Let D(x) = vmatrix x & (1+x) & e^x-1 1 & 1 & 1 x & x & 2x vmatrix and L = _ x 0 D(x) x^3 . Then the value of 12L is equal to :

Step-by-step solution

Given D(x) = vmatrix x & (1+x) & e^x-1 1 & 1 & 1 x & x & 2x vmatrix . Applying the column operation C₁ C₁ - C₂ , we get: D(x) = vmatrix x - (1+x) & (1+x) & e^x-1 0 & 1 & 1 0 & x & 2x vmatrix Expanding the determinant along the first column: D(x) = ( x - (1+x)) vmatrix 1 & 1 x & 2x vmatrix - 0 + 0 D(x) = ( x - (1+x))(2x - x) = x( x - (1+x)) Now, substitute D(x) into the limit: L = _ x 0 D(x) x^3 = _ x 0 x( x - (1+x)) x^3 = _ x 0 x - (1+x) x^2 This is a 0 0 indeterminate form. Using L'Hôpital's rule: L = _ x 0 x - 1