Practice Questions

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

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

Q76.The differential equation of the family of circles passing through the origin and having centre at the line y = x is : (1) (x2 −y2 + 2xy)dx = (x2 −y2 −2xy)dy (2) (x2 + y2 + 2xy)dx = (x2 + y2 −2xy)dy (3) (x2 + y2 −2xy)dx = (x2 + y2 + 2xy)dy (4) (x2 −y2 + 2xy)dx = (x2 −y2 + 2xy)dy π

Q76.A function y = f(x) satisfies f(x) sin 2x + sin x −(1 + cos2 x)f ′(x) = 0 with condition f(0) = 0. Then f( π2 ) is equal to (1) 1 (2) 0 (3) −1 (4) 2 → → →

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 integral ∫π/40 3 sin136x+5sincosx x (1) 3π −50 loge 2 + 20 loge 5 (2) 3π −25 loge 2 + 10 loge 5 (3) 3π −10 loge(2√2) + 10 loge 5 (4) 3π −30 loge 2 + 20 loge 5

Q76.If sin( xy ) = loge x + α2 is the solution of the differential equation x cos( xy ) dxdy = y cos( xy ) + x and y(1) = π3 , then α2 is equal to (1) 3 (2) 12 (3) 4 (4) 9 −−−

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

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.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.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 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 (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 the area of the region enclosed by the curves y = 3x, 2y = 27 −3x and y = 3x −x√x be A . Then 10A is equal to (1) 172 (2) 162 (3) 154 (4) 184

Q76.Let 𝛼 be a non-zero real number. Suppose 𝑓: 𝑅→𝑅 is a differentiable function such that 𝑓0 = 1 and 𝑥→−∞𝑓𝑥=lim 1. If 𝑓'𝑥= 𝛼𝑓𝑥+ 3, for all 𝑥∈𝑅, then 𝑓−log𝑒2 is equal to ________. JEE Main 2024 (01 Feb Shift 2) JEE Main Previous Year Paper (1) 1 (2) 5 (3) 9 (4) 7

Q76.If y = y ( x ) is the solution curve of the differential equation x2 - 4dy - y2 - 3ydx = 0, x > 2, y(4) = 3 and 2 the slope of the curve is never zero, then the value of y ( 10 ) equals : 3 3 (1) 1 (2) 1 + 2√2 1 + ( 8 ) 4 3 3 (3) (4) 1 1 - 2√2 1 - ( 8 ) 4

Q76.The area (in sq. units) of the region described by {(x, y) : y2 ≤2x, and y ≥4x −1} is (1) 11 (2) 8 32 9 (3) 11 (4) 9 12 32

Q77.If y = y(x) is the solution of the differential equation dydx + 2y = sin(2x), y(0) = 43 , then y ( π8 ) is equal to: JEE Main 2024 (05 Apr Shift 1) JEE Main Previous Year Paper (1) eπ/8 (2) eπ/4 (3) e−π/4 (4) e−π/8

Q77.The set of all α, for which the vectors →a = αt^i + 6^j −3^k and →b = t^i −2^j −2αt^k are inclined at an obtuse angle for all t ∈R, is (1) (−43 , 1) (2) [0, 1) (3) (−43 , 0] (4) (−2, 0] L1 : →r = (2 + λ)^i + (1 −3λ)^j + (3 + 4λ)^k, λ ∈R m