CAT 2014 | Question: 11

Question

If $y = f(x)$ and $f(x) = (1 - x) / (1 + x)$, which of the following is true? 

1. $f(2x) = f(x) – 1$ 
2. $x = f(2y) - 1$ 
3. $f(1/x) = f(x)$ 
4. $x = f(y)$

Answer 1

Given that, 

• $y=f(x) \longrightarrow (1)$
• $f(x)= \frac{(1-x)}{(1+x)} \longrightarrow (2)$

Now, we can check all of the options.

$\text{A}. f(2x) = f(x)-1$

Take left hand side term.

$f(2x) = \frac{(1-2x)}{(1+2x)}$

We can do rationalization.

$\Rightarrow f(2x) = \frac{(1-2x)}{(1+2x)} \times \frac{(1-2x)}{(1-2x)}$

$\Rightarrow f(2x) = \frac{(1-2x)^{2}}{1-4x^{2}} \longrightarrow (3)$

Take right hand side term.

$f(x)-1 = \frac{1-x}{1+x}-1$

$\Rightarrow f(x)-1 = \frac{1-x-1-x}{1+x} = \frac{-2x}{1+x} \longrightarrow (4)$

$\boxed{\text{LHS} \neq  \text{RHS}}$

So, option $\text{A}$ is not correct

$\text{B}. x=f(2y)-1$

$\Rightarrow x= f(2f(x))-1$

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

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

$\Rightarrow x= \frac{1-(\frac{2-2x}{1+x})}{1+(\frac{2-2x}{1+x})}-1$

$\Rightarrow x= \frac{\frac{1+x-2+2x}{1+x}}{\frac{1+x+2-2x}{1+x}}-1$

$\Rightarrow x= \frac{(3x-1)}{(1+x)} \times \frac{(1+x)}{(3-x)}$

$\Rightarrow x= \frac{3x-1-3+x}{3-x}$

$\Rightarrow \boxed{x= \frac{4x-4}{3-x}(\text{False})}$

So, option $\text{B}$ is not correct.

100. $f(\frac{1}{x}) = f(x)$

$\Rightarrow \frac{1-\frac{1}{x}}{1+\frac{1}{x}} = \frac{1-x}{1+x}$

$\Rightarrow \frac{\frac{x-1}{x}}{\frac{x+1}{x}}= \frac{1-x}{1+x}$

$\Rightarrow \boxed{\frac{x-1}{x+1} \neq \frac{1-x}{1+x} \text{(False)}}$

So, option $\text{C}$ is not correct.

500. $x=f(y)$

$\Rightarrow x= f(f(x))$

$\Rightarrow x= f(\frac{1-x}{1+x})$

$\Rightarrow x= \frac{1-(\frac{1-x}{1+x})}{1+(\frac{1-x}{1+x})}$

$\Rightarrow x=\dfrac{\frac{1+x-1+x}{1+x}}{\frac{1+x+1-x}{1+x}}$

$\Rightarrow x=\frac{2x}{2}$

$\Rightarrow \boxed{x=x \text{(True)}}$

So, option $\text{D}$ is not correct​​​​​​​.

Correct Answer $: \text{D}$

Short Method: We can put the value of $x=1,2,3, \cdots $ and verify each of the options.

$\text{A}) f(2x) = f(x)-1$

$x=1: f(2) = f(1)-1$

$\Rightarrow \frac{1-2}{1+2} = \frac{1-1}{1+!}-1$

$\Rightarrow \frac{-1}{3} = \frac{0}{2}-1$

$\Rightarrow \boxed{\frac{-1}{3} = -1\text{(False)}}$

$\text{B}) x=f(2y)-1$

$x=1,y=1: 1=f(2)-1$

$\Rightarrow 1=\frac{-1}{3}-1$

$\Rightarrow 1=\frac{-1-4}{3}$

$\Rightarrow 1=\boxed{\frac{-5}{4}\text{(False)}}$

$\text{C}) f(\frac{1}{x}) = f(x)$

$x=1: \; f(\frac{1}{1}) = f(1)$

$\Rightarrow \boxed{f(1) = f(1) \; \text{(True)}}$

$x=2: \; f(\frac{1}{2}) = f(2)$

$\Rightarrow \frac{1}{2} = \frac{1-2}{1+2}$

$\Rightarrow \frac{1-\frac{1}{2}}{1+\frac{1}{2}} = \frac{1-2}{1+2}$

$\Rightarrow \frac{\frac{2-1}{2}}{\frac{2+1}{2}} = \frac{-1}{3}$

$\Rightarrow \boxed{\frac{1}{3} = \frac{-1}{3}\text{(False)}}$