Question

If $f\left(x\right)=(1+\frac{1}{x})(1+\frac{1}{x+1})(1+\frac{1}{x+2}).......(1+\frac{1}{x+n})$
$g\left(x\right)=(1-\frac{1}{x})(1-\frac{1}{x-1})(1-\frac{1}{x-2}).......(1-\frac{1}{x-n})$ .Find f(x)+g(x).

• A. x + n + 1/x – n – 1
• B. 1
• C. 2

Answer 1

The correct option is C 2

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

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

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

Shortcut:- Single Substitution

Put x=1, n=0

f(x)= 2

g(x)=0 $\Rightarrow$ f(x) +g(x) =2

put x=1 and n=0 and look for 2 in the answer options.