Practice Questions

Q74.The value of n→∞∑nlim k=1 (n2+k2)(n2+3k2)n3 is : (1) (2√3+3)π (2) 13π 24 8(4√3+3) (3) 13(2√3−3)π (4) π 8 8(2√3+3)

Q74.The integral ∫ x8 - x2dx 1 is equal to : x12 + 3x6 + 1tan-1x3 + x3 (1) 1 13 (2) 1 12 logtan-1x3 + x3 + C logetan-1x3 + x3 + C 1 1 3 + + C (3) logetan-1x3 + x3 + C (4) logetan-1x3 x3 𝜋 𝑑𝑥

Q75.If ∫10 √3+x+√1+x1 (1) 4 (2) 10 (3) 7 (4) 8

Q75.Let f(x) = −2 ≤x ≤0 and h(x) = f(|x|) + |f(x)| . Then ∫2−2 h(x)dx {−2,x −2, 0 < x ≤2 (1) 1 (2) 6 (3) 4 (4) 2

Q75.If the area of the region {(x, y) : x2a ≤y ≤1x , 1 ≤x ≤2, 0 < a < 1} is (loge 2) −17 then the value of 7a −3 is equal to: (1) 0 (2) 2 (3) -1 (4) 1 dy

Q75.The value of the integral ∫2−1 loge (x + √x2 + 1)dx JEE Main 2024 (09 Apr Shift 2) JEE Main Previous Year Paper (1) √5 −√2 + loge ( 7+4√51+√2 ) (2) √5 −√2 + loge ( 9+4√51+√2 ) + loge (3) √2 −√5 + loge ( 7+4√51+√2 ) (4) √2 −√5 ( 9+4√51+√2 )

Q75.Let 𝑓, 𝑔: 0, ∞→𝑅 be two functions defined by 𝑓𝑥= 𝑥𝑡−𝑡2𝑒−𝑡2𝑑𝑡 and 𝑔𝑥= 𝑥2 𝑡 12𝑒−𝑡2𝑑𝑡. Then the ∫−𝑥 ∫0 value of 9𝑓√log𝑒9 + 𝑔√log𝑒9 is equal to (1) 6 (2) 9 (3) 8 (4) 10

Q75.The area (in square units) of the region bounded by the parabola y2 = 4(x −2) and the line y = 2x −8. (1) 8 (2) 9 (3) 6 (4) 7

Q75.If the value of the integral ∫1−1 cos1+3xαx (1) π (2) π 3 6 (3) π (4) π 4 2

Q75.The area enclosed between the curves y = x|x| and y = x −|x| is : (1) 4 (2) 1 3 (3) 2 (4) 8 3 3

Q75.The solution curve, of the differential equation 2y dydx + 3 = 5 dydx , passing through the point (0, 1) is a conic, whose vertex lies on the line: JEE Main 2024 (09 Apr Shift 1) JEE Main Previous Year Paper (1) 2x + 3y = 9 (2) 2x + 3y = −9 (3) 2x + 3y = −6 (4) 2x + 3y = 6

Q75.Let 𝑦= 𝑓( 𝑥) be a thrice differentiable function in ( - 5, 5 ) . Let the tangents to the curve 𝑦= 𝑓( 𝑥) at ( 1, f ( 1 ) ) and ( 3, f ( 3 ) ) make angles 𝜋 and 𝜋 respectively with positive x-axis. If 6 4, 3 2 27 ∫1 𝑓'𝑡 + 1𝑓"𝑡𝑑𝑡= 𝛼+ 𝛽√3 where 𝛼, 𝛽 are integers, then the value of 𝛼+ 𝛽 equals JEE Main 2024 (30 Jan Shift 2) JEE Main Previous Year Paper (1) -14 (2) 26 (3) -16 (4) 36 39 , then 𝑓𝑥- 𝑐𝑥𝑑𝑥=

Q75.For 0 < a < 1, the value of the integral ∫0 1 - 2𝑎cos𝑥+ 𝑎2 is : (1) 𝜋2 (2) 𝜋2 𝜋+ 𝑎2 𝜋- 𝑎2 𝜋 𝜋 (3) (4) 1 - 𝑎2 1 + 𝑎2 JEE Main 2024 (27 Jan Shift 2) JEE Main Previous Year Paper

Q75.The solution curve of the differential equation 𝑦 𝑑𝑥 1, 𝑥> 0, 𝑦> 0 passing through the 𝑑𝑦= 𝑥log𝑒𝑥- log𝑒𝑦+ point ( 𝑒, 1 ) is 𝑦 𝑦 (1) log𝑒 𝑥= 𝑥 (2) log𝑒 𝑥= 𝑦2 (3) 𝑥 𝑦 (4) 𝑥 𝑦+ 1 log𝑒 𝑦= 2log𝑒 𝑦=

Q75.The area of the region in the first quadrant inside the circle x2 + y2 = 8 and outside the parabola y2 = 2x is equal to : (1) π 2 −13 (2) π −13 (3) π 2 −23 (4) π −23

Q75.Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e−a + 4a2 + a −1. Then the differential equation, whose general solution is y = c1f(x) + c2 , where c1 and c2 are arbitrary constants, is d2y dy (1) (8ex −1) = 0 (2) (8ex −1) + dx d2y −dydx = 0 dx2 dx2 (3) (8ex + 1) + dxdy = 0 dx2 d2y −dydx = 0 (4) (8ex + 1) dx2d2y

Q75.For x ∈(−π2 , π2 ), if y(x) = ∫ cosecxcosecx+sinsec x+tan xx sin2 x dx and limπ = 0 then y( π4 ) is equal to x→( 2 )−y(x) (1) tan−1( √21 ) (2) 21 tan−1( √21 ) (3) −1 2 ) √2 tan−1( √21 ) (4) √21 tan−1(−1

Q75.If ∫ 3 3 √sin3 x cos3 x sin(x−θ) constant, then AB is equal to (1) 4 cosec (2θ) (2) 4 sec θ (3) 2 sec θ (4) 8 cosec (2θ) JEE Main 2024 (29 Jan Shift 2) JEE Main Previous Year Paper

Q75.The value of ∫π−π 2y(1+sin1+cos2 yy) (1) 2π2 (2) π22 (3) π (4) π2 2 dx is equal to :

Q76.Let 𝑓: 𝑅→𝑅 be defined 𝑓𝑥= 𝑎𝑒2𝑥+ 𝑏𝑒𝑥+ 𝑐𝑥. If 𝑓(0) = - 1, 𝑓'log𝑒2 = 21 and ∫0log4 2 the value of |𝑎+ 𝑏+ 𝑐| equals: (1) 16 (2) 10 (3) 12 (4) 8 2

Q76.One of the points of intersection of the curves y = 1 + 3x −2x2 and y = x1 is ( 21 , 2). Let the area of the region enclosed by these curves be 1 (l√5 + m) −n loge(1 + √5), where l, m, n ∈N. Then l + m + n is 24 equal to (1) 29 (2) 31 (3) 30 (4) 32

Q76.The area enclosed by the curves 𝑥𝑦+ 4𝑦= 16 and 𝑥+ 𝑦= 6 is equal to: (1) 28 −30log𝑒2 (2) 30 −28log𝑒2 (3) 30 −32log𝑒2 (4) 32 −30log𝑒2 2

Q76.Let y = y(x) be the solution of the differential equation sec xdy + {2(1 −x) tan x + x(2 −x)}dx = 0 such that y(0) = 2. Then y(2) is equal to : (1) 2 (2) 2{1 −sin(2)} (3) 2{sin(2) + 1} (4) 1

Q76.The area (in square units) of the region enclosed by the ellipse x2 + 3y2 = 18 in the first quadrant below the line y = x is (1) √3π −34 (2) √3π + 1 (3) √3π (4) √3π + 34

Q76.The area of the region enclosed by the parabola 𝑦= 4𝑥−𝑥2 and 3𝑦= 𝑥−42 is equal to 32 (1) (2) 4 9 14 (3) 6 (4) 3