Question

Let $f:x,y,z\to (a,b,c)$ be a one-one function. It is known that only one of the following statements is true:

(i) $f\left(x\right)\ne b$
(ii)$f\left(y\right)=b$
(iii)$f\left(z\right)\ne a$

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

Answer 1

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

When (i) is true, then $f\left(x\right)\ne b,f\left(y\right)\ne b,f\left(z\right)=a$

$\Rightarrow$ Two ordered pair function is possible $f\left(x\right)=a,f\left(y\right)=c,f\left(z\right)=a$ or $f\left(x\right)=c,f\left(y\right)=a,f\left(z\right)=a$

But given $f$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then $f\left(y\right)=b,f\left(z\right)=a,f\left(x\right)=b$. This is also not possible.

Clearly if (iii) is true then it is satisfying every condition.
Hence ordered pair of $f$ is $\left\{\right(x,a),(y,b),(z,c\left)\right\}$