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ExamsJEE MainMaths

Find the value of the determinant | (a^x + a^-x)² (a^x - a^-x)² 1 | | (b^x + b^-x)² (b^x - b^-x)² 1 | | (c^x + c^-x)² (c^x - c^-x)² 1 |

  1. 0
  2. 2abc
  3. a²b²c²
  4. None of these

Correct answer: 0

Solution

For each row, (a^x + a^-x)^2 - (a^x - a^-x)^2 = 4*(a^x)*(a^-x) = 4, a constant. So column1 - column2 = 4 in every row, making it proportional to the column of 1's. Hence the determinant is 0.

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