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Let f be an injective map. with domain (...

Let f be an injective map. with domain (x, y, z and range (1, 2, 3), such that exactly one following statements is correct and the remaining are false : `f(x)=1 , f(y) != 1, f(z) !=2` The value of `f^-1(1)` is

A

x

B

y

C

z

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Following cases arise,
CASE I- When f(x)= 1 is true
In this case, the remaining two statements are false,
`therefore f(y)=1 and f(z)=2`
This means that x, and y have the same image, So, f(x) is not an injection, which is a contradiction.
Hence, f(x)=1 is not true.
CASE-II- When `f(y)ne1` is true
If `f(y) and f(z)=2`
This shows that x and y both are not mapped to 1. So, either both are associated to 3 or both are associted to 2 or one of them is mapped to 3 and other to 2. In all the three cases, f cannot be an injective map.
So, `f(y) ne 1` is not true.
CASE-III- When `f(z) ne 2` is true
If `f(z)ne 2 ` is true, then the remaining statements are false.
`therefore f(x)ne 1 and f(y)=1`
But, f is an injective map.
Thus, we have
f(y)=1, f(z)=3 and f(x)=2
Hence, `f^(-1)(1)=y`.
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