The values of α, for which |1, 3/2, α+3/2; 1, 1/3, α+1/3; 2 α+3, 3 α+1, 0|=0, lie in the interval

Solution

Maths sir · 5 steps

The third column is almost C₂ + α C₁. Subtract that off and two zeros appear for free.

1. Spot the column. In rows 1 and 2, the third entry is α times the first entry plus the second entry. So apply C₃ → C₃ - C₂ - α C₁, which does not change the determinant.
2. New third column. Rows 1 and 2 become 0. Row 3 becomes 0 - (3α + 1) - α(2α + 3) = -(2α² + 6α + 1).
3. Expand along C3. Only the bottom entry survives. Its cofactor is |1, 3/2; 1, 1/3| = 1/3 - 3/2 = -7/6, which is not zero.

   7/6(2α² + 6α + 1) = 0 ⇒ α = (-3 ± √7)/2

4. Locate the roots. With √7 ≈ 2.65, the roots are α ≈ -0.18 and α ≈ -2.82. Both are negative and both are greater than -3.
5. So. Both values of α lie in (-3, 0), which is option (c).

The idea to keep. Before expanding a determinant full of a parameter, check whether one column is a combination of the others. A column operation can empty it out and leave a one-term expansion.

Answer from the NTA final answer key. Solution by PrepWiser.