Practice Questions
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Q71.The value of 1+2β3+4+5β6+β¦+(3nβ2)+(3nβ1)β3n lim is nββ β2n4+4n+3ββn4+5n+4 (1) β2+1 + 2 (2) 3(β2 1) (3) 3 + 2 (β2 1) (4) 2β23
Q71.Let the tangents at the points A(4, β11) and B(8, β5) on the circle x2 + y2 β3x + 10y β15 = 0 , intersect at the point C . Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to (1) 3β3 (2) 2β13 4 (3) β13 (4) 2β13 3 Q72. 1βcos(x2β4px+q2+8q+16) β§ , x β 2p Let x = 2 be a root of the equation x2 + px + q = 0 and f(x) = (xβ2p)4 . Then β¨ β© 0, x = 2p xβ2p+[f(x)]lim where [β ] denotes greatest integer function, is (1) 2 (2) 1 (3) 0 (4) β1
Q71.The set of values of a for which xβa([xlim β5] β[2x + 2]) = 0 , where, [ΞΆ] denotes the greatest integer less than or equal to ΞΆ is equal to (1) (β7. 5, β6. 5) (2) (β7. 5, β6. 5] (3) [β7. 5, β6. 5] (4) [β7. 5, β6. 5)
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.Let π denote the set of all real values of π such that the system of equations ππ₯+ π¦+ π§= 1 π₯+ ππ¦+ π§= 1 π₯+ π¦+ ππ§= 1 is inconsistent, then βπβππ2 + π is equal to (1) 2 (2) 12 (3) 4 (4) 6 - 1
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. (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 π΄= 2, 3, 4 and π΅= 8, 9, 12. Then the number of elements in the relation π = π1, π1, π2, π2 βπ΄Γ π΅, π΄Γ π΅: π1 divides π2 and π2 divides π1 is (1) 36 (2) 24 (3) 18 (4) 12 Q72. 5! 6! 7! 1 If π΄= 6! 7! 8! , then adj adj 2π΄ is equal to 5!6!7! 7! 8! 9! (1) 220 (2) 28 (3) 212 (4) 216
Q72.Let the system of linear equations π₯+ π¦+ ππ§= 2 2π₯+ 3π¦- π§= 1 3π₯+ 4π¦+ 2π§= π have infinitely many solutions. Then the system π+ 1 π₯+ 2π- 1 π¦= 7 2π+ 1π₯+ π+ 5π¦= 10 has : (1) infinitely many solutions (2) unique solution satisfying π₯- π¦= 1 (3) no solution (4) unique solution satisfying π₯+ π¦= 1
Q72.The range of ππ₯= 4sin-1 π₯2 is π₯2 + 1 (1) [0, 2π] (2) [0, π] (3) [0, 2π) (4) [0, π) π 4 π-π₯tan 50 π₯ππ₯ Q73. π-π4 + β«0 The value of π β«04 π-π₯(tan49π₯+ tan51π₯)ππ₯ (1) 51 (2) 50 (3) 25 (4) 49 JEE Main 2023 (13 Apr Shift 2) JEE Main Previous Year Paper
Q72.Let π be the set of all solutions of the equation cos-12π₯- 2cos-1β1 - π₯2 = π, π₯β-1 2, 12. Then βπ₯βπ2sin-1π₯2 is equal to -2π (1) 0 (2) 3 (3) π- sin-1β3 (4) π- 2sin-1β3 4 4
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. nββ{(2 1 1 1 1 1 1 (1) 1 (2) 0 (3) β2 (4) 1 β2
Q72.The converse of ((~p) β§q) βr is (1) ((~p) β¨q) βr (2) (~r) βp β§q (3) (~r) β((~p) β§q) (4) (p β¨(~q)) β(~r)
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.If the domain of the function f(x) = loge(4x2 + 11x + 6) + sinβ1(4x + 3) + cosβ1( 10x+63 ) is (Ξ±, Ξ²] , then 36|Ξ± + Ξ²| is equal to (1) 54 (2) 72 (3) 63 (4) 45
Q72.The number of values of r β{p, q, ~p, ~q} for which ((p β§q) β(r β¨q) β§((p β§r) βq) is a tautology, is : (1) 1 (2) 2 (3) 4 (4) 3
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. xβ0((lim 1βcos2(3x)cos3(4x) )( (loge(2x+1))5sin3(4x) )) is equal to (1) 15 (2) 9 (3) 18 (4) 24
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.The statement (p β§(~q)) β(p β(~q)) is (1) equivalent to (~p) β¨(~q) (2) a tautology (3) equivalent to p β¨q (4) a contradiction
Q72.If πΌπ₯= β«πsin2π₯cosπ₯ sin2π₯- sinπ₯ππ₯ and πΌ0 = 1, then πΌ π is equal to 3 (1) -1 34 (2) 1 34 2π 2π 3 (3) -π 4 (4) π 34
Q72.If the domain of the function ππ₯= where π₯ is greatest integer β€π₯, is [2, 6 ) , then its range is 1 + π₯2, 5 2 9 27 18 9 5 2 (1) 26, 5 - 29, 109, 89, 53 (2) 26, 5 (3) 5 2 - 9 27 18 9 (4) 5 2 37, 5 29, 109, 89, 53 37, 5 3
Q72.Which of the following statements is a tautology? (1) p β(p β§(p βq)) (2) (p β§q) β(~(p) βq) (3) (p β§(p βq)) β~q (4) p β¨(p β§q)