Set, Relations and FunctionsPYQ May 18Question 1957 of 217
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Let N\displaystyle N be the set of all natural numbers; E\displaystyle E be the set of all even natural numbers then the function f:N→E\displaystyle f: N \to E defined f(x)=2x,x∈N\displaystyle f(x) = 2x, x \in N

Options

AOne-One into
BMany-One into
COne-One Onto
DMany-One Onto
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Correct Answer

✅ Option c — One-One Onto

All Options:

  • AOne-One into
  • BMany-One into
  • COne-One Onto
  • DMany-One Onto

Detailed Solution & Explanation

We are given the function f:N→E\displaystyle f: N \to E defined by f(x)=2x\displaystyle f(x) = 2x, where N={1,2,3,… }\displaystyle N = \{1, 2, 3, \dots\} is the set of natural numbers and E={2,4,6,… }\displaystyle E = \{2, 4, 6, \dots\} is the set of even natural numbers.
Let's test the properties of f\displaystyle f:
1. **One-One (Injective)**:
Let x1,x2∈N\displaystyle x_1, x_2 \in N such that f(x1)=f(x2)\displaystyle f(x_1) = f(x_2):
2x1=2x2  ⟹  x1=x22x_1 = 2x_2 \implies x_1 = x_2
Since the images are equal only when the inputs are equal, the function is **One-One**.
2. **Onto (Surjective)**:
Let y∈E\displaystyle y \in E (an even natural number). By definition, y\displaystyle y can be written as y=2k\displaystyle y = 2k for some natural number k∈N\displaystyle k \in N.
We need to find x∈N\displaystyle x \in N such that f(x)=y\displaystyle f(x) = y:
2x=y  ⟹  x=y22x = y \implies x = \frac{y}{2}
Since y\displaystyle y is an even natural number, y/2\displaystyle y/2 is always a natural number. Thus, for every element y∈E\displaystyle y \in E, there exists a pre-image x=y/2∈N\displaystyle x = y/2 \in N such that f(x)=y\displaystyle f(x) = y. Therefore, the function is **Onto**.
Since f\displaystyle f is both One-One and Onto, it is a **One-One Onto** function (Bijective).
*Note: The mathematically correct option is Option C. The textbook answer key incorrectly lists Option B (Many-One into) as the correct choice.*
Hence, **Option C** is the correct answer.

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