Practice Questions

Q68.Let 𝑃 be a parabola with vertex 2, 3 and directrix 2𝑥+ 𝑦= 6. Let an ellipse 𝐸: 𝑥2 + 𝑦2 = 1, 𝑎> 𝑏 𝑎2 𝑏2 1 of eccentricity pass through the focus of the parabola 𝑃. Then the square of the length of the latus rectum √2 of 𝐸, is (1) 385 (2) 347 8 8 512 656 (3) (4) 25 25

Q69.Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and . The number of one-to-one functions from A to the + f : A →Z be the function f(x) = [log2 (x2 [ x35 ])] range of f is (1) 25 (2) 24 (3) 20 (4) 120

Q70.If A is a square matrix of order 3 such that det(A) = 3 and det (adj (−4 adj (−3 adj (3 adj ((2 A)−1))))) = 2m3n , then m + 2n is equal to : (1) 2 (2) 3 (3) 6 (4) 4 JEE Main 2024 (06 Apr Shift 2) JEE Main Previous Year Paper

Q71.Let A = [ 10 21 ] and B = I + adj(A) + (adj A)2 + … + (adj A)10 . Then, the sum of all the elements of the matrix B is: (1) -124 (2) 22 (3) -88 (4) -110

Q71.The integral ∫3/41/4 cos (2 (1) 1/2 (2) −1/2 (3) −1/4 (4) 1/4

Q72.If 𝑓𝑥= 4𝑥+ 3 , 𝑥≠2 and ( 𝑓𝑜𝑓) ( 𝑥) = 𝑔( 𝑥) , where 𝑔: 𝑅- 2 →𝑅- 2 then ( 𝑔𝑜𝑔𝑜𝑔) ( 4 ) is equal to 6𝑥- 4 3 3 3, 19 19 (1) - (2) 20 20 (3) -4 (4) 4 Q73. 𝑔𝑥, 𝑥≤0 Let 𝑔𝑥 be a linear function and 𝑓𝑥= 1 , is continuous at 𝑥= 0. If 𝑓'1 = 𝑓−1, then the value of 1 + 𝑥 𝑥, 𝑥> 0 2 + 𝑥 𝑔3 is JEE Main 2024 (31 Jan Shift 1) JEE Main Previous Year Paper 1 4 1 4 (1) 3log𝑒 1 (2) 3log𝑒 9 + 1 9𝑒 3 4 4 (3) log𝑒 9 −1 (4) log𝑒 1 9𝑒 3 𝑥𝑦𝑥- 1𝑥- 2

Q72.If the domain of the function 𝑓𝑥= √𝑥2 −25 + + 2𝑥−15 is −∞, 𝛼∪𝛽, ∞, then 𝛼2 + 𝛽3 is equal to: 4 −𝑥2 log10𝑥2 (1) 140 (2) 175 (3) 150 (4) 125

Q72.The function f: R->R, f(x) = x2+2x−15 , x ∈R is x2−4x+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 ≠0 x then

Q72.Consider the function f: ( 0, 2 ) →R defined by f(x) = x + 2 and the function g ( x ) defined by 2 x min{f ( t ) }, 0 < t ≤x and 0 < x ≤1 gx = 3 . Then + x, 1 < x < 2 2 (1) g is continuous but not differentiable at x = 1 (2) g is not continuous for all x ∈( 0, 2 ) (3) g is neither continuous nor differentiable at x = 1(4) g is continuous and differentiable for all x ∈( 0, 2 )

Q73.Let g : R →R be a non constant twice differentiable such that g′( 21 ) = g′( 23 ). If a real valued function f is defined as f(x) = 12 [ g(x) + g(2 −x)], then (1) f ′′(x) = 0 for atleast two x in (0, 2) (2) f ′′(x) = 0 for exactly one x in (0, 1) (3) f ′′(x) = 0 for no x in (0, 1) (4) f ′( 23 ) + f ′( 21 ) = 1

Q73.If f(x) = {x30 sin, x (= 0 (1) f ′′ ( π2 ) = 24−π22π (2) f ′′ ( π2 ) = 12−π22π (3) f ′′(0) = 1 (4) f ′′(0) = 0

Q73.If the function 𝑓: −∞, −1 →𝑎, 𝑏 defined by 𝑓𝑥= 𝑒𝑥3 −3𝑥+ 1 is one-one and onto, then the distance of the point 𝑃2𝑏+ 4, 𝑎+ 2 from the line 𝑥+ 𝑒−3𝑦= 4 is: (1) 2√1 + 𝑒6 (2) 4√1 + 𝑒6 (3) 3√1 + 𝑒6 (4) √1 + 𝑒6 JEE Main 2024 (31 Jan Shift 2) JEE Main Previous Year Paper

Q73.Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS . Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a + b)2 is equal to : (1) 72 (2) 60 (3) 64 (4) 80

Q73.Let 𝑓: 𝑅→𝑅 be defined as 𝑎−𝑏cos2𝑥 ; 𝑥< 0 𝑥2 𝑓𝑥= 𝑥2 + 𝑐𝑥+ 2; 0 ≤𝑥≤1 2𝑥+ 1; 𝑥> 1 If 𝑓 is continuous everywhere in 𝑅 and 𝑚 is the number of points where 𝑓 is NOT differential then 𝑚 + 𝑎 + 𝑏 + 𝑐 equals: JEE Main 2024 (01 Feb Shift 1) JEE Main Previous Year Paper (1) 1 (2) 4 (3) 3 (4) 2 1

Q74.If the value of the integral ∫ −π2 2 ( x21+πxcos x 1+sin2 x π 1+e(sin x)2023 )dx (1) 3 (2) −32 (3) 2 (4) 32 JEE Main 2024 (29 Jan Shift 1) JEE Main Previous Year Paper

Q74.If 5𝑓𝑥+ 4𝑓 𝑥= 𝑥2 −2, ∀𝑥≠0 and 𝑦= 9𝑥2𝑓𝑥, then 𝑦 is strictly increasing in: (1) 0, 1 ∪1 ∞ (2) −1 0 ∪1 ∞ √5 √5, √5, √5, (3) −1 0 ∪0, 1 (4) −∞, 1 ∪0, 1 √5, √5 √5 √5 𝜋 Q75. 4 𝑥𝑑𝑥 The value of the integral ∫ equals: 0 sin42𝑥+ cos42𝑥 (1) √2𝜋2 (2) √2𝜋2 8 16 (3) √2𝜋2 (4) √2𝜋2 32 64

Q75.For x ∈(−π2 , π2 ), if y(x) = ∫ cosecxcosecx+sinsec x+tan xx sin2 x dx and limπ = 0 then y( π4 ) is equal to x→( 2 )−y(x) (1) tan−1( √21 ) (2) 21 tan−1( √21 ) (3) −1 2 ) √2 tan−1( √21 ) (4) √21 tan−1(−1

Q75.The solution curve, of the differential equation 2y dydx + 3 = 5 dydx , passing through the point (0, 1) is a conic, whose vertex lies on the line: JEE Main 2024 (09 Apr Shift 1) JEE Main Previous Year Paper (1) 2x + 3y = 9 (2) 2x + 3y = −9 (3) 2x + 3y = −6 (4) 2x + 3y = 6

Q75.The value of the integral ∫2−1 loge (x + √x2 + 1)dx JEE Main 2024 (09 Apr Shift 2) JEE Main Previous Year Paper (1) √5 −√2 + loge ( 7+4√51+√2 ) (2) √5 −√2 + loge ( 9+4√51+√2 ) + loge (3) √2 −√5 + loge ( 7+4√51+√2 ) (4) √2 −√5 ( 9+4√51+√2 )

Q75.Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e−a + 4a2 + a −1. Then the differential equation, whose general solution is y = c1f(x) + c2 , where c1 and c2 are arbitrary constants, is d2y dy (1) (8ex −1) = 0 (2) (8ex −1) + dx d2y −dydx = 0 dx2 dx2 (3) (8ex + 1) + dxdy = 0 dx2 d2y −dydx = 0 (4) (8ex + 1) dx2d2y

Q75.Let 𝑦= 𝑓( 𝑥) be a thrice differentiable function in ( - 5, 5 ) . Let the tangents to the curve 𝑦= 𝑓( 𝑥) at ( 1, f ( 1 ) ) and ( 3, f ( 3 ) ) make angles 𝜋 and 𝜋 respectively with positive x-axis. If 6 4, 3 2 27 ∫1 𝑓'𝑡 + 1𝑓"𝑡𝑑𝑡= 𝛼+ 𝛽√3 where 𝛼, 𝛽 are integers, then the value of 𝛼+ 𝛽 equals JEE Main 2024 (30 Jan Shift 2) JEE Main Previous Year Paper (1) -14 (2) 26 (3) -16 (4) 36 39 , then 𝑓𝑥- 𝑐𝑥𝑑𝑥=

Q75.If ∫ 3 3 √sin3 x cos3 x sin(x−θ) constant, then AB is equal to (1) 4 cosec (2θ) (2) 4 sec θ (3) 2 sec θ (4) 8 cosec (2θ) JEE Main 2024 (29 Jan Shift 2) JEE Main Previous Year Paper

Q76.If (a, b) be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1), and I1 = ∫ba xsin(4x −x2) dx, I2 = ∫ba sin(4x −x2) dx , then 36 I1I2 is equal to : (1) 72 (2) 88 (3) 80 (4) 66

Q76.Suppose the solution of the differential equation (2+α)x−βy+2 represents a circle passing through dx = βx−2αy−(βγ−4α) origin. Then the radius of this circle is : (1) 2 (2) √17 (3) 1 (4) √17 2 2 → →

Q77.Let →a, b and→cbe three non-zero vectors such that b and→care non-collinear if →a+ 5b is collinear with →c,→b + 6→cis collinear with →a and →a+ α→b + β→c= →0, then α + β is equal to (1) 35 (2) 30 (3) −30 (4) −25