Question

Let $f:\{x,y,z\}\to \{a,b,c\}$ be a one-one function and only one of the conditions $\left(i\right)f\left(x\right)\ne b,\left(ii\right)f\left(y\right)=b,\left(iii\right)f\left(z\right)\ne a$ is true then the function $f$ is given by the set

• A. $\left\{\right(x,a),(y,b),(z,c\left)\right\}$
• B. $\left\{\right(x,a),(y,c),(z,b\left)\right\}$
• C. $\left\{\right(x,b),(y,a),(z,c\left)\right\}$
• D. $\left\{\right(x,c),(y,b),(z,a\left)\right\}$

Answer 1

The correct option is C $\left\{\right(x,b),(y,a),(z,c\left)\right\}$

$f:\{x,y,z\}\to \{a,b,c\}$ is a one-one function

$\Rightarrow$ each element in $\{x,y,z\}$ will have exactly one image in $\{a,b,c\}$

and no two elements of $\{x,y,z\}$ will have same image in $\{a,b,c\}$

Coming to the given 3 conditions, only one is true.

1) if $f\left(x\right)\ne b$ is true then $f\left(y\right)=b$ is false which makes $f\left(z\right)\ne a$ true $⟹f\left(x\right)\ne b$ is false.

2) if $f\left(y\right)=b$ is true then $f\left(x\right)\ne b$ will also be true $⟹f\left(y\right)=b$ is false

$\therefore f\left(z\right)\ne a$ is the true condition and remainig two are false conditions.

$\therefore f\left(x\right)=b,f\left(y\right)=a,f\left(z\right)=c$

hence $f=\{(x,b),(y,a),(z,c)\}$