If _ x 1 (3x^2 - 4x + 1) - x^2 + 1 2x^3 - 7x^2 + ax + b = -2 , then the quadratic equation having roots a and b is

Step-by-step solution

Given limit is _ x 1 (3x^2 - 4x + 1) - x^2 + 1 2x^3 - 7x^2 + ax + b = -2 Let N(x) = (3x^2 - 4x + 1) - x^2 + 1 and D(x) = 2x^3 - 7x^2 + ax + b . As x 1 , N(1) = (0) - 1 + 1 = 0 . For the limit to exist and be finite, the denominator must also approach 0 as x 1 . D(1) = 2(1)^3 - 7(1)^2 + a(1) + b = 0 a + b = 5 Applying L'Hopital's rule, we differentiate the numerator and the denominator: N'(x) = (6x - 4) (3x^2 - 4x + 1) - 2x D'(x) = 6x^2 - 14x + a As x 1 , N'(1) = 2 (0) - 2 = 0 . Since N'(1) = 0 and the limit is a no