Question

Let\(f:\left\{ {x,y,z} \right\} \to \left\{ {1,2,3} \right\}\)be a one-one mapping such that only one of the following three statements is true and the remaining two are false: \(\)\(f(x) \ne 2\),\(f(y) = 2\)and\(f(z) \ne 1\), then
A. \(f(x) > f(y) > f(z)\)
B. \(f(x) < f(y) < f(z)\)
C. \(f(y) < f(x) < f(z)\)
D. \(f(y) < f(z) < f(x)\)

Answer 1

Hint: One-one mapping means for one value of image there should be one pre-image and many-one mapping means one image may have one or many pre-images.

Complete step by step solution: According to the question, only one condition should be true and the other should be false so there comes 3 cases wherein
1st \[f(x) \ne 2\]is true and the other two are false,
2nd \[f(y) = 2\]is true and the other two are false,
3rd \[f(z) \ne 1\]is true and the other two are false. Now checking all three cases
Case 1st \[f(x) \ne 2\]is true and \[f(y) = 2\],\[f(z) \ne 1\]is false
\[f(x)\]is one-one function and value of \[f:\left\{ {x,y,z} \right\} \to \left\{ {1,2,3} \right\}\](from the question) means,
 \[f(x)\]=1 or 3, \[f(y) \ne 2\]and \[f(z) = 1\] to be a one-one function, and we get \[f(z) = 1\], \[f(y) = 3\]and \[f(x) = 3\]from the above condition, So clearly we say that it is a many-one function not satisfied for one-one criteria.
Case 2nd \[f(y) = 2\]is true and \[f(x) \ne 2\],\[f(z) \ne 1\]is false. So values we get are \[f(y) = 2\],\[f(x) = 2\]and\[f(z) = 1\], clearly see that one-one condition does not follow. Now
Case 3rd \[f(z) \ne 1\]is true and \[f(x) \ne 2\],\[f(y) = 2\]is false,
 means \[f(z)\]= 2 or 3 ,\[f(x) = 2\]and \[f(y)\]= 1 or 3 to be a one-one function criteria values of
\[f(z) = 3\],\[f(x) = 2\]and \[f(y) = 1\],Now from an arrangement we get \[f(y) < f(x) < f(z)\].

Thus, Option (C) is correct.

Note: Knowing the conditions of one-one function and many-one function will be helpful in the examination. Students without knowing the problem condition and trying to solve the questions create/make mistakes in the solution.