Question

If $f\left(x\right)=\frac{1}{x^2+4x+4}-\frac{4}{x^4+4x^3+4x^2}+\frac{4}{x^3+2x^2}$ , then $f\left(\frac{1}{2}\right)$ is equal to

• A. 1
• B. 2
• C. -1
• D. 3
• E. 4

Answer 1

Answer: (E)

We have,
$f(x)=\frac{1}{x^2+4x+4}-\frac{4}{x^4+4x^3+4x^2}+\frac{4}{x^3+2x^2}$
$=\frac{1}{(x+2{)}^2}-\frac{4}{x^2(x+2{)}^2}+\frac{4}{x^2(x+2)}$
$=\frac{x^2-4+4(x+2)}{(x+2{)}^2\cdot x^2}$
$=\frac{x^2-4+4x+8}{(x+2{)}^2\cdot x^2}$
$=\frac{x^2+4x+4}{(x+2{)}^2\cdot x^2}$
$=\frac{(x+2{)}^2}{(x+2{)}^2\cdot x^2}$
$=\frac{1}{x^2}$
$\therefore f(x)=\frac{1}{x^2}$
$\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{{\left(\frac{1}{2}\right)}^2}=4$