Practice Questions
10,171 questions across 23 years of JEE Main β find and practise any topic!
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Q82.The slope of the line touching both the parabolas y2 = 4x and x2 = β32y is (1) 1 (2) 2 8 3 (3) 1 (4) 3 2 2 x dx, is equal to
Q82.For the curve y = 3 sin ΞΈ cos ΞΈ, x = eΞΈ sin ΞΈ, 0 β€ΞΈ β€Ο, the tangent is parallel to x-axis when ΞΈ is: (1) 3Ο (2) Ο 4 2 (3) Ο (4) Ο 4 6
Q82.Let f and g be two differentiable functions on R such that f β²(x) > 0 and gβ²(x) < 0 for all x βR. Then for all x : (1) f(g(x)) > f(g(x β1)) (2) f(g(x)) > f(g(x + 1)) (3) g(f(x)) > g(f(x β1)) (4) g(f(x)) < g(f(x + 1)) JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper
Q82.If non-zero real numbers b and c are such that min f(x) > max g(x), where f(x) = x2 + 2bx + 2c2 and g(x) = βx2 β2cx + b2, (x βR); then cb lies in the interval , (1) (β2, β) (2) [ 12 1 ) β2 , β2] (3) (0, 12 ) (4) [ β21
Q83.The integral β«(1 + x β1x )ex+ 1 JEE Main 2014 (06 Apr) JEE Main Previous Year Paper (1) (x + 1)ex+ x1 + c (2) βxex+ x1 + c (3) (x β1)ex+ x1 + c (4) xex+ x1 + c Ο x 2 β4 sin x2 dx equals
Q83. sin2 x cos2 x The integral dx is equal to: β« 2 (sin3 x + cos3 x) 1 1 (1) + c (2) β + x (1+cot3 x 3(1+tan3 c) (3) sin3 x + c (4) β cos3 x + c (1+cos3 x 3(1+sin3 x
Q83.The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius = β3 is: (1) 3 4 β3Ο (2) 83 β3Ο (3) 4Ο (4) 2Ο > 0) is equal to:
Q84.The integral β«x cosβ1 ( 1+x21βx2 )dx(x (1) βx + (1 + x2) tanβ1 x + c (2) x β(1 + x2) cotβ1 x + c (3) βx + (1 + x2) cotβ1 x + c (4) x β(1 + x2) tanβ1 x + c
Q84.Let, the function F be defined as F(x) = β«x1 ett dt, x > 0, then the value of the integral β«x1 t+aet dt, where a > 0, is (1) ea[F(x) βF(1 + a)] (2) eβa[F(x + a) βF(a)] (3) ea[F(x + a) βF(1 + a)] (4) eβa[F(x + a) βF(1 + a)]
Q84.The integral β« β1 + 4 sin2 0 (1) 4β3 β4 (2) 4β3 β4 βΟ3 (3) Ο β4 (4) 2Ο3 β4 β4β3
Q84.If [ ] denotes the greatest integer function, then the integral β«Ο0 [cos xdx is equal to: (1) Ο (2) 0 2 (3) β1 (4) βΟ2
Q85.If for a continuous function f(x), β«tβΟ(f(x) + xdx) = Ο2 βt2 , for all t β₯βΟ, then f (βΟ3 ) is equal to: (1) Ο (2) Ο2 (3) Ο (4) Ο 3 6
Q85.The area of the region (in square units ) above the x-axis bounded by the curve y = tan x, 0 β€x β€Ο2 and the tangent to the curve at x = Ο4 is (1) 2 1 (log 2 β12 ) (2) 12 (1 + log 2) (3) 1 2 (1 βlog 2) (4) 12 (log 2 + 12 )
Q85.Let A = {(x, y) : y2 β€4x, y β2x β₯β4}. The area of the region A in square units is (1) 10 (2) 8 (3) 9 (4) 11
Q85.If for n β₯1, Pn = β«e1 (log xn)dx, then P10 β90P8 is equal to: (1) β9 (2) 10e (3) β9e (4) 10 Ξ¦, is given by
Q86.If the differential equation representing the family of all circles touching x-axis at the origin is (x2 βy2) dxdy = g(x)y, then g(x) equals (1) 1 x2 (2) 2x 2 (3) 1 x (4) 2x2 2 β β β
Q86.The general solution of the differential equation, sin 2x βy = 0, is : dx ( dy ββtan x) (1) yβtan x = x + c (2) yβcot x = tan x + c (3) yβtan x = cot x + c (4) yβcot x = x + c
Q86.If the general solution of the differential equation yβ² = xy + Ξ¦ ( xy ), for some function y ln |cx| = x, where c is an arbitrary constant, then Ξ¦(2) is equal to: (1) 4 (2) 1 4 (3) β4 (4) β14 β
Q86.Let the population of rabbits surviving at a time t be governed by the differential equation dp(t) . If p(0) = 100, then p(t) equals dt = 12 {p(t) β400} (1) 600 β 500 e 2t (2) 400 β300 e βt2 (3) 400 β 300 et/2 (4) 300 β200 e βt2 2 β β β β a b then Ξ» is equal to
Q86.If dxdy + ytan x=sin 2x and y(0) = 1, then y(Ο) is equal to (1) β1 (2) 5 (3) 1 (4) β5 β β β
Q87.If x = 3Λi β6Λj βΛk , y = Λi + 4Λj β3Λk and βz= 3Λi β4Λj β12Λk, then the magnitude of the projection of x Γβy on βzis (1) 14 (2) 12 (3) 15 (4) 10
Q87.If Γβb βb Γβc c = Ξ» [βa Γβa] [ c] (1) 0 (2) 1 (3) 2 (4) 3 yβ3
Q87.If βa = 2, b = 3 and 2βaβ b = 5, then 2βa+ b equals : (1) 5 (2) 7 (3) 17 (4) 1 yβ2
Q87.If |βc|2 = 60 and βc Γ (^i + 2^j + 5^k) = 0, then a value of βc β (β7^i + 2^j + 3^k) is: (1) 4β2 (2) 12 (3) 24 (4) 12β2 yβ2
Q88.A symmetrical form of the line of intersection of the planes x = ay + b and z = cy + d is (1) xβb a = yβ11 = zβdc (2) xβbβaa = yβ11 = zβdβcc (3) xβa b = yβ01 = zβcd (4) xβbβab = yβ10 = zβdβcd