbers only one of which is positive

Q86.Let →a, b and →cbe three unit vectors, out of which vectors b and →care non-parallel. If α and β are the angles → → → b = 21 b, then |α −β| is equal to : which vector →a makes with vectors b and →crespectively and →a× ( ×→c) (1) 90o (2) 60o (3) 45o (4) 30o y−2

Q86.Let 𝑦= 𝑦( 𝑥) be the solution of the differential equation, 𝑥2 + 1 2 𝑑𝑦 2𝑥(𝑥2 + 1)𝑦= 1 such that 𝑦0 = 0 . 𝑑𝑥+ If 𝑦1 = 𝜋 then the value of 𝑎 is √𝑎 32, (1) 1 (2) 1 (3) 1 (4) 1 16 2 4

Q87.Let →a = ^i + 2^j + 4^k,→b = ^i + λ^j + 4^k and →c = 2^i + 4^j + (λ2 −1)^k be coplanar vectors. Then the non-zero vector →a × →c is: (1) −10^i −5^j (2) −14^i −5^j (3) −14^i + 5^j (4) −10^i + 5^j

Q87.Two lines x−3 1 = y+13 = z−6−1 and x+57 = y−2−6 = z−34 intersect at the point R. The reflection of R in the xy - plane has coordinates: (1) (2,-4,-7) (2) (2,4,7) (3) (2,-4,7) (4) (-2,4,7)