Practice Questions
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Q72.A variable line L passes through the point (3, 5) and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle OAB, where O is the origin, is : (1) 30 (2) 25 (3) 40 (4) 35
Q72.Let π: π βπ and π: π βπ be defined as ππ₯= logππ₯, π₯> 0 and ππ₯= π₯, π₯β₯0 . Then, πππ: π βπ is: πβπ₯, π₯β€0 ππ₯, π₯< 0 (1) one-one but not onto (2) neither one-one nor onto (3) onto but not one-one (4) both one-one and onto
Q72.Let f : [β1, 2] βR be given by f(x) = 2x2 + x + [x2] β[x], where [t] denotes the greatest integer less than or equal to t. The number of points, where f is not continuous, is : (1) 5 (2) 6 (3) 3 (4) 4
Q72.Given that the inverse trigonometric function assumes principal values only. Let x, y be any two real numbers in [β1, 1] such that cosβ1 x βsinβ1 y = Ξ±, βΟ2 β€Ξ± β€Ο. Then, the minimum value of x2 + y2 + 2xy sin Ξ± is (1) 0 (2) -1 (3) 1 2 (4) β12 72xβ9xβ8x+1
Q72.Let the sum of the maximum and the minimum values of the function f(x) = 2x2+3x+82x2β3x+8 be mn , where gcd(m, n) = 1. Then m + n is equal to : (1) 195 (2) 201 (3) 217 (4) 182 2x , x < 0
Q72.The number of critical points of the function f(x) = (x β2)2/3(2x + 1) is (1) 1 (2) 2 (3) 0 (4) 3 6
Q72.Let the range of the function f(x) = 2+sin 3x+cos1 3x , x βR be [a, b]. If Ξ± and Ξ² are respectively the A.M. and the G.M. of a and b, then Ξ±Ξ² is equal to (1) Ο (2) βΟ (3) 2 (4) β2
Q72.Let a and b be real constants such that the function π defined by ππ₯= π₯2 + 3π₯+ π, π₯β€1 be differentiable ππ₯+ 2, π₯> 1 2 on π . Then, the value of β«-2 ππ₯ππ₯ equals 15 19 (1) (2) 6 6 (3) 21 (4) 17
Q72.If π= sinβ1sin5 and π= cosβ1cos5, then π2 + π2 is equal to (1) 4π2 + 25 (2) 8π2 β40π+ 50 (3) 4π2 β20π+ 50 (4) 25
Q72.Let y = loge( 1βx21+x2 ), (1) 732 (2) 746 (3) 742 (4) 736
Q72.If the function f(x) = sin 3x+Ξ± sin xβΞ² cos 3x , x βR , is continuous at x = 0 , then f(0) is equal to : x3 (1) 2 (2) -2 (3) 4 (4) -4
Q72.If the domain of the function f(x) = cosβ1( 2β|x|4 ) equal to : (1) 12 (2) 9 (3) 11 (4) 8
Q72.Consider the function f : [ 12 , 1] βR defined by f(x) = 4β2x3 β3β2x β1. Consider the statements (I) The curve y = f(x) intersects the x-axis exactly at one point (II) The curve y = f(x) intersects the x-axis at x = cos 12Ο Then (1) Only (II) is correct (2) Both (I) and (II) are incorrect (3) Only (I) is correct (4) Both (I) and (II) are correct
Q72.The function f: R->R, f(x) = x2+2xβ15 , x βR is x2β4x+9 (1) one-one but not onto. (2) both one-one and onto. (3) onto but not one-one. (4) neither one-one nor onto. JEE Main 2024 (06 Apr Shift 1) JEE Main Previous Year Paper 1 ), x β 0 x then
Q72.Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and hβ²(0) = hβ²(1) = 2. If g(x) = h (ex)eh(x) , then gβ²(0) is equal to: (1) 5 (2) 4 (3) 8 (4) 3
Q73.If loge y = 3 sinβ1 x, then (1 βx2)yβ²β² βxyβ² at x = 12 is equal to (1) 3eΟ/6 (2) 9eΟ/2 (3) 3eΟ/2 (4) 9eΟ/6 y β₯0, y(0) = 0. Then at x = 2, yβ²β² + y + 1 is equal to
Q73.If the function f(x) = 2x3 β9x2 + 12a2x + 1, a > 0 has a local maximum at x = Ξ± and a local minimum at x = Ξ±2 , then Ξ± and Ξ±2 are the roots of the equation : JEE Main 2024 (08 Apr Shift 2) JEE Main Previous Year Paper (1) x2 β6x + 8 = 0 (2) x2 + 6x + 8 = 0 (3) 8x2 + 6x β1 = 0 (4) 8x2 β6x + 1 = 0 = Ο6 . Then eΞ± and eβΞ± are the roots of the equation :
Q73. x2 β§ 1βcos where Ξ±, Ξ² βR. If f is continuous at Let f : R βR be a function given by f(x) = β¨ Ξ±, x = 0, Ξ²β1βcos x β© x , x > 0 x = 0, then Ξ±2 + Ξ²2 is equal to : (1) 3 (2) 12 (3) 48 (4) 6 JEE Main 2024 (04 Apr Shift 1) JEE Main Previous Year Paper
Q73.If f(x) = {x30 sin, x (= 0 (1) f β²β² ( Ο2 ) = 24βΟ22Ο (2) f β²β² ( Ο2 ) = 12βΟ22Ο (3) f β²β²(0) = 1 (4) f β²β²(0) = 0
Q73.The function f(x) = 2x + 3x 23 , x βR, has (1) exactly one point of local minima and no point of (2) exactly one point of local maxima and no point local maxima of local minima (3) exactly one point of local maxima and exactly (4) exactly two points of local maxima and exactly one point of local minima one point of local minima
Q73.Let π: π βπ be defined as πβπcos2π₯ ; π₯< 0 π₯2 ππ₯= π₯2 + ππ₯+ 2; 0 β€π₯β€1 2π₯+ 1; π₯> 1 If π is continuous everywhere in π and π is the number of points where π is NOT differential then π + π + π + π equals: JEE Main 2024 (01 Feb Shift 1) JEE Main Previous Year Paper (1) 1 (2) 4 (3) 3 (4) 2 1
Q73.Suppose f(x) = (2x+2βx) tan xβtanβ1(x2βx+1) . Then the value of f β²(0) is equal to (7x2+3x+1)3 (1) Ο (2) 0 (3) βΟ (4) Ο2 Ο + = 4 ( Ο + a) β2, then the value of a is
Q73.If the function f(x) = ( x1 ) 2x; x > 0 attains the maximum value at x = 1e then : (1) eΟ < Οe (2) eΟ > Οe (3) (2e)Ο > Ο(2e) (4) e2Ο < (2Ο)e 1
Q73.Let ππ₯= 2π₯2 + 5π₯- 3, π₯βπ . If π and π denote the number of points where π is not continuous and not differentiable respectively, then π+ π is equal to: (1) 5 (2) 2 (3) 0 (4) 3
Q73.Let g : R βR be a non constant twice differentiable such that gβ²( 21 ) = gβ²( 23 ). If a real valued function f is defined as f(x) = 12 [ g(x) + g(2 βx)], then (1) f β²β²(x) = 0 for atleast two x in (0, 2) (2) f β²β²(x) = 0 for exactly one x in (0, 1) (3) f β²β²(x) = 0 for no x in (0, 1) (4) f β²( 23 ) + f β²( 21 ) = 1