Atomos Test Generator Free PYQs ExamsMy practice Log in / Sign up free

Determinants

JEE Main 2025 (Online) 29th January Morning ShiftMCQMedium📈 Topic in 19% papers+4−1
Ques. 1JEE Main 2025, 29 January (Morning Shift)

Let M and m respectively be the maximum and the minimum values of f(x) = vmatrix 1 + 2 x & 2 x & 4 4x \ 2 x & 1 + 2 x & 4 4x \ 2 x & 2 x & 1 + 4 4x vmatrix, x R Then M4 - m4 is equal to :

  1. 1280
  2. 1040
  3. 1215
  4. 1295

Answer

(1) 1280

Solution

Applying C1 C1 + C2 + C3, every entry of C1 becomes 1 + 2 x + 2 x + 4 4x = 2 + 4 4x, so

f(x) = (2 + 4 4x) vmatrix 1 & 2 x & 4 4x \ 1 & 1 + 2 x & 4 4x \ 1 & 2 x & 1 + 4 4x vmatrix

Then R2 R2 - R1, R3 R3 - R1 gives

f(x) = (2 + 4 4x) vmatrix 1 & 2 x & 4 4x \ 0 & 1 & 0 \ 0 & 0 & 1 vmatrix = 2 + 4 4x

-1 4x 1, M = 6 and m = -2.

M4 - m4 = 1296 - 16 = 1280

Answer: (1)

Keep practising — Determinants

Ques. 2JEE Main 2025, 29 January (Evening Shift)

Let , ( ) be the values of m, for which the equations x + y + z = 1, x + 2y + 4z = m and x + 4y + 10z = m2 have infinitely many solutions. Then the value of n=110(n + n ) is equal to :

  1. 3410
  2. 560
  3. 3080
  4. 440
Ques. 3JEE Main 2025, 24 January (Morning Shift)

If the system of equations 2x-y+z=4 5x+ y+3z=12 100x-47y+ z=212 has infinitely many solutions, then -2 is equal to

  1. 56
  2. 59
  3. 57
  4. 55
Ques. 4JEE Main 2025, 24 January (Evening Shift)

For some a,b, let f(x)= vmatrixa+ xx&1&b\ &1+ xx&b\ &1&b+ xx vmatrix, x 0, x 0f(x)= + a+ b. Then ( + + )2 is equal to :

  1. 25
  2. 16
  3. 9
  4. 36
Ques. 5JEE Main 2025, 24 January (Evening Shift)

If the system of equations x+2y-3z=2 2x+ y+5z=5 14x+3y+ z=33 has infinitely many solutions, then + is equal to :

  1. 13
  2. 10
  3. 12
  4. 11
Ques. 6JEE Main 2025, 23 January (Morning Shift)

If the system of equations ( -1)x+( -4)y+ z=5 x+( -1)y+( -4)z=7 ( +1)x+( +2)y-( +2)z=9 has infinitely many solutions, then 2+ is equal to

  1. 20
  2. 10
  3. 6
  4. 12
Ques. 7JEE Main 2025, 23 January (Morning Shift)

If A, B, and (adj (A-1 )+adj (B-1 ) ) are non-singular matrices of same order, then the inverse of A (adj (A-1 )+adj (B-1 ) )-1B, is equal to

  1. AB-1|A|+ BA-1|B|
  2. adj (B-1 )+adj (A-1 )
  3. AB-1+A-1B
  4. 1|AB| (adj(B)+adj(A) )
Ques. 8JEE Main 2025, 23 January (Evening Shift)

The system of equations x+y+z=6, x+2y+5z=9, x+5y+ z= , has no solution if

  1. =17, =18
  2. =17, 18
  3. =15, 17
  4. 17, 18
Ques. 9JEE Main 2025, 22 January (Morning Shift)

Let A be a square matrix of order 3 such that (A)=-2 and (3 adj (-6 adj(3A) ) )=2m+n 3mn, m>n. Then 4m+2n is equal to .

Ques. 10JEE Main 2025, 22 January (Evening Shift)

If the system of linear equations : x+y+2z=6 2x+3y+az=a+1 -x-3y+bz=2b where a,b , has infinitely many solutions, then 7a+3b is equal to :

  1. 12
  2. 9
  3. 22
  4. 16
Ques. 11JEE Main 2025, 22 January (Evening Shift)

For a 3 3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3 3 matrix such that |A|= 12 and trace (A)=3. If B=adj(adj(2A)), then the value of |B|+ trace (B) equals :

  1. 56
  2. 132
  3. 174
  4. 280

All Determinants PYQs →

PDFDownload Determinants PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Mathematics chapters