Let a function f(x) be defined as f(x) = cases a e^x + b x + c x x x & , x 0 L & , x = 0 cases If f(x) is differentiable at x = 0 and f'(0) = 2 , then the value of a - b - c is equal to :

Step-by-step solution

Since f(x) is differentiable at x=0 , it must be continuous at x=0 . Therefore, _ x 0 f(x) = f(0) = L . Using Maclaurin series expansions for the numerator: a e^x + b x + c x = a (1 + x + x^2 2 + x^3 6 ) + b (1 - x^2 2 ) + c (x - x^3 6 ) = (a + b) + (a + c)x + ( a 2 - b 2 )x^2 + ( a 6 - c 6 )x^3 + The denominator is x x = x (x - x^3 6 + ) = x^2 - x^4 6 + For the limit to exist finitely, the coefficients of x^0 and x^1 in the numerator must be zero: a + b = 0 b = -a a + c = 0 c = -a Substituting b and c , the x^2 co