Question

Let f(x) is a quadratic function such that f(0)=1, f'(0)=1 and ∫f(x)x2(x−1)2dx is a rational function then value of |f'(1)| is

Answer 1

Let $g\left(x\right)=\int \frac{f\left(x\right)dx}{x^3(x-1{)}^2}=\int (\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x-1}+\frac{E}{(x-1{)}^2})dx$

Since g(x) is a rational function hence A=D=0

So $g\left(x\right)=\int (\frac{B}{x^2}+\frac{C}{x^3}+\frac{3}{(x-1{)}^2})dx\dots \left(2\right)$

Comparing (1) & (2)

$\frac{f\left(x\right)}{x^3(x-1{)}^2}=\frac{Bx(x-1{)}^2+C(x-1{)}^{+}E\left(x^3\right)}{x^3(x-1{)}^2}$

$f\left(x\right)=(B+E)x^3+(C-2B)x^2+(B-2C)+x+C$

Since f(x) is quadratic

So $B+E=0,f\left(0\right)=1\Rightarrow C=1$

$f^{\prime }\left(x\right)=2x(C-2B)+(B-2C)$

$f^{\prime }\left(0\right)=B-2C=1\Rightarrow B=3$

So $f^{\prime }\left(1\right)=2(C-2B)+(B-2C)=-3B=-9$

So $\left(f^{\prime }\right(1\left)\right)=9$