wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let f be an injective function with domain {x,y,z}and range {1,2,3} such that exactly one of the follwowing statements is correct and the remaining are false :

f(x)=1,f(y)1,
f(z)2,
then the value of f1(1) is

A
x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
z
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B y
It gives three cases
Case (1) When f(x)=1 is true.
In this case remaining two are false
f(y)=1 and f(z)=2
This means x and y have the same image, so f(x) is not an injective, which is a contradiction.

Case (2) when f(y)1 is true
If f(y)1is true then the remaining statements are false.
f(x)1 and f(z)=2
i.e., both x and y are not mapped to 1.
So, either both associate to 2 or 3,
Thus, it is not injective.

Case (3) When f(z)2 is true
If f(z)2 is true then remaining statements are fales
If f(x)1 and f(y)=1
But f is injective
Thus, we have f(x)=2,f(y)=1 and f(z)=3
Hence, f1(1)=y

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Inverse of a Function
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon