If _ x 0 ( x)- x a x^3+b x^5+c = -1 12 , then

Step-by-step solution

Given, _ x 0 ( x)- x a x^3+b x^5+c =- 1 12 ....(i) 0- 0 a(0)^3+b(0)^5+c =- 1 12 0 c =- 1 12 c=0 Applying L' Hospital's rule in Eq.(i),we get _ x 0 ( x) x- x 3 a x^2+5 b x^4+0 =- 1 12 _ x 0 x( ( x)-1) x^2(3 a+5 b x) =- 1 12 _ x 0 x 2 ^2 ( x 2 ) x^2(3 a+5 b x) =- 1 12 0 3 a+5 b 0 _ x 0 2 ^2 ( x 2 ) 4 ( x 2 )^2 ( x x )^2 =- 1 12 1 3 a+0 1 2 1 1 =- 1 12 ( _ 0 =1 ) 1 6 a =- 1 12 6 a=-12 a=-2 Hence, a=-2, b R and c=0