Practice Questions

Q70.If the inverse trigonometric functions take principal values, then cos−1( 103 cos(tan−1( 43 )) + 25 sin(tan−1( 43 ))) is equal to (1) 0 (2) π4 (3) π (4) π 3 6

Q70.Let R1 and R2 be two relations defined on R by aR1b ⇔ab ≥0 and a R2b ⇔a ≥b, then JEE Main 2022 (27 Jul Shift 1) JEE Main Previous Year Paper (1) R1 is an equivalence relation but not R2 (2) R2 is an equivalence relation but not R1 (3) both R1 and R2 are equivalence relations (4) neither R1 nor R2 is an equivalence relation

Q70.The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to (1) 133 (2) 19 104 103 (3) 18 (4) 271 103 104

Q71.Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos-1𝑥- 2sin-1𝑥= cos-12𝑥 is equal to (1) 0 (2) 1 (3) 1 (4) -1 2 2

Q71.The system of equations -𝑘𝑥+ 3𝑦- 14𝑧= 25 -15𝑥+ 4𝑦- 𝑘𝑧= 3 -4𝑥+ 𝑦+ 3𝑧= 4 Question: is consistent for all 𝑘 in the set (1) 𝑅 (2) 𝑅- -11, 13 (3) 𝑅- -13 (4) 𝑅- -11, 11 - 1 4

Q71.Let A = [aij] be a square matrix of order 3 such that aij = 2j−i , for all i, j = 1, 2, 3 . Then, the matrix A2 + A3 + … + A10 is equal to (1) ( 310−12 )A (2) ( 310+12 )A (3) ( 310+32 )A (4) ( 310−32 )A

Q71.If y = tan−1(sec x3 −tan x3), π2 < x3 < 3π2 , then (1) xy′′ + 2y′ = 0 (2) x2y′′ −6y + 3π2 = 0 (3) x2y′′ −6y + 3π = 0 (4) xy′′ −4y′ = 0

Q71.The number of distinct real roots of x4 −4x + 1 = 0 is (1) 0 (2) 1 (3) 2 (4) 4

Q71.From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is: (1) 15√3 (2) 20√3 (3) 20 + 10√3 (4) 30 Q72. 2 −1 Let A = − . . . − 5C5(adj A)5 , then the sum of . If B = I −5C1(adj A) + 5C2(adj A)2 (0 2 ) all elements of the matrix B is: (1) −5 (2) −6 (3) −7 (4) −8

Q71.The set of all values of k for which (tan−1 x)3 + (cot−1 x)3 = kπ3, x ∈R, is the interval (1) [ 321 , 87 ) (2) ( 241 , 1613 ) (3) [ 481 , 1613 ] (4) [ 321 , 89 ) x2−9 ) is

Q71.If the function f(x) = loge(1−x+x2)+loge(1+x+x2) −π π sec x−cos x , x ∈( 2 , 2 ) −{0} is continuous at x = 0 , then k is equal { k , x = 0 to: (1) 1 (2) −1 (3) e (4) 0 are continuous on R, then and g(x) =

Q71.The domain of the function f(x) = sin−1[2x2 −3] + log2(log (x2 −5x + 5)), where 2 integer function, is 2 , 5+√52 ) 2 , 5−√52 (1) (−√5 ) (2) ( 5−√5 (3) (1, 5−√52 ) (4) [1, 5+√52 )

Q71.Let A = (−21 −52 ). Let α, β ∈R be such that αA2 + βA = 2I . Then α + β is equal to (1) −10 (2) −6 (3) 6 (4) 10

Q71.If the absolute maximum value of the function 𝑓𝑥= x2 - 2x + 7e4x3 - 12x2 - 180x + 31in the interval -3, 0 is 𝑓𝛼, then (1) 𝛼= 0 (2) 𝛼= - 3 (3) 𝛼∈-1, 0 (4) 𝛼∈-3, - 1

Q71.Let 𝑓: 𝑁→𝑅 be a function such that 𝑓𝑥+ 𝑦= 2 𝑓𝑥 𝑓𝑦 for natural numbers 𝑥 and 𝑦. If 𝑓1 = 2, then the 10 512 value of 𝛼 for which ∑𝑘= 1 𝑓𝛼+ 𝑘= 3 220 - 1 holds, is (1) 3 (2) 4 (3) 5 (4) 6

Q71.Let f : R →R be defined as f(x) = x −1 and g : R →{1, −1} →R be defined as g(x) = x2 . Then the x2−1 function fog is: (1) One-one but not onto (2) onto but not one-one (3) Both one-one and onto (4) Neither one-one nor onto

Q71.The function f(x) = xex(1−x), x ∈R, is (1) increasing in (−12 , 1) (2) decreasing in ( 12 , 2) (3) increasing in (−1, −12 ) (4) decreasing in (−12 , 12 )

Q71.If 0 < 𝑥< 1 and sin-1𝑥 = cos-1𝑥 , then a value of sin 2𝜋𝛼 is √2 𝛼 𝛽 𝛼+ 𝛽 (1) 4√1 - 𝑥2 1 - 2𝑥2 (2) 4𝑥√1 - 𝑥2 1 - 2𝑥2 (3) 2𝑥√1 - 𝑥2 1 - 4𝑥2 (4) 4√1 - 𝑥2 1 - 4𝑥2

Q71.The domain of the function 𝑓𝑥= sin-1 𝑥2 - 3𝑥+ 2 is 𝑥2 + 2𝑥+ 7 (1) [1, ∞) (2) ( - 1, 2] (3) [ - 1, ∞) (4) ( - ∞, 2]

Q71.If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is (1) 11. 5 (2) 10. 5 (3) 12 (4) 11

Q71. a −1 0 Let f(x) = ax a −1 , a ∈R. Then the sum of the squares of all the values of a for ax2 ax a 2f ′(10) −f ′(5) + 100 = 0 is (1) 117 (2) 106 (3) 125 (4) 136 is

Q72.The sum of the absolute maximum and absolute minimum values of the function f(x) = tan−1(sin x −cos x) in the interval [0, π] is (1) 0 (2) tan−1( √21 ) −π4 12 (3) cos−1( √31 ) −π4 (4) −π dt, n = 1, 2, 3, … . Then

Q72.The lengths of the sides of a triangle are 10 + x2 , 10 + x2 and 20 −2x2 . If for x = k, the area of the triangle is maximum, then 3k2 is equal to (1) 5 (2) 12 (3) 10 (4) 20 d3f dx = f(x)ex + C , where C is a constant, then at x = 1 is equal to Q73. ∫ (x2+1)ex dx3 (x+1)2 (1) 3 (2) 3 4 8 (3) −32 (4) 78 dx is equal to

Q72. log𝑒1 + 5𝑥- log𝑒1 + 𝛼𝑥 if 𝑥≠0 Let the function 𝑓𝑥= 𝑥 be continuous at 𝑥= 0. Then 𝛼 is equal to 10 if 𝑥= 0 (1) 10 (2) -10 (3) 5 (4) -5

Q72.The number of real values of λ, such that the system of linear equations 2x −3y + 5z = 9 x + 3y −z = −18 3x −y + (λ2 −|λ|)z = 16 has no solutions, is (1) 0 (2) 1 (3) 2 (4) 4 JEE Main 2022 (25 Jul Shift 2) JEE Main Previous Year Paper