Two cubic function f(x)=x^3+a x^2+b x+c and g(x)=c x^3+b x^2+a x+1 satisfy the following. (i) f(3)=0, g(4)=0 (ii) The value of _ x p f(x) g(x) exists for all p R- 4 Find the value of _ x -1 f(x)+g(x) x+1 .

Step-by-step solution

f(x)=x^3+a x^2+b x+c=0 Put x=1 / t , we get c t^3+b t^2+a t+1=0=g(t) Hence, roots of f(x) and g(x) are reciprocal to each other. Also, g(1 / 3)=0 hence for _ x p f(x) g(x) exists for all p R- 4 f(1 / 3) is also 0 . Roots of f(x) are x^3+a x^2+b x+c=0 < _ 1 / 4 ^3 Now, aligned & _ x -1 f(x)+g(x) x+1 =3(c+1)-(a+b) & x^3+a x^2+b x+c=0 < array l 1 / 4 1 / 3 array aligned aligned -a & =3+ 1 4 + 1 3 = 43 12 a= -43 12 b & = 3 4 + 1 12 +1= 11 6 = 22 12 c & = -1 4 aligned Now, _ x -1 f(x)+g(x) x+1 =3(c+1)-(a+b)=3 ( -1 4 +1