Question

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Answer 1

Given :

$\sf\:f(x)=\dfrac{x}{x-1}=\dfrac{1}{y}$

To Find :

f(y) equals to

Solution :

We have ,

$\sf\:f(x)=\dfrac{x}{x-1}$

$\sf\implies\:f(y)=\dfrac{y}{y-1}....(1)$

Given :

$\sf\:f(x)=\dfrac{x}{x-1}=\dfrac{1}{y}$

$\sf\implies\dfrac{x}{x-1}=\dfrac{1}{y}$

$\sf\implies\:xy=x-1$

$\sf\implies\:y=\dfrac{x-1}{x}....(2)$

Now put the value of y in Equation (1) then ,

$\sf\:f(y)=\dfrac{y}{y-1}$

$\sf\implies\:f(y)=\dfrac{\frac{x-1}{x}}{\frac{x-1}{x}-1}$

$\sf\implies\:f(y)=\dfrac{\frac{x-1}{x}}{\frac{x-1-x}{x}}$

$\sf\implies\:f(y)=\dfrac{\frac{x-1}{x}}{\frac{-1}{x}}$

$\sf\implies\:f(y)=\dfrac{x-1}{x}\times(-x)$

$\sf\implies\:f(y)=-(x-1)$

$\sf\implies\:f(y)=1-x$

Therefore,

$\tt\implies\green{f(y)=1-x}$

Correct option d)

Answer 2

Question

If $f(x)\:=\:\dfrac{x}{x\:-\:1}\:=\:\dfrac{1}{y}$

What is the value of f (y) ?

Answer

$\:\sf\underline\green{Given}\:$

$f(x)\:=\:\dfrac{x}{x\:-\:1}\:=\:\sf\dfrac{1}{y}$

$\:\sf\underline\green{To\:Find}\:$

The value of f (y)

$\:\sf\underline\green{Solution}\:$

According to the statement .

$f(x)\:=\:\dfrac{x}{x\:-\:1}$

Then we can frame an another equation from this that is

$f(y)\:=\:\dfrac{y}{y\:-\:1}$

Let this be equation 1 .

Also it is given that ,

$f(x)\:=\:\dfrac{x}{x\:-\:1}\:=\:\dfrac{1}{y}$

Cross multiplying the terms ,

$f(x)\:=\:\:x\:\times\:y\:=\:1\:\times\:(x\:-\:1)$

$xy\:=\:x\:-\:1$

Dividing the both sides by x ,

$\dfrac{xy}{x}\:=\:\dfrac{x\:-\:1}{x}$

Hence the second equation formed is ,

$y\:=\:\dfrac{x\:-\:1}{x}$

Now substituting the value of y in first equation we get ,

$\large\:f(y)\:=\:\dfrac{y}{y\:-\:1}$

$\large\:f(y)\:=\:\dfrac{\frac{x\:-\:1}{x}}{\frac{x\:-\:1}{x}\:-\:1}$

$\large\:f(y)\:=\:\dfrac{\frac{x\:-\:1}{x}}{\frac{x\:-\:1\:-\:x}{x}}$

$\large\:f(y)\:=\:\dfrac{\frac{x\:-\:1}{x}}{\frac{-\:1}{x}}$

$\large\:f(y)\:=\:\dfrac{x\:-\:1}{x}\:\times\:\dfrac{-\:x}{1}$

$\large\:f(y)\:=\:-(\:x\:-\:1\:)$

$\large\:f(y)\:=\:1\:-\:x$

$\large{\boxed{\purple{Required\:Answer\:is\:f(y)\:=\:1\:-\:x}}}$