Question

If $f\left(x\right)=\frac{x}{x+1},$ determine the value of g(2) if, $g\left(x\right)=f\left[\frac{1}{x+1}\right]-\frac{1}{f(x+1)}.$

• A. 1/12
• B. 13/12
• C. -13/12
• D. -1/12

Answer 1

The correct option is C

-13/12

Method :1 Conventional Approach

$f\left[\frac{1}{x+1}\right]=\frac{\left[\frac{1}{x+1}\right]}{\left[\right\{\frac{1}{x+1}\}+1]}=\frac{1}{x+2}$

$\frac{1}{f(x+1)}=\frac{1}{\left[\frac{x+1}{x+2}\right]}=\frac{x+2}{x+1}$

$g\left(x\right)=f\left[\frac{1}{x+1}\right]-\frac{1}{f(x+1)}=\left[\frac{1}{x+2}\right]-\left[\frac{x+2}{x+1}\right]$

$g\left(2\right)=\left(\frac{1}{4}\right)-\left(\frac{4}{3}\right)=-\frac{13}{12}$

Hence option (c)

SHORTCUT : Direct Substitution

When x=2
$G\left(2\right)=f\left(\frac{1}{3}\right)-\frac{1}{f\left(3\right)}=\left(\frac{1}{4}\right)-\frac{4}{3}=-\frac{13}{12}.$