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The value of ∣xx+1x−1x∣\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} is:

Solution

✅ Correct Option: 2

Expanding the 2×22 \times 2 determinant, the value is x⋅x−(x+1)(x−1)=x2−(x2−1)=1x \cdot x - (x+1)(x-1) = x^2 - (x^2-1) = 1.

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