Practice Questions

Q74.The value of k ∈N for which the integral In = ∫10 (1 −xk) ndx, (1) 14 (2) 8 (3) 10 (4) 7

Q74.The parabola y2 = 4x divides the area of the circle x2 + y2 = 5 in two parts. The area of the smaller part is equal to: (1) 1 3 + 5 sin−1 ( √52 ) (2) 31 + √5 sin−1 ( √52 ) (3) 3 2 + 5 sin−1 ( √52 ) (4) 32 + √5 sin−1 ( √52 )

Q74.Let ∫logeα 4 √ex−1dx (1) x2 + 2x −8 = 0 (2) x2 −2x −8 = 0 (3) 2x2 −5x + 2 = 0 (4) 2x2 −5x −2 = 0

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

Q74.The value of 1 1 2𝑥3 −3𝑥2 −𝑥+ 1 3𝑑𝑥 is equal to: ∫0 (1) 0 (2) 1 (3) 2 (4) -1 𝜋 Q75. 3 If ∫ cos4𝑥𝑑𝑥= 𝑎𝜋+ 𝑏√3, where 𝑎 and 𝑏 are rational numbers, then 9𝑎+ 8𝑏 is equal to: 0 (1) 2 (2) 1 3 (3) 3 (4) 2

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

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

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 (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.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 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 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.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

Q76.Let y = y(x) be the solution of the differential equation (1 + y2)etan xdx + cos2 x (1 + e2 tan x)dy = 0, y(0) = 1. Then y ( π4 ) is equal to (1) 2 (2) 2 e e2 (3) 1 (4) 1 e e2

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.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.Let 𝑦= 𝑦( 𝑥) be the solution of the differential equation 𝑑𝑦 tan𝑥+ 𝑦 𝜋 𝑑𝑥= sin𝑥sec𝑥- sin𝑥tan𝑥, 𝑥∈0, 2 satisfying the 𝜋 𝜋 condition 𝑦 = 2. Then, 𝑦 is 4 3 2 + log𝑒3 (1) √32 + log𝑒√3 (2) √32 (3) √31 + 2log𝑒3 (4) √32 + log𝑒3 →

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

Q76.Let y = y(x) be the solution curve of the differential equation sec y dydx + 2x sin y = x3 cos y, y(1) = 0. Then y(√3) is equal to : (1) π (2) π 3 6 (3) π (4) π 12 4

Q76.The solution of the differential equation (x2 + y2)dx −5xy dy = 0, y(1) = 0, is : (1) x2 −2y2 6 = x (2) x2 −4y2 6 = x (3) x2 −4y2 5 = x2 (4) x2 −2y2 5 = x2 →

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