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  4. If f(x) = (1/x2 + 4x + 4) - (4/x4 + 4x3+4x2) + (4/x3 + 2x2) , then f((1/2)) is equal to

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Q. 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

KEAMKEAM 2017Relations and Functions

Solution:

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