Practice Questions
3,340 questions across 23 years of JEE Main β find and practise any topic!
Found 3,340 results
Q70.Let A and B be two 3 Γ 3 matrices such that AB = I and |A| = 18 then |adj(Badj(2A))| is equal to (1) 128 (2) 32 (3) 64 (4) 102
Q70.The number of values of Ξ± for which the system of equations x + y + z = Ξ± Ξ±x + 2Ξ±y + 3z = β1 x + 3Ξ±y + 5z = 4 is inconsistent, is (1) 0 (2) 1 (3) 2 (4) 3
Q70.The negation of the Boolean expression ~πβ§πβ~πβ¨π is logically equivalent to (1) πβπ (2) πβπ (3) ~πβπ (4) ~πβπ
Q70.If the system of equations π₯+ π¦+ π§= 6 2π₯+ 5π¦+ πΌπ§= π½ π₯+ 2π¦+ 3π§= 14 has infinitely many solutions, then πΌ+ π½ is equal to (1) 8 (2) 36 (3) 44 (4) 48
Q70.The value of nββ6lim tan{βnr=1 tanβ1( r2+3r+31 )} is equal to (1) 1 (2) 2 (3) 3 (4) 6
Q70.The total number of functions, π: 1, 2, 3, 4 β1, 2, 3, 4, 5, 6 such that π1 + π2 = π3, is equal to (1) 60 (2) 90 (3) 108 (4) 126
Q70.If the inverse trigonometric functions take principal values, then cosβ1( 103 cos(tanβ1( 43 )) + 25 sin(tanβ1( 43 ))) is equal to (1) 0 (2) Ο4 (3) Ο (4) Ο 3 6
Q70.Let A = (Ξ±4 β2Ξ² ) (1) β18 (2) 18 (3) β50 (4) 50 1 [t] is the greatest
Q70.If cosβ1( 2y ) = loge ( x5 ) 5, |y| < 2, then (1) x2yβ²β² + xyβ² β25y = 0 (2) x2yβ²β² βxyβ² β25y = 0 (3) x2yβ²β² βxyβ² + 25y = 0 (4) x2yβ²β² + xyβ² + 25y = 0
Q70.Let A and B be two 3 Γ 3 non-zero real matrices such that AB is a zero matrix. Then (1) The system of linear equations AX = 0 has a (2) The system of linear equations AX = 0 has unique solution infinitely many solutions (3) B is an invertible matrix (4) adj(A) is an invertible matrix
Q70.The ordered pair (a, b), for which the system of linear equations 3x β2y + z = b 5x β8y + 9z = 3 2x + y + az = β1 has no solution, is (1) (3, 13 ) (2) (β3, 31 ) (3) (β3, β13 ) (4) (3, β13 )
Q71.Let A = (β21 β52 ). Let Ξ±, Ξ² βR be such that Ξ±A2 + Ξ²A = 2I . Then Ξ± + Ξ² is equal to (1) β10 (2) β6 (3) 6 (4) 10
Q71.If the function f(x) = loge(1βx+x2)+loge(1+x+x2) βΟ Ο sec xβcos x , x β( 2 , 2 ) β{0} is continuous at x = 0 , then k is equal { k , x = 0 to: (1) 1 (2) β1 (3) e (4) 0 are continuous on R, then and g(x) =
Q71.Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos-1π₯- 2sin-1π₯= cos-12π₯ is equal to (1) 0 (2) 1 (3) 1 (4) -1 2 2
Q71.The number of distinct real roots of x4 β4x + 1 = 0 is (1) 0 (2) 1 (3) 2 (4) 4
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.Let A = [aij] be a square matrix of order 3 such that aij = 2jβi , for all i, j = 1, 2, 3 . Then, the matrix A2 + A3 + β¦ + A10 is equal to (1) ( 310β12 )A (2) ( 310+12 )A (3) ( 310+32 )A (4) ( 310β32 )A
Q71.If the absolute maximum value of the function ππ₯= x2 - 2x + 7e4x3 - 12x2 - 180x + 31in the interval -3, 0 is ππΌ, then (1) πΌ= 0 (2) πΌ= - 3 (3) πΌβ-1, 0 (4) πΌβ-3, - 1
Q71.If y = tanβ1(sec x3 βtan x3), Ο2 < x3 < 3Ο2 , then (1) xyβ²β² + 2yβ² = 0 (2) x2yβ²β² β6y + 3Ο2 = 0 (3) x2yβ²β² β6y + 3Ο = 0 (4) xyβ²β² β4yβ² = 0
Q71.From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60Β°. The pole subtends an angle 30Β° at the top of the tower. Then the height of the tower is: (1) 15β3 (2) 20β3 (3) 20 + 10β3 (4) 30 Q72. 2 β1 Let A = β . . . β 5C5(adj A)5 , then the sum of . If B = I β5C1(adj A) + 5C2(adj A)2 (0 2 ) all elements of the matrix B is: (1) β5 (2) β6 (3) β7 (4) β8
Q71.The system of equations -ππ₯+ 3π¦- 14π§= 25 -15π₯+ 4π¦- ππ§= 3 -4π₯+ π¦+ 3π§= 4 Question: is consistent for all π in the set (1) π (2) π - -11, 13 (3) π - -13 (4) π - -11, 11 - 1 4
Q71. a β1 0 Let f(x) = ax a β1 , a βR. Then the sum of the squares of all the values of a for ax2 ax a 2f β²(10) βf β²(5) + 100 = 0 is (1) 117 (2) 106 (3) 125 (4) 136 is
Q71.The domain of the function ππ₯= sin-1 π₯2 - 3π₯+ 2 is π₯2 + 2π₯+ 7 (1) [1, β) (2) ( - 1, 2] (3) [ - 1, β) (4) ( - β, 2]
Q71.The domain of the function f(x) = sinβ1[2x2 β3] + log2(log (x2 β5x + 5)), where 2 integer function, is 2 , 5+β52 ) 2 , 5ββ52 (1) (ββ5 ) (2) ( 5ββ5 (3) (1, 5ββ52 ) (4) [1, 5+β52 )
Q71.Let f : R βR be defined as f(x) = x β1 and g : R β{1, β1} βR be defined as g(x) = x2 . Then the x2β1 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