Question

If functions $f\left(x\right)$ and $g\left(x\right)$ are defined on $R\to R$ such that
$f\left(x\right)=x+3,x$ $\in$ rational
$=4x,x$ $\in$ irrational
$g\left(x\right)=x+\sqrt{5}$, x$\in$ irrational
$=-x,x$ $\in$ rational
then $(f-g)\left(x\right)$ is

• A. one-one & onto
• B. neither one-one nor onto
• C. one-one but not onto
• D. onto but not one-one

Answer 1

Answer: (B)

neither one-one nor onto

$f\left(x\right)=x+3,xϵ$ rational

$=4x,xϵ$ irrational

$g\left(x\right)=x+\sqrt{5},xϵ$ irrational

$=-x,xϵ$ rational

$(f-g)\left(x\right)=2x+3,xϵ$ rational

$+3x-\sqrt{5}xϵ$ irrational

For one-one;

We know that one one function is a $f^{\underset{–}{n}}$ for which every element of the range of the function corresponds to exactly one element of the domain.

But in this case this is not true as for all rational nos. the$f^{\underset{–}{n}}$ is one one but for irrational $f^{\underset{–}{n}}$, every elements of the range does not corresponds to exactly one element of the domain.

$(f-g)\left(x\right)\ne (f-g)\left(x^{\prime }\right)$ when $xϵ$ rational and

$xϵ$ irrational

for onto:

The $f^{\underset{–}{n}}(f-g)\left(x\right)$ does not cover the whole range of the function.