Practice Questions

Q71.Let the mean of the data x 1 3 5 7 9 Frequency (f) 4 24 28 α 8 be 5. If m and σ2 are respectively the mean deviation about the mean and the variance of the data, then 3α m+σ2 is equal to _______. JEE Main 2023 (13 Apr Shift 1) JEE Main Previous Year Paper

Q71.Let 𝐴= 𝑑= 𝐴≠0 and 𝐴- d Adj 𝐴= 0. Then 𝑝 𝑞, (1) 1 + 𝑑2 = 𝑚+ 𝑞2 (2) 1 + 𝑑2 = 𝑚+ 𝑞2 (3) 1 + 𝑑2 = 𝑚2 + 𝑞2 (4) 1 + 𝑑2 = 𝑚2 + 𝑞2

Q71.Points P(−3, 2), Q(9, 10) and R(α, 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 −ky = 1 , then k is equal to _____ .

Q71. (√3x+1+√3x−1) 6 +(√3x+1−√3x−1) 6 lim 6 6 x3 x→∞ (x+√x2−1) +(x−√x2−1) (1) is equal to 272 (2) is equal to 9 (3) does not exist (4) is equal to 27

Q71.If the system of equations 2𝑥+ 𝑦- 𝑧= 5 2𝑥- 5𝑦+ 𝜆𝑧= 𝜇 𝑥+ 2𝑦- 5𝑧= 7 has infinitely many solutions, then ( 𝜆+ 𝜇) 2 + ( 𝜆- 𝜇) 2 is equal to (1) 904 (2) 916 (3) 912 (4) 920

Q71.The ordinates of the points P and Q on the parabola with focus (3, 0) and directrix x = −3 are in the ratio β2 3 : 1 . If R(α, β) is the point of intersection of the tangents to the parabola at P and Q, then α is equal to

Q71.Let P( 2√3√7 √7 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 𝑓𝑥= 𝑥2 - 𝑥+ -𝑥+ 𝑥, where 𝑥∈ℝ and 𝑡 denotes the greatest integer less than or equal to 𝑡. Then, 𝑓 is (1) continuous at 𝑥= 0, but not continuous at 𝑥= 1 (2) continuous at 𝑥= 1, but not continuous at 𝑥= 0 (3) continuous at 𝑥= 0 and 𝑥= 1 (4) not continuous at 𝑥= 0 and 𝑥= 1 1

Q71.Let the tangent to the parabola y2 = 12x at the point (3, α) be perpendicular to the line 2x + 2y = 3 . Then the square of distance of the point (6, −4) from the normal to the hyperbola α2x2 −9y2 = 9α2 at its point (α −1, α + 2) is equal to .............

Q71.The value of 1+2−3+4+5−6+…+(3n−2)+(3n−1)−3n lim is n→∞ √2n4+4n+3−√n4+5n+4 (1) √2+1 + 2 (2) 3(√2 1) (3) 3 + 2 (√2 1) (4) 2√23

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

Q72. n→∞{(2 1 1 1 1 1 1 (1) 1 (2) 0 (3) √2 (4) 1 √2

Q72.The equation 𝑥2 – 4𝑥+ [𝑥] + 3 = 𝑥[𝑥], where [𝑥] denotes the greatest integer function, has: (1) exactly two solutions in ( - ∞, ∞) (2) no solution (3) a unique solution in ( - ∞, 1 ) (4) a unique solution in ( - ∞, ∞) Q73. 𝑥2sin1 𝑥≠0 Let 𝑓𝑥= 𝑥; , then at 𝑥= 0 0; 𝑥= 0 (1) 𝑓 is continuous but not differentiable (2) 𝑓 is continuous but 𝑓' is not continuous (3) both 𝑓 and 𝑓' are continuous (4) 𝑓' is continuous but not differentiable

Q72.If 𝐼𝑥= ∫𝑒sin2𝑥cos𝑥 sin2𝑥- sin𝑥𝑑𝑥 and 𝐼0 = 1, then 𝐼 𝜋 is equal to 3 (1) -1 34 (2) 1 34 2𝑒 2𝑒 3 (3) -𝑒 4 (4) 𝑒 34

Q72.If the domain of the function f(x) = loge(4x2 + 11x + 6) + sin−1(4x + 3) + cos−1( 10x+63 ) is (α, β] , then 36|α + β| is equal to (1) 54 (2) 72 (3) 63 (4) 45

Q72.Let p and q be two statements. Then ~(p ∧(p →~q) is equivalent to (1) p ∨(p ∧(~q)) (2) p ∨((~p) ∧q) (3) (~p) ∨q (4) p ∨(p ∧q)

Q72.If the domain of the function 𝑓𝑥= where 𝑥 is greatest integer ≤𝑥, is [2, 6 ) , then its range is 1 + 𝑥2, 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.The number of values of r ∈{p, q, ~p, ~q} for which ((p ∧q) ⇒(r ∨q) ∧((p ∧r) ⇒q) is a tautology, is : (1) 1 (2) 2 (3) 4 (4) 3

Q72.The statement (p ∧(~q)) ⇒(p ⇒(~q)) is (1) equivalent to (~p) ∨(~q) (2) a tautology (3) equivalent to p ∨q (4) a contradiction

Q72.Which of the following statements is a tautology? (1) p →(p ∧(p →q)) (2) (p ∧q) →(~(p) →q) (3) (p ∧(p →q)) →~q (4) p ∨(p ∧q)

Q72.Among the two statements (S1) : (p ⇒q) ∧(p ∧(~q)) is a contradiction and (S2) : (p ∧q) ∨((~p) ∧q) ∨(p ∧(~q)) ∨((~p) ∧(~q)) is a tautology (1) only (S2) is true (2) only (S1) is true (3) both are false (4) both are true

Q72.Let 5𝑓𝑥+ 4𝑓 𝑥= 𝑥+ 3, 𝑥> 0 . Then 18 ∫1 𝑓𝑥𝑑𝑥 is equal to (1) 5 loge2 + 3 (2) 10 loge2 + 6 (3) 10 loge2 - 6 (4) 5loge2 - 3 ∞ 3 𝑥- 3

Q72.Let 𝑆 be the set of all solutions of the equation cos-12𝑥- 2cos-1√1 - 𝑥2 = 𝜋, 𝑥∈-1 2, 12. Then ∑𝑥∈𝑆2sin-1𝑥2 is equal to -2𝜋 (1) 0 (2) 3 (3) 𝜋- sin-1√3 (4) 𝜋- 2sin-1√3 4 4

Q72. x→0((lim 1−cos2(3x)cos3(4x) )( (loge(2x+1))5sin3(4x) )) is equal to (1) 15 (2) 9 (3) 18 (4) 24

Q72.Consider the following statements: P : I have fever Q : I will not take medicine R : I will take rest The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to: JEE Main 2023 (30 Jan Shift 2) JEE Main Previous Year Paper (1) ((~P) ∨~Q) ∧((~P) ∨R) (2) ((~P) ∨−Q) ∧((~P) ∨~R) (3) (P ∨Q) ∧((~P) ∨R) (4) (P ∨~Q) ∧(P ∨~R)