Exercise 1.2 Class 12 Maths Relations and Functions

Question 1

Show that \(f: R_* \to R_*\) defined by \(f(x) = 1/x\) is one-one and onto (\(R_*\) is set of non-zero real numbers). Is the result true if domain is replaced by N with co-domain \(R_*\)?

Case 1: Domain \(R_*\), Codomain \(R_*\)

One-One (Injectivity):
Let \(f(x_1) = f(x_2)\).
\(\frac{1}{x_1} = \frac{1}{x_2} \implies x_1 = x_2\).
Hence, f is one-one.
Onto (Surjectivity):
Let \(y \in R_*\) (Codomain). Then \(y = \frac{1}{x} \implies x = \frac{1}{y}\).
Since \(y \neq 0\), \(x\) is a real number and \(x \neq 0\), so \(x \in R_*\) (Domain).
Hence, f is onto.

Case 2: Domain N, Codomain \(R_*\)

One-One:
\(f(x_1) = f(x_2) \implies \frac{1}{x_1} = \frac{1}{x_2} \implies x_1 = x_2\). True.
Onto:
Let \(y = 1.2\) (which is in \(R_*\)).
\(x = \frac{1}{1.2} = \frac{5}{6}\).
But \(\frac{5}{6} \notin N\) (Domain).
So, not all elements in codomain have a pre-image in N.
Hence, f is not onto.

Question 2

Check injectivity and surjectivity:

The graph above helps visualize that for real numbers or integers, \(x^2\) hits the same value twice (e.g., at -2 and 2).

(i) \(f: N \to N, f(x) = x^2\)

One-One: Yes. For \(x_1, x_2 \in N\), \(x_1^2 = x_2^2 \implies x_1 = x_2\) (Since x is positive).
Onto: No. \(2 \in N\) (Codomain) but \(\sqrt{2} \notin N\) (Domain).

(ii) \(f: Z \to Z, f(x) = x^2\)

One-One: No. \(f(-1) = 1\) and \(f(1) = 1\). Different inputs, same output.
Onto: No. \(-2 \in Z\) (Codomain) has no pre-image (square can’t be negative).

(iii) \(f: R \to R, f(x) = x^2\)

One-One: No. \(f(-1) = 1, f(1) = 1\).
Onto: No. Range is \([0, \infty)\). Negative numbers in codomain are not covered.

(iv) \(f: N \to N, f(x) = x^3\)

One-One: Yes. \(x_1^3 = x_2^3 \implies x_1 = x_2\).
Onto: No. \(2 \in N\) (Codomain) is not a perfect cube of any natural number.

(v) \(f: Z \to Z, f(x) = x^3\)

One-One: Yes. Cubes preserve sign, so distinct integers have distinct cubes.
Onto: No. \(2 \in Z\) is not a cube of any integer.

Question 3

Prove that Greatest Integer Function \(f: R \to R, f(x) = [x]\) is neither one-one nor onto.

Injectivity (One-One)

Let \(x_1 = 1.2\) and \(x_2 = 1.9\).
\(f(1.2) = [1.2] = 1\)
\(f(1.9) = [1.9] = 1\)
Since distinct inputs give the same output, f is not one-one.

Surjectivity (Onto)

The range of \(f(x) = [x]\) is the set of Integers (Z).
However, the codomain is Real numbers (R).
Take \(y = 0.7 \in R\). There is no \(x\) such that \([x] = 0.7\) (output must be integer).
Hence, f is not onto.

Question 4

Show that Modulus Function \(f: R \to R, f(x) = |x|\) is neither one-one nor onto.

[Image of modulus function graph]

Proof

Not One-One: \(f(-1) = |-1| = 1\) and \(f(1) = |1| = 1\).
Since \(f(-1) = f(1)\) but \(-1 \neq 1\), it is not injective.
Not Onto: The value of \(|x|\) is always non-negative. The range is \([0, \infty)\).
Negative numbers in the codomain R (e.g., -5) have no pre-image.

Question 5

Show that Signum Function is neither one-one nor onto.

Definition: \(f(x) = 1\) if \(x>0\), \(0\) if \(x=0\), \(-1\) if \(x<0\).

Not One-One: \(f(2) = 1\) and \(f(3) = 1\). Many inputs map to 1.
Not Onto: Range is only \(\{-1, 0, 1\}\). Elements like \(2\) in the codomain R have no pre-image.

Question 6

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)}. Show f is one-one.

The images are:
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)

Since distinct elements in A have distinct images in B (4, 5, 6 are all different), the function is one-one.

Question 7

State whether function is one-one, onto or bijective.

(i) \(f: R \to R, f(x) = 3 – 4x\)

One-One: \(3 – 4x_1 = 3 – 4x_2 \implies -4x_1 = -4x_2 \implies x_1 = x_2\). (Yes).
Onto: Let \(y = 3 – 4x \implies x = \frac{3-y}{4}\). For any \(y \in R\), \(x\) is defined and real. (Yes).

Conclusion: Bijective.

(ii) \(f: R \to R, f(x) = 1 + x^2\)

One-One: No. \(f(1) = 2, f(-1) = 2\).
Onto: No. \(x^2 \ge 0 \implies 1+x^2 \ge 1\). Range is \([1, \infty)\). Negative numbers in codomain not covered.

Conclusion: Neither one-one nor onto.

Question 8

Show that \(f: A \times B \to B \times A\) defined by \(f(a, b) = (b, a)\) is bijective.

One-One: Let \(f(a_1, b_1) = f(a_2, b_2)\).
\((b_1, a_1) = (b_2, a_2)\).
By definition of ordered pairs, \(b_1 = b_2\) and \(a_1 = a_2\).
Thus \((a_1, b_1) = (a_2, b_2)\). Yes.
Onto: Let \((b, a)\) be any element in \(B \times A\).
There exists an element \((a, b) \in A \times B\) such that \(f(a, b) = (b, a)\). Yes.

Conclusion: Bijective.

Question 9

Is \(f: N \to N\) defined by \(f(n) = \frac{n+1}{2}\) (if n odd), \(\frac{n}{2}\) (if n even) bijective?

Analysis

One-One:
Let \(n=1\) (odd). \(f(1) = \frac{1+1}{2} = 1\).
Let \(n=2\) (even). \(f(2) = \frac{2}{2} = 1\).
Since \(f(1) = f(2)\) but \(1 \neq 2\), it is not one-one.
Onto:
For any \(m \in N\), take \(n = 2m\) (even). \(f(2m) = \frac{2m}{2} = m\).
Every natural number has a pre-image. It is onto.

Conclusion: Not bijective (Only onto).

Question 10

\(A = R – \{3\}, B = R – \{1\}\). \(f: A \to B\) defined by \(f(x) = \frac{x-2}{x-3}\). Is f one-one and onto?

One-One

\(\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}\)
\((x_1-2)(x_2-3) = (x_2-2)(x_1-3)\)
\(x_1x_2 – 3x_1 – 2x_2 + 6 = x_1x_2 – 3x_2 – 2x_1 + 6\)
\(-3x_1 – 2x_2 = -3x_2 – 2x_1\)
\(-x_1 = -x_2 \implies x_1 = x_2\).
Yes, one-one.

Onto

Let \(y = \frac{x-2}{x-3}\). Solve for x.
\(y(x-3) = x-2 \implies xy – 3y = x – 2\)
\(xy – x = 3y – 2 \implies x(y-1) = 3y – 2\)
\(x = \frac{3y-2}{y-1}\).
For any \(y \in B\) (i.e., \(y \neq 1\)), \(x\) is defined.
We must check if \(x=3\). \(\frac{3y-2}{y-1} = 3 \implies 3y-2 = 3y-3 \implies -2 = -3\) (Impossible).
So \(x\) is never 3, thus \(x \in A\).
Yes, onto.

Question 11

\(f: R \to R, f(x) = x^4\).

One-One: \(f(1) = 1, f(-1) = 1\). (No).
Onto: Range is \([0, \infty)\). Negatives not covered. (No).

Answer: (D) Neither one-one nor onto.

Question 12

\(f: R \to R, f(x) = 3x\).

One-One: \(3x_1 = 3x_2 \implies x_1 = x_2\). (Yes).
Onto: \(y = 3x \implies x = y/3\). Real for all y. (Yes).

Answer: (A) One-one onto.

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