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 :
- 1280
- 1040
- 1215
- 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)