Practice Questions
10,171 questions across 23 years of JEE Main β find and practise any topic!
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Q67.If x3dy + xy β dx = x2dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to : (1) βe (2) 1 + βe 2 2 (3) 3 2 βe (4) 23 + βe
Q68.Let f(x) = |x β2| and g(x) = f(f(x)), x β[0, 4]. Then β«30 (g(x) βf(x)) (1) 1 (2) 0 (3) 1 (4) 3 2 2
Q68.Let y = y(x) be the solution of the differential equation, 2+siny+1 x . dxdy = βcos and dy at x = Ο is b, then the ordered pair (a, b) is equal to dx (1) (2, 32 ) (2) (1, β1) (3) (1, 1) (4) (2, 1)
Q68.The solution of the differential equation β + 3 = 0 is dx loge(y+3x) (where C is a constant of integration) (1) x β12 (loge(y + 3x))2 = C (2) x βloge(y + 3x) = C (3) y + 3x β12 (loge x)2 = C (4) x β2 loge(y + 3x) = C
Q68.If a curve y = f(x) , passing through the point (1, 2), is the solution of the differential equation 2x2dy = (2xy + y2)dx, then f( 21 ) is equal to JEE Main 2020 (02 Sep Shift 2) JEE Main Previous Year Paper (1) 1 (2) 1 1+loge 2 1βloge 2 (3) 1 + loge 2 (4) 1+logeβ1 2
Q68.If y = ( 2Ο β1) then the function p(x) is equal to : (1) cot x (2) cosec x (3) sec x (4) tan x
Q68.If y = y(x) is the solution of the differential equation 5+ex2+y β dydx + ex = 0 satisfying y(0) = 1 then value of y(loge 13) is (1) 1 (2) β1 (3) 0 (4) 2
Q68.A vector βa = Ξ±Λi + 2Λj + Ξ²Λk(Ξ±, Ξ² βR) lies in the plane of the vectors, b = Λi + Λj and βc= Λi βΛj + 4Λk. If βa β bisects the angle between b and βc, then (1) βaβ Λi + 3 = 0 (2) βaβ Λi + 1 = 0 (3) βaβ Λk + 2 = 0 (4) βaβ Λk + 4 = 0
Q68.The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, β2, 3) and (1, 1, 0) lies on the plane (1) 2 x + y βz = 1 (2) x βy β2 z = 1 (3) x β2 y + z = 1 (4) x + 2 y βz = 1 + + +
Q68.If dy = xy ; y(1) = 1; then a value of x satisfying y(x) = e is: dx x2+y2 e (1) 1 β3e (2) 2 β2 (3) β2e (4) β3e
Q68.Let y = y(x) be the solution curve of the differential equation, (y2 βx) dxdy = 1 , satisfying y(0) = 1 . This curve intersects the Xβaxis at a point whose abscissa is (1) 2 βe (2) βe (3) 2 (4) 2 + e β β β β β
Q68.If f '(x) = tanβ1(sec x + tan x), βΟ2 < x < Ο2 and f(0) = 0 , then f(1) is equal to: (1) Ο+1 (2) 1 4 4 (3) Οβ1 (4) Ο+2 4 4
Q68.The general solution of the differential equation β1 + x2 + y2 + x2y2 + xy dxdy = 0 (where C is a constant of integration) + C (1) β1 + y2 + β1 + x2 = 12 loge( β1+x2+1β1+x2β1 ) + C (2) β1 + y2 ββ1 + x2 = 12 loge( β1+x2+1β1+x2β1 ) + C (3) β1 + y2 + β1 + x2 = 12 loge( β1+x2β1β1+x2+1 ) (4) 1 β1+x2+1 + C β1 + y2 ββ1 + x2 = 2 loge( β1+x2β1 )
Q68.Let βa = Λi β2Λj + Λk and b = Λi βΛj + Λk, be two vectors. If βc, is a vector such that b Γβc= b Γβa and βcβ βa = 0, β then βcβ b, is equal to. (1) β32 (2) 21 (3) β12 (4) β1
Q68.Let the volume of a parallelepiped whose coterminous edges are given by u = Λi + Λj + Ξ»Λk,βv = Λi + Λj + 3Λk and β β β w = 2Λi + Λj + Λk be 1 cu. unit. If ΞΈ be the angle between the edges u and w, then the value of cos ΞΈ can be (1) 7 (2) 7 6β6 6β3 (3) 5 (4) 5 7 3β3 yβ8
Q69.Let D be the centroid of the triangle with vertices (3, β1) , (1, 3) and (2, 4) . Let P be the point of intersection of the lines x + 3y β1 = 10 and 3x βy + 1 = 0 . Then, the line passing through the points D and P also passes through the point: (1) (β9, β6) (2) (9,7) (3) (7,6) (4) (β9, β7)
Q69.Let y = y(x) be the solution of the differential equation cos x dxdy + 2y sin x = sin 2x, x β(0, Ο2 ) If y(Ο/3) = 0, then y(Ο/4) is equal to : (1) 2 ββ2 (2) 2 + β2 (3) β2 β2 (4) 1 β1 β2
Q69.The distance of the point (1, β2, 3) from the plane x βy + z = 5 measured parallel to the line x2 = 3y = β6z is : (1) 7 (2) 1 5 (3) 1 (4) 7 7
Q69.If the volume of a parallelopiped, whose coterminous edges are given by the vectors βa = Λi + Λj + nΛk , β b = 2Λi + 4Λj β nΛk and,βc= Λi + nΛj + 3Λk (n β₯0) is 158 cubic units, then : β (1) βaβ βc= 17 (2) b β βc= 10 (3) n = 7 (4) n = 9
Q69.Let y = y(x) be the solution of the differential equation, xyβ² βy = x2(x cos x + sin x), x > 0. If y(Ο) = Ο, then yβ²β²( Ο2 ) + y( Ο2 ) is equal to : (1) 2 + Ο2 (2) 1 + Ο2 + Ο24 (3) 2 + Ο2 + Ο24 (4) 1 + Ο2 b whereβa = xΛi β2Λj + 3Λk, βb = β2Λi + xΛj βΛk and
Q69.Let βa, b and βc, be three unit vectors such that βa+ b +βc= 0. If Ξ» =βaβ b + b β βc+βcβ βa and β β β β , is equal to. d =βaΓ b + b Γβc+βcΓβa, then the order pair, (Ξ», d) 3 β , 3βaΓβc) (1) ( 2 (2) (β3 2 , 3βcΓ b) 2 , 3b (3) ( 3 β (4) β Γβc) (β3 2 , 3βaΓ b)
Q69.The shortest distance between the lines xβ3 3 = β1 = zβ31 and x+3β3 = y+72 = zβ64 is (1) 2β30 (2) 72 β30 (3) 3β30 (4) 3
Q69.The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point (1) (0, 6, β2) (2) (β2, 0, 1) (3) (0, β6, 2) (4) (2, 0, β 1)
Q69.The lines βr= (Λi βΛj) l(2Λi Λk) and βr= (2Λi βΛj) m(Λi + Λj βΛk) (1) Do not intersect for any values of l and m (2) Intersect for all values of l and m (3) Intersect when l = 2 and m = 21 (4) Intersect when l = 1 and m = 2
Q69.Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6) .Then the image of R in the plane P is (1) (6, 5, 2) (2) (6, 5, β2) (3) (4, 3, 2) (4) (3, 4, β2)