Board 2025 Class 12 Maths Q19 Solution: Is f(x) = 3x − 5 a bijection from Z to Z?

The question

Select the correct answer from the options (A), (B), (C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and the 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.

Assertion (A) : Let Z be the set of integers. A function $f : Z \to Z$ defined as $f(x) = 3x - 5$ ⁠, $\forall x \in Z$ is a bijective.

Reason (R) : A function is a bijective if it is both surjective and injective.

Exam answer

One-one: let $x_1, x_2 \in Z$ ⁠.

\begin{aligned} f(x_1) &= f(x_2) \\ \Rightarrow 3x_1 - 5 &= 3x_2 - 5 \\ \Rightarrow x_1 &= x_2 \end{aligned}

So f is one-one.

Why?

To prove one-one, assume two inputs give the same output and show the inputs are equal. Add 5 to both sides, then divide by 3.

\begin{aligned} 3x_1 - 5 &= 3x_2 - 5 \\ 3x_1 - 5 + 5 &= 3x_2 - 5 + 5 \\ 3x_1 &= 3x_2 \\ \frac{3x_1}{3} &= \frac{3x_2}{3} \\ x_1 &= x_2 \end{aligned}

Onto: for $y \in Z$ ⁠, $3x - 5 = y$ gives $x = \frac{y + 5}{3}$ ⁠, which need not be an integer.

For $0 \in Z$ ⁠, $3x - 5 = 0$ gives $x = \frac{5}{3} \notin Z$ ⁠, so f is not onto.

Why?

Onto means every element of the codomain Z has a pre-image in the domain Z. One integer with no pre-image is enough to break it, and 0 is one.

Solving for x, one move per line:

\begin{aligned} 3x - 5 &= y \\ 3x &= y + 5 \\ x &= \frac{y + 5}{3} \end{aligned}

With $y = 0$ ⁠:

\begin{aligned} x &= \frac{0 + 5}{3} = \frac{5}{3} \end{aligned}

$\frac53 = 1\frac23$ lies between 1 and 2, so it is not an integer.

In fact the values of f are $\dots, -5, -2, 1, 4, \dots$ ⁠: only the integers that leave remainder 1 on division by 3.

Watch out: as a function from the real numbers to the real numbers, $3x - 5$ is a bijection. Here the domain and codomain are Z, and that changes the answer.

So f is not a bijection, and Assertion (A) is false.

Reason (R) is true: it is the definition of a bijective function.

Why?

Bijective means one-one (injective) and onto (surjective) together, which is exactly what the Reason says.

The Assertion is false and the Reason is true. That is code (D), whatever the link between them.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Where this comes from

Chapter 1, Relations and Functions: Bijective functions.

Question type: Prove one-one and onto

How to spot it: click to see

Type 5. Prove one-one and onto: the bijective function

Spot it: "show that $f$ is one-one and onto", or "is $f$ bijective?", with a formula like $f(x) = 3 - 4x$ ⁠.

Here: $3x_1 - 5 = 3x_2 - 5$ gives $x_1 = x_2$ ⁠, so $f$ is one-one. For onto, $x = \frac{y + 5}{3}$ fails the domain check in $Z$ ⁠: $y = 0$ needs $x = \frac53$ ⁠. So the Assertion is false.

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Q19: Is f(x)=3x−5f(x) = 3x - 5f(x)=3x−5 a bijection from Z to Z?

Q19: Is $f(x) = 3x - 5$ a bijection from Z to Z?