Question

Direction: Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.
Assertion (A) : A function f : N $\to$ N given by f(x) = $x^3$ + 2, $\forall$ x $\in$ N is one-one but not onto.
Reason (R) : Since $\forall$ y $\in$ N (Codomain), there does not exist x = $(y - 2)^{1/3}$ in N (Domain) such that f(x) = $x^3$ + 2 = y.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

Pay extra close attention to mathematical quantifiers like $\forall$ (for all) and $\exists$ (there exists). A statement containing $\forall$ is completely invalidated if you can find even one single exception where the property fails to hold true.

Answer 1

The Correct Option is C

Concept: Let a function be defined as $f: X \to Y$, where $X$ is the domain and $Y$ is the codomain.
• One-to-one (Injective): A function $f$ is one-one if distinct elements in the domain have distinct images in the codomain. Mathematically, $f(x_1) = f(x_2) \implies x_1 = x_2$.
• Onto (Surjective): A function $f$ is onto if every element in the codomain $Y$ has at least one pre-image in the domain $X$. That is, for every $y \in Y$, there must exist an $x \in X$ such that $f(x) = y$. In this specific problem, both the domain and the codomain are the set of natural numbers $\mathbb{N} = \{1, 2, 3, 4, \dots\}$.

Step 1: Detailed mathematical analysis of Assertion (A).
The given function is $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x^3 + 2$. Part I: Testing for One-One Functionality
Let $x_1, x_2 \in \mathbb{N}$ such that their functional outputs are equal: \[ f(x_1) = f(x_2) \] Substitute the algebraic definition of the function into the equation: \[ x_1^3 + 2 = x_2^3 + 2 \] Subtracting 2 from both sides of the equality yields: \[ x_1^3 = x_2^3 \] Taking the real cube root on both sides gives: \[ x_1 = x_2 \] Since $f(x_1) = f(x_2)$ uniquely dictates that $x_1 = x_2$, the function is verified to be strictly one-one (injective). Part II: Testing for Onto Functionality
Let $y$ be an arbitrary element belonging to the codomain $\mathbb{N}$. We set up the relation to solve for the pre-image $x$: \[ y = x^3 + 2 \quad \implies \quad x^3 = y - 2 \quad \implies \quad x = (y - 2)^{1/3} \] For $f$ to be onto, this computed value of $x$ must be a natural number ($x \in \mathbb{N}$) for every natural number $y \in \mathbb{N}$. Let us test a counterexample from the codomain, say $y = 1 \in \mathbb{N}$: \[ x = (1 - 2)^{1/3} = (-1)^{1/3} = -1 \notin \mathbb{N} \] Let us test another element from the codomain, say $y = 4 \in \mathbb{N}$: \[ x = (4 - 2)^{1/3} = (2)^{1/3} \approx 1.26 \notin \mathbb{N} \] Since there are many elements in the codomain (such as $1, 2, 4, 5$, etc.) that do not possess a corresponding valid pre-image $x$ in the domain $\mathbb{N}$, the function is not onto (not surjective). Combining both findings, the function is one-one but not onto. Thus, Assertion (A) is completely TRUE.

Step 2: Detailed mathematical analysis of Reason (R).
Let us inspect the phrasing of Reason (R): "Since $\forall y \in \mathbb{N}$ (Codomain), there does not exist $x = (y - 2)^{1/3}$ in $\mathbb{N}$". The quantifier used here is the universal quantifier $\forall$ (which stands for "for all" or "every"). The statement asserts that for every single value of $y \in \mathbb{N}$, a natural pre-image $x$ fails to exist. Let us verify if this claim holds true universally by substituting specific values of $y$: Consider $y = 3 \in \mathbb{N}$ (Codomain): \[ x = (3 - 2)^{1/3} = (1)^{1/3} = 1 \] Since $1 \in \mathbb{N}$ (Domain), a valid natural pre-image does exist for $y = 3$ because $f(1) = 1^3 + 2 = 3$. Consider $y = 10 \in \mathbb{N}$ (Codomain): \[ x = (10 - 2)^{1/3} = (8)^{1/3} = 2 \] Since $2 \in \mathbb{N}$ (Domain), a valid natural pre-image does exist for $y = 10$ because $f(2) = 2^3 + 2 = 10$. Because there are values of $y \in \mathbb{N}$ for which a natural number $x$ does exist, the sweeping declaration that there is no pre-image $\forall y \in \mathbb{N}$ is logically incorrect. The statement is flawed due to the misuse of the universal quantifier. To be correct, it should state that there exist some values of $y$ for which no pre-image exists. Consequently, Reason (R) is mathematically FALSE.