Practice Questions
1,770 questions across 23 years of JEE Main β find and practise any topic!
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Q70.The equations of sides AB and AC of a triangle ABC are (Ξ» + 1)x + Ξ»y = 4 and Ξ»x + (1 βΞ»)y + Ξ» = 0 respectively. Its vertex A is on the yβaxis and its orthocentre is (1, 2). The length of the tangent from the point C to the part of the parabola y2 = 6x in the first quadrant is (1) β6 (2) 2β2 (3) 2 (4) 4 JEE Main 2023 (24 Jan Shift 2) JEE Main Previous Year Paper
Q70.Let πΌ be a root of the equation π- ππ₯2 + π- ππ₯+ π- π= 0 where π, π, π are distinct real numbers such that πΌ2 πΌ1 π- π2 π- π2 π- π2 the matrix 1 1 1 is singular. Then the value of is π- ππ- π+ π- ππ- π+ π- ππ- π π π π (1) 6 (2) 3 (3) 9 (4) 12
Q70.Let A be a point on the x-axis. Common tangents are drawn from A to the curves x2 + y2 = 8 and y2 = 16x . If one of these tangents touches the two curves at Q and R, then (QR)2 is equal to (1) 64 (2) 76 (3) 81 (4) 72
Q70.A triangle is formed by X -axis, Y -axis and the line 3x + 4y = 60 . Then the number of points P(a, b) which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is _____ .
Q70.If the radius of the largest circle with centre (2, 0) inscribed in the ellipse x2 + 4y2 = 36 is r, then 12 r2 is equal to (1) 115 (2) 92 (3) 69 (4) 72
Q70.Let B and C be the two points on the line y + x =0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y β2x = 2 such that ΞABC is an equilateral triangle. Then, the area of the ΞABC is (1) 3β3 (2) 2β3 (3) 8 (4) 10 β3 β3
Q70.Let π΄ be a 2 Γ 2 matrix with real entries such that π΄' = πΌπ΄+ 1, where πΌββ- -1, 1., If det π΄2 - π΄= 4, the sum of all possible values of πΌ is equal to 3 (1) 0 (2) 2 (3) 2 (4) 5 2
Q71.Let f, g and h be the real valued functions defined on R as x , x β 0 sin(x+1) |x| (x+1) , x β β1 f(x) = , g(x) = and h(x) = 2[x] βf(x), where [x] is the greatest integer { 1, x = 0 { 1, x = β1 β€x. Then the value of lim g(h(x β1)) is xβ1 (1) 1 (2) sin(1) (3) β1 (4) 0
Q71.Let the tangent to the parabola y2 = 12x at the point (3, Ξ±) be perpendicular to the line 2x + 2y = 3 . Then the square of distance of the point (6, β4) from the normal to the hyperbola Ξ±2x2 β9y2 = 9Ξ±2 at its point (Ξ± β1, Ξ± + 2) is equal to .............
Q71.Let the eccentricity of an ellipse x2 + y2 = 1 is reciprocal to that of the hyperbola 2x2 β2y2 = 1 . If the a2 b2 ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____. JEE Main 2023 (06 Apr Shift 2) JEE Main Previous Year Paper lim 2 β2 3 2 β2 5 . . . 2 β2 2n+1 )(2 ). (2 )} is equal to
Q71.A triangle is formed by the tangents at the point (2, 2) on the curves y2 = 2x and x2 + y2 = 4x, and the line x + y + 2 = 0. If r is the radius of its circumcircle, then r2 is equal to
Q71.Let the system of linear equations βx + 2y β9z = 7 βx + 3y + 7z = 9 β2x + y + 5z = 8 β3x + y + 13z = Ξ» has a unique solution x = Ξ±, y = Ξ², z = Ξ³ . Then the distance of the point (Ξ±, Ξ², Ξ³) from the plane 2x β2y + z = Ξ» is (1) 11 (2) 7 (3) 9 (4) 13
Q71.Let P( 2β3β7 β7 perpendicular and pass through the origin. If 1 + 1 = pq , where p and q are coprime, then p + q is (PQ)2 (RS)2 equal to (1) 147 (2) 143 (3) 137 (4) 157
Q71. (β3x+1+β3xβ1) 6 +(β3x+1ββ3xβ1) 6 lim 6 6 x3 xββ (x+βx2β1) +(xββx2β1) (1) is equal to 272 (2) is equal to 9 (3) does not exist (4) is equal to 27
Q71.Let R be the focus of the parabola y2 = 20x and the line y = mx + c intersect the parabola at two points P and Q. Let the points G(10, 10) be the centroid of the triangle PQR . If c βm = 6 , then PQ2 is (1) 296 (2) 325 (3) 317 (4) 346
Q71.Points P(β3, 2), Q(9, 10) and R(Ξ±, 4) lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S . If S lies on the line 2x βky = 1 , then k is equal to _____ .
Q72.If Ξ± > Ξ² > 0 are the roots of the equation ax2 + bx + 1 = 0 , and 1 1βcos(x2+bx+a) 2 1 1 k is equal to lim ( 2(1βΞ±x)2 ) = k ( Ξ² β1Ξ± ), then xβ1Ξ± (1) 2Ξ² (2) Ξ± (3) 2Ξ± (4) Ξ²
Q72.If the tangent at a point P on the parabola y2 = 3x is parallel to the line x + 2y = 1 and the tangents at the x2 y2 points Q and R on the ellipse 4 + 1 = 1 are perpendicular to the line x βy = 2, then the area of the triangle PQR is: (1) 9 (2) 5β3 β5 (3) 3 2 β5 (4) 3β5
Q72.The equation π₯2 β 4π₯+ [π₯] + 3 = π₯[π₯], where [π₯] denotes the greatest integer function, has: (1) exactly two solutions in ( - β, β) (2) no solution (3) a unique solution in ( - β, 1 ) (4) a unique solution in ( - β, β) Q73. π₯2sin1 π₯β 0 Let ππ₯= π₯; , then at π₯= 0 0; π₯= 0 (1) π is continuous but not differentiable (2) π is continuous but π' is not continuous (3) both π and π' are continuous (4) π' is continuous but not differentiable
Q72.The equations of two sides of a variable triangle are x = 0 and y = 3 , and its third side is a tangent to the parabola y2 = 6x . The locus of its circumcentre is : (1) 4y2 β18y β3x β18 = 0 (2) 4y2 + 18y + 3x + 18 = 0 (3) 4y2 β18y + 3x + 18 = 0 (4) 4y2 β18y β3x + 18 = 0 JEE Main 2023 (25 Jan Shift 2) JEE Main Previous Year Paper
Q72. nββ{(2 1 1 1 1 1 1 (1) 1 (2) 0 (3) β2 (4) 1 β2
Q72.Let π: 2, 4 ββ be a differentiable function such that π₯logππ₯π'π₯+ logππ₯ππ₯+ ππ₯β₯1, π₯β2, 4 with π2 = 2 and 1 π4 = 2. Consider the following two statements: (A) ππ₯β€1, for all π₯β2, 4 (B) ππ₯β₯1 / 8, for all π₯β2, 4 Then, (1) Neither statement ( π΄) nor statement ( π΅) is (2) Only statement ( π΅) is true true (3) Both the statements ( π΄) and ( π΅) are true (4) Only statement ( π΄) is true β1 + π2π₯ππ₯ is equal to
Q72.Let 5ππ₯+ 4π π₯= π₯+ 3, π₯> 0 . Then 18 β«1 ππ₯ππ₯ is equal to (1) 5 loge2 + 3 (2) 10 loge2 + 6 (3) 10 loge2 - 6 (4) 5loge2 - 3 β 3 π₯- 3
Q73.Let π΄= {π₯ββ: π₯+ 3 + π₯+ 4 β€3}, π΅= π₯ββ: 3π₯βπ= 1 10π < 3-3π₯, where [π‘] denotes greatest integer function. Then, (1) π΅βπΆ, π΄β π΅ (2) π΄β©π΅= π (3) π΄βπ΅, π΄β π΅ (4) π΄= π΅
Q73.Let [x] denote the greatest integer function and f(x) = max{1 + x + [x], 2 + x, x + 2[x]}, 0 β€x β€2 , where f is not continuous and n be the number of points in (0, 2), where f is not differentiable. Then (m + n)2 + 2 is equal to (1) 2 (2) 11 (3) 6 (4) 3 Ξ±, Ξ² > 0 , then Ξ±4 βΞ²4 is equal to dx = Ξ±1 loge( Ξ±+1Ξ² ),