Question

Let $f\left(x\right)$ is a quadratic function such that $f\left(0\right)=1,f\left(1\right)=7,f(-1)=1$ and $\int \frac{f\left(x\right)dx}{x^2(x+1{)}^3}$ is a rational function. Find the value of $f^{\prime }\left(0\right)$.

• A. $3$
• B. $1$
• C. $2$
• D. $6$

Answer 1

The correct option is A $3$

Given:
$f\left(x\right)$ is a quadratic function such that

$f\left(0\right)=1,f\left(1\right)=7,f(-1)=1$ and

$\int \frac{f\left(x\right)dx}{x^2{(x+1)}^3}$ is a rational function
Let $f\left(x\right)=a{x}^2+bx+c$

Let $f\left(0\right)=1\Rightarrow c=1$

$f\left(1\right)=7\Rightarrow a+b+c=7\Rightarrow a+b=\mathrm{6.......}\left(1\right)$

$f(-1)=1\Rightarrow a-b+1=1\Rightarrow a-b=\mathrm{0.....}\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right),$

$\Rightarrow a=3,b=3,$ and $c=1$

Also, $f\left(x\right)=3x^2+3x+1$

$\Rightarrow f^{\prime }\left(x\right)=(6x+3)$

$\Rightarrow f^{\prime }\left(0\right)=6\left(0\right)+3=3$

Hence, the answer is $3$.