RankLab

Practice Questions

10,171 questions across 23 years of JEE Main β€” find and practise any topic!

Found 10,171 results

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

202430 Jan Shift 1Differential Equations
MathsMedium

Q77.Let A(2, 3, 5) and C(βˆ’3, 4, βˆ’2) be opposite vertices of a parallelogram ABCD if the diagonal βˆ’β†’ BD = Λ†i + 2Λ†j + 3Λ†k then the area of the parallelogram is equal to (1) 1 2 √410 (2) 21 √474 (3) 1 2 √586 (4) 21 √306 β†’ β†’ β†’

202430 Jan Shift 1Vectors
MathsMedium

Q77.The temperature 𝑇𝑑 of a body at time 𝑑= 0 is 160Β° 𝐹 and it decreases continuously as per the differential 𝑑𝑇 equation 𝑑𝑑= βˆ’πΎπ‘‡βˆ’80, where 𝐾 is positive constant. If 𝑇15 = 120Β° 𝐹, then 𝑇45 is equal to (1) 85Β° 𝐹 (2) 95Β° 𝐹 (3) 90Β° 𝐹 (4) 80Β° 𝐹

202431 Jan Shift 2Differential Equations
MathsMedium

Q77.Let β†’a = 4^i βˆ’^j + ^k,β†’b = 11^i βˆ’^j + ^k and β†’c be a vector such that (β†’a + β†’b) Γ— β†’c = β†’c Γ— (βˆ’2β†’a + 3β†’b). If (2β†’a + 3β†’b) β‹…β†’c = 1670, then |β†’c|2 is equal to : (1) 1609 (2) 1618 (3) 1600 (4) 1627 β†’

202408 Apr Shift 2Vectors
MathsMedium

Q77.The set of all Ξ±, for which the vectors β†’a = Ξ±t^i + 6^j βˆ’3^k and β†’b = t^i βˆ’2^j βˆ’2Ξ±t^k are inclined at an obtuse angle for all t ∈R, is (1) (βˆ’43 , 1) (2) [0, 1) (3) (βˆ’43 , 0] (4) (βˆ’2, 0] L1 : β†’r = (2 + Ξ»)^i + (1 βˆ’3Ξ»)^j + (3 + 4Ξ»)^k, Ξ» ∈R m

202408 Apr Shift 1Differential Equations
MathsMedium

Q77.Let β†’a = 2^i + ^j βˆ’^k, b = ((β†’aΓ— (^i + ^j)) Γ—^i) Γ—^i. Then the square of the projection of β†’a on b is : (1) 1 (2) 2 3 3 (3) 2 (4) 1 5 β†’

202406 Apr Shift 2Vectors
MathsMedium

Q77.Let x = x(t) and y = y(t) be solutions of the differential equations dxdt + ax = 0 and dydt + by = 0 respectively, a, b ∈R. Given that x(0) = 2 ; y(0) = 1 and 3 y(1) = 2 x(1), the value of t, for which x(t) = y(t), is : (1) log 2 2 (2) log4 3 3 4 2 (3) log3 4 (4) log 3 β†’ β†’ andβ†’cbe the vector such that β†’aΓ—β†’c= b and β†’aβ‹…β†’c= 3, then

202427 Jan Shift 1Differential Equations
MathsMedium

Q77.Let three vectors β†’a = Ξ±^i + 4^j + 2^k, b = 5^i + 3^j + 4^k,β†’c= x^i + y^j + z^k form a triangle such that β†’c = β†’a βˆ’β†’b and the area of the triangle is 5√6. If Ξ± is a positive real number, then |β†’c|2 is equal to: (1) 16 (2) 14 (3) 12 (4) 10 β†’ β†’βˆ’βˆ’βˆ’

202409 Apr Shift 1Differential Equations
MathsMedium

Q77.Consider a π›₯𝐴𝐡𝐢 where 𝐴1, 3, 2, π΅βˆ’2, 8, 0 and 𝐢3, 6, 7. If the angle bisector of ∠𝐡𝐴𝐢 meets the line 𝐡𝐢 at 𝐷, then the length of the projection of the vector →𝐴𝐷 on the vector →𝐴𝐢 is: (1) 37 (2) √38 2√38 2 39 (3) (4) √19 2√38

202401 Feb Shift 2Vectors
MathsMedium

Q77.Let β†’π‘Ž= ^𝑖+ 𝛼 ^𝑗+ 𝛽 ^π‘˜ , 𝛼, π›½βˆˆπ‘…. Let a vector →𝑏 be such that the angle between β†’π‘Ž and →𝑏 is πœ‹ and →𝑏 = 6, If 4 β†’π‘ŽΒ· →𝑏= 3√2, then the value of 𝛼2 + 𝛽2 | β†’π‘ŽΓ— →𝑏|2 is equal to (1) 90 (2) 75 (3) 95 (4) 85 2 is equal to

202430 Jan Shift 2Vectors
MathsMedium

Q77.Consider three vectors β†’a,β†’b, β†’c. Let |β†’a| = 2, |β†’b| = 3 and β†’a = β†’b Γ— β†’c. If Ξ± ∈[0, 3 ] is the angle between the vectors β†’b and β†’c, then the minimum value of 27|β†’c βˆ’β†’a|2 is equal to: (1) 110 (2) 124 (3) 121 (4) 105

202405 Apr Shift 2Differential Equations
MathsMedium

Q77.If y = y(x) is the solution of the differential equation dydx + 2y = sin(2x), y(0) = 43 , then y ( Ο€8 ) is equal to: JEE Main 2024 (05 Apr Shift 1) JEE Main Previous Year Paper (1) eΟ€/8 (2) eΟ€/4 (3) eβˆ’Ο€/4 (4) eβˆ’Ο€/8

202405 Apr Shift 1Differential Equations
MathsMedium

Q77.The position vectors of the vertices A, B and C of a triangle are 2 ^i - 3 ^j + 3 ^k, 2 ^i + 2 ^j + 3 ^k and - ^i + ^j + 3 ^k respectively. Let 𝑙 denotes the length of the angle bisector AD of ∠BAC where D is on the line segment BC, then 2𝑙2 equals : (1) 49 (2) 42 (3) 50 (4) 45

202427 Jan Shift 2Vectors
MathsMedium

Q77.Let β†’π‘Ž= 3 ^𝑖+ ^π‘—βˆ’2 ^π‘˜, 𝑏= 4 ^𝑖+ ^𝑗+ 7 ^π‘˜ and →𝑐= ^π‘–βˆ’3 ^𝑗+ 4 ^π‘˜ be three vectors. If a vectors →𝑝 satisfies →𝑝× →𝑏= →𝑐× →𝑏 and β†’π‘β‹…β†’π‘Ž= 0, then →𝑝⋅ ^π‘–βˆ’ ^π‘—βˆ’ ^π‘˜ is equal to (1) 24 (2) 36 (3) 28 (4) 32

202431 Jan Shift 1Vectors
MathsMedium

Q77.Let y = y(x) be the solution of the differential equation (1 + x2) dxdy + y = etanβˆ’1 x , y(1) = 0. Then y(0) is (1) 2 1 (eΟ€/2 βˆ’1) (2) 21 (1 βˆ’eΟ€/2) (3) 4 1 (1 βˆ’eΟ€/2) (4) 14 (eΟ€/2 βˆ’1)

202406 Apr Shift 1Differential Equations
MathsMedium

Q77.Let OA→ =→a, OB→ = 12→a+ 4→b and OC→ = →b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then area of the areaquadrilateralof S OABC is equal to _____ (1) 6 (2) 10 (3) 7 (4) 8

202429 Jan Shift 2Vectors
MathsMedium

Q77.Let 𝑦= 𝑦π‘₯ be the solution of the differential equation 𝑑𝑦 2π‘₯π‘₯+ 𝑦3 βˆ’π‘₯π‘₯+ π‘¦βˆ’1, 𝑦0 = 1. Then, 1 + 𝑦1 𝑑π‘₯= √2 √2 equals: (1) 4 (2) 3 4 + βˆšπ‘’ 3 βˆ’βˆšπ‘’ 2 1 (3) (4) 1 + βˆšπ‘’ 2 βˆ’βˆšπ‘’

202401 Feb Shift 1Definite Integration & Area
MathsMedium

Q78.The distance of the point 𝑄( 0, 2, – 2 ) form the line passing through the point 𝑃( 5, – 4, 3 ) and perpendicular to the lines β†’π‘Ÿ= βˆ’3 ^𝑖+ 2 ^π‘˜+ πœ†2 ^𝑖+ 3 ^𝑗+ 5 ^π‘˜, πœ†βˆˆβ„ and β†’π‘Ÿ= ^π‘–βˆ’2 ^𝑗+ ^π‘˜+ πœ‡βˆ’ ^𝑖+ 3 ^𝑗+ 2 ^π‘˜, πœ‡βˆˆβ„ is (1) √86 (2) √20 (3) √54 (4) √74

202431 Jan Shift 13D Geometry
MathsMedium

Q78.If the mirror image of the point 𝑃( 3, 4, 9 ) in the line π‘₯βˆ’1 = 𝑦+ 1 = π‘§βˆ’2 is 𝛼, 𝛽, 𝛾, then 14𝛼+ 𝛽+ 𝛾 is: 3 2 1 (1) 102 (2) 138 (3) 108 (4) 132 π‘₯+ 3 π‘¦βˆ’4 𝑧+ 1

202401 Feb Shift 23D Geometry
MathsMedium

Q78.If β†’a = Λ†i + 2Λ†j + Λ†k, b = 3(Λ†i βˆ’Λ†j + Λ†k) is equal to Γ— βˆ’ b β†’aβ‹…((β†’c β†’ β†’ b) βˆ’β†’c) (1) 32 (2) 24 (3) 20 (4) 36

202427 Jan Shift 1Vectors
MathsMedium

Q78.If the shortest distance between the lines is √n L2 : β†’r = 2(1 + ΞΌ)^i + 3(1 + ΞΌ)^j + (5 + ΞΌ)^k, ΞΌ ∈R , where gcd(m, n) = 1, then the value of m + n equals (1) 390 (2) 384 (3) 377 (4) 387

202408 Apr Shift 1Vectors
MathsMedium

Q78.Let y = y(x) be the solution of the differential equation (2x loge x) dxdy + 2y = x3 loge x, x > 0 and y (eβˆ’1) = 0. Then, y(e) is equal to (1) βˆ’3e (2) βˆ’32e (3) βˆ’23e (4) βˆ’2e

202406 Apr Shift 1Differential Equations
MathsMedium

Q78.Let a unit vector which makes an angle of 60∘ with 2^i + 2^j βˆ’^k and angle 45∘ with ^i βˆ’^k be C. Then β†’ is : C + + (βˆ’12^i 1 ^j βˆ’βˆš23 ^k) 3√2 (1) √2 + βˆ’ + 3 + 21 )^i 1 )^j + √23 )^k ^i βˆ’12 ^k (2) ( √31 ( √31 3√2 ( √31 2√2 (3) √2 ^i + + 3 3√2 1 ^j βˆ’12 ^k (4) βˆ’βˆš23 ^i + √23 ^j + ( 21 3 )^k

202404 Apr Shift 1Vectors
MathsMedium

Q78.Let β†’a = 2^i + 5^j βˆ’^k,β†’b = 2^i βˆ’2^j + 2^k andβ†’cbe three vectors such that (β†’c +^i) Γ— (β†’a + β†’b +^i) = β†’a Γ— (β†’c +^i). If β†’a β‹…β†’c = βˆ’29, then β†’c β‹…(βˆ’2^i + ^j + ^k) is equal to: (1) 15 (2) 12 (3) 10 (4) 5

202405 Apr Shift 2Vectors
MathsMedium

Q78.Let β†’a = aiΛ†i + a2Λ†j + a3Λ†k and b = b1Λ†i + b2Λ†j + b3Λ†k be two vectors such that β†’a = 1;β†’aβ‹… b = 2 and b = 4. If Γ— βˆ’3b, then the angle between b and β†’cis equal to : β†’c= 2(β†’a β†’ β†’ β†’ b) JEE Main 2024 (30 Jan Shift 1) JEE Main Previous Year Paper (1) cosβˆ’1( √32 ) (2) cosβˆ’1(βˆ’1√3 ) 2 ) 3 (3) cosβˆ’1(βˆ’βˆš32 ) (4) cosβˆ’1(

202430 Jan Shift 1Vectors
MathsMedium

Showing 1501–1525 of 10,171