let (a, b, c ) be real numbers such that (2 a+3 b+6 c=0 ) and (g(x) a x^2+b x+c=0 ) has atleast one root in the interval ((1,2) ). If a function (f:[1,2] R ) for which Rolle's mean value theorem holds is such that (f(x) ) is a primitive of (g(x) ), then (f(x)= )

Step-by-step solution

Given, (2 a+3 b+6 c=0 )...(i) and (g(x)=a x^2+b x+c=0 ) According to given information, ( aligned & f(x)= g(x) d x= (a x^2+b x+c ) d x & f(x)= a 3 x^3+ b 2 x^2+c x aligned ) ( aligned Now, f(2) & = a 3 (2)^3+ b 2 (2)^2+c(2) & = 8 3 a+2 b+2 c (ii) aligned ) ( aligned and f( l ) & = a 3 ( l )^3+ b 2 ( l )^2+c( l ) & = a 3 + b 2 +c (iii) aligned ) Here, (f(2)=f(1) ) ( 8 3 a+2 b+2 c= a 3 + b 2 +c ) ( 14 a+9 b+6 c=0 )...(iv) By solving Eqs. (i) and (iv), we get ( aligned a & =3, b=-6 and c=2 f(x) & = 3 3 x^3- 6 2 x^2+2