Practice Questions

Q72. sin x −ex if x ≤0 ⎧ Let a function f : R →R be defined as, f(x) = a + [−x] if 0 < x < 1 ⎨ ⎩ 2x −b if x ≥1 JEE Main 2021 (20 Jul Shift 1) JEE Main Previous Year Paper Where [x] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to: (1) 4 (2) 3 (3) 2 (4) 5 Q73. ⎧ 1 , if i = j Let A = [aij] be a 3 × 3 matrix, where aij = −x , if |i −j| = 1 ⎨ ⎩2x + 1 , otherwise Let a function f : R →R be defined as f(x) =det (A). Then the sum of maximum and minimum values of f on R is equal to: (1) −2027 (2) 2788 (3) 27 20 (4) −8827

Q72.The function 𝑓𝑥= 𝑥3 - 6𝑥2 + 𝑎𝑥+ 𝑏 is such that 𝑓2 = 𝑓4 = 0. Consider two statements: 𝑆1 there exists 𝑥1, 𝑥2 ∈2, 4, 𝑥1 < 𝑥2, such that 𝑓'𝑥1 = - 1 and 𝑓'𝑥2 = 0 . 𝑆2 there exists 𝑥3, 𝑥4 ∈2, 4, 𝑥3 < 𝑥4, such that 𝑓 is decreasing in 2, 𝑥4, increasing in 𝑥4, 4 and 2𝑓'𝑥3 = √3𝑓𝑥4 then (1) 𝑆1 is true and 𝑆2 is false (2) both 𝑆1 and 𝑆2 are false (3) both 𝑆1 and 𝑆2 are true (4) 𝑆1 is false and 𝑆2 is true JEE Main 2021 (01 Sep Shift 2) JEE Main Previous Year Paper Q73. 𝜋 sec2𝑥𝑓(𝑥)d𝑥 4 ∫2 Let f : R →R be a continuous function. Then lim 𝜋2 is equal to: 𝑥→𝜋/ 4 𝑥2 - 16 (1) 𝑓( 2 ) (2) 2𝑓( √2 ) (3) 2𝑓( 2 ) (4) 4𝑓( 2 )

Q72.If lim sin−1 x−tan−1 x is equal to L, then the value of (6L + 1) is x→0 3x3 (1) 1 (2) 1 6 2 (3) 6 (4) 2 JEE Main 2021 (18 Mar Shift 1) JEE Main Previous Year Paper Q73. 1 2 0 2 −1 5 Let A + 2B = ⎡ 6 −3 3⎤ and 2A −B = ⎡2 −1 6⎤ . If Tr(A) denotes the sum of all diagonal elements −5 3 1 0 1 2 ⎣ ⎦ ⎣ ⎦ of the matrix A, then Tr (A)−Tr (B) has value equal to (1) 1 (2) 2 (3) 0 (4) 3

Q73.Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A × B. Then : JEE Main 2021 (25 Feb Shift 2) JEE Main Previous Year Paper (1) y = 273x (2) 2y = 273x (3) 2y = 91x (4) y = 91x

Q73.Let 𝑓𝑥= 3sin4𝑥+ 10sin3𝑥+ 6sin2𝑥- 3, 𝑥∈- 6, 2. Then, 𝑓 is : (1) increasing in -𝜋 𝜋 (2) decreasing in 0, 𝜋 6, 2 2 𝜋 𝜋 (3) increasing in - 6, 0 (4) decreasing in - 6, 0

Q73.If [x] be the greatest integer less than or equal to x, then 100∑ [ (−1)nn2 ] n=8 (1) 0 (2) 4 (3) −2 (4) 2

Q73.For x > 0 , if f(x) = ∫x1 (1+t)loge t (1) 0 (2) 21 (3) −1 (4) 1 x ∈R. Then f(x) equals :

Q73.The function f(x) = x2 −2x −3 ⋅e9x2−12x+4 is not differentiable at exactly : (1) Four points (2) Two points (3) three points (4) one point 1 1+ xaQ74. , x < 0 ⎧ x loge( 1−xb ) If the function f(x) = k , x = 0 ⎨ cos2 x−sin2 x−1 , x > 0 ⎩ √x2+1−1 is continuous at x = 0, then a1 + 1b + k4 is equal to : (1) 4 (2) 5 (3) −4 (4) −5

Q73.The maximum slope of the curve y = 21 x4 −5x3 + 18x2 −19x occurs at the point (1) (3, 212 ) (2) (2, 2) (3) (2, 9) (4) (0, 0)

Q73.If cot−1(α) = cot−1 2 + cot−1 8 + cot−1 18 + cot−1 32 + … . upto 100 terms, then α is: JEE Main 2021 (17 Mar Shift 1) JEE Main Previous Year Paper (1) 1. 01 (2) 1. 00 (3) 1. 02 (4) 1. 03

Q73.An angle of intersection of the curves, 𝑥2 + 𝑦2 = 1 and 𝑥2 + 𝑦2 = 𝑎𝑏, 𝑎> 𝑏, is : 𝑎2 𝑏2 (1) tan-12√𝑎𝑏 (2) tan-1𝑎+ 𝑏 √𝑎𝑏 (3) tan-1𝑎- 𝑏 (4) tan-1 𝑎- 𝑏 √𝑎𝑏 2√𝑎𝑏

Q73.If Rolle's theorem holds for the function f(x) = x3 −ax2 + bx −4, x ∈[1, 2] with f ′( 43 ) = 0 , then ordered pair (a, b) is equal to : (1) (−5, −8) (2) (−5, 8) (3) (5, 8) (4) (5, −8) dθ is (where c is a constant of integration)

Q73.If the tangent to the curve 𝑦= 𝑥3 at the point 𝑃𝑡, 𝑡3 meets the curve again at 𝑄, then the ordinate of the point which divides 𝑃𝑄 internally in the ratio 1: 2 is: (1) 0 (2) -2𝑡3 (3) -𝑡3 (4) 2𝑡3

Q73.Let f : R →R be defined as f(x) = { −43 x3 +3xex2x2 + 3x,, xx >≤00 . Then f is increasing function in the interval (1) (−12 , 2) (2) (0, 2) (3) (−1, 23 ) (4) (−3, −1) , α ∈R where [x] is the greatest integer less than or equal to x, then the value of

Q73.Consider the function f : R →R defined by f(x) = { (2 −sin(0, x1 )) x , xx =≠00 (1) monotonic on (−∞, 0) ∪(0, ∞) (2) not monotonic on (−∞, 0) and (0, ∞) (3) monotonic on (0, ∞) only (4) monotonic on (−∞, 0) only

Q73.Let [t] denote the greatest integer less than or equal to t. Let f(x) = x −[x], g(x) = 1 −x + [x], and h(x) = min{f(x), g(x)}, x ∈[−2, 2]. Then h is : (1) continuous in [−2, 2] but not differentiable at (2) Continous in [−2, 2] but not differentiable at more than four points in (−2, 2) exactly three poionts in (−2, 2) (3) not continuous at exactly four points in [−2, 2] (4) not continuous at exactly three points in [−2, 2] is

Q73.The value of the integral, ∫31 [x2 −2x −2]dx, where [x] denotes the greatest integer less than or equal to x, is (1) −4 (2) −5 (3) −√2 −√3 + 1 (4) −√2 −√3 −1

Q73.Let the functions f : R →R and g : R →R be defined as : + 2, x < 0 x < 1 f(x) = and g(x) = {xx2, x ≥0 {x3,3x −2, x ≥1 Then, the number of points in R where (fog)(x) is NOT differentiable is equal to : (1) 3 (2) 1 (3) 0 (4) 2

Q73.Let f : R →R be defined as f(x + y) + f(x −y) = 2f(x)f(y), f( 21 ) = −1. Then the value of ∑20k=1 sin(k) sin(k+f(k))1 is equal to : (1) cosec2 (21) cos(20) cos(2) (2) sec2(1) sec(21) cos(20) (3) cosec2 (1) cosec (21) sin(20) (4) sec2(21) sin(20) sin(2) . Then which of

Q73.Let θ ∈(0, π2 ). If the system of linear equations (1 + cos2 θ)x + sin2 θy + 4 sin 3θz = 0 cos2 θx + (1 + sin2 θ)y + 4 sin 3θz = 0 cos2 θx + sin2 θy + (1 + 4 sin 3θ)z = 0 has a non-trivial solution, then the value of θ is: (1) 4π (2) 5π 9 18 (3) 7π (4) π 18 18 = tan−1 0 < x < 1. Then: x

Q73.Let M and m respectively be the maximum and minimum values of the function f(x) = tan−1(sin x + cos x) in [0, π2 ]. Then the value of tan(M −m) is equal to: (1) 2 −√3 (2) 3 −2√2 (3) 3 + 2√2 (4) 2 + √3

Q73.Consider the integral I = ∫100 [x]e[x]ex−1 value of I is equal to : (1) 9(e −1) (2) 45(e + 1) (3) 45(e −1) (4) 9(e + 1)

Q73.If [x] denotes the greatest integer less than or equal to x, then the value of the integral ∫π/2−π/2[[x] −sin x]dx is equal to: (1) −π (2) π (3) 0 (4) 1

Q73.A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is (1) 10 (2) 5 2+3√3 3+√3 (3) 10 (4) 5 3+2√3 2+√3 + … + n2

Q73.Let f : R −{3} →R −{1} be defined by f(x) = x−3x−2 . Let g : R →R be given as g(x) = 2x −3 . Then, the sum of all the values of x for which f −1(x) + g−1(x) = 132 is equal to (1) 7 (2) 2 (3) 5 (4) 3