Class 12-NCERT Solutions-Chapter-1-Relations and Functions-Ex 1.2

Solution:

Part 1: Domain and Codomain are \( \mathbf{R}_* \).

• One-one: Let \( f(x_1) = f(x_2) \).
  \( \Rightarrow \frac{1}{x_1} = \frac{1}{x_2} \)
  \( \Rightarrow x_1 = x_2 \).
  Thus, f is one-one.
• Onto: Let \( y \in \mathbf{R}_* \) be any element in the codomain.
  We need \( f(x) = y \Rightarrow \frac{1}{x} = y \Rightarrow x = \frac{1}{y} \).
  Since \( y \neq 0 \), \( x = \frac{1}{y} \) is a non-zero real number (\( x \in \mathbf{R}_* \)).
  Thus, f is onto.

Part 2: Domain is \( \mathbf{N} \), Codomain is \( \mathbf{R}_* \).

• One-one: It remains one-one (subset of previous domain).
• Onto: Consider \( y = 1.2 \) (or any non-integer like 2.5) in the codomain \( \mathbf{R}_* \).
  \( f(x) = 1.2 \Rightarrow \frac{1}{x} = 1.2 \Rightarrow x = \frac{1}{1.2} = \frac{5}{6} \).
  But \( \frac{5}{6} \notin \mathbf{N} \).
  So, not every element in the codomain has a pre-image in the domain \( \mathbf{N} \).

Therefore, the result is not true if the domain is replaced by \( \mathbf{N} \) (it becomes not onto).

Solution:

(i) \( f(x) = x^2 \) on \( \mathbf{N} \):

• Injective (One-one): \( x_1^2 = x_2^2 \Rightarrow x_1 = x_2 \) (since natural numbers are positive). Yes.
• Surjective (Onto): 2 is in codomain \( \mathbf{N} \). \( x^2 = 2 \Rightarrow x = \sqrt{2} \notin \mathbf{N} \). No.

(ii) \( f(x) = x^2 \) on \( \mathbf{Z} \):

• Injective: \( f(1) = 1^2 = 1 \) and \( f(-1) = (-1)^2 = 1 \). Not injective.
• Surjective: -2 is in codomain \( \mathbf{Z} \). \( x^2 = -2 \) has no solution. No.

(iii) \( f(x) = x^2 \) on \( \mathbf{R} \):

• Injective: \( f(1) = f(-1) = 1 \). Not injective.
• Surjective: -2 is in codomain \( \mathbf{R} \). Square of a real number cannot be negative. No.

(iv) \( f(x) = x^3 \) on \( \mathbf{N} \):

• Injective: \( x_1^3 = x_2^3 \Rightarrow x_1 = x_2 \). Yes.
• Surjective: 2 is in codomain. \( x^3 = 2 \Rightarrow x = \sqrt[3]{2} \notin \mathbf{N} \). No.

(v) \( f(x) = x^3 \) on \( \mathbf{Z} \):

• Injective: Yes, cube of distinct integers is distinct.
• Surjective: 2 is in codomain. No integer cubed gives 2. No.

Solution:

• One-one: Consider \( x_1 = 1.2 \) and \( x_2 = 1.9 \).
  \( f(1.2) = [1.2] = 1 \)
  \( f(1.9) = [1.9] = 1 \)
  Since distinct elements have the same image, f is not one-one.
• Onto: The range of \( f(x) = [x] \) is the set of all integers \( \mathbf{Z} \).
  However, the codomain is \( \mathbf{R} \).
  Elements like \( 0.7 \) in the codomain have no pre-image.
  Thus, f is not onto.

Solution:

• One-one: \( f(-1) = |-1| = 1 \) and \( f(1) = |1| = 1 \).
  Since \( f(-1) = f(1) \) but \( -1 \neq 1 \), f is not one-one.
• Onto: The codomain is \( \mathbf{R} \) (all real numbers).
  The value of \( |x| \) is always non-negative.
  Negative numbers in the codomain (e.g., -5) do not have a pre-image.
  Thus, f is not onto.

Solution:

• One-one: \( f(1) = 1 \) and \( f(2) = 1 \).
  Distinct elements map to same image. Not one-one.
• Onto: The range of f is \( \{-1, 0, 1\} \).
  The codomain is \( \mathbf{R} \).
  Elements like 2, 3, 1.5 in the codomain are not in the range. Not onto.

Solution:

The images of elements are:

\( f(1) = 4 \)

\( f(2) = 5 \)

\( f(3) = 6 \)

Since distinct elements in the domain (1, 2, 3) have distinct images in the codomain (4, 5, 6), the function is one-one.

Solution:

(i) \( f(x) = 3 - 4x \)

• One-one: \( 3 - 4x_1 = 3 - 4x_2 \Rightarrow -4x_1 = -4x_2 \Rightarrow x_1 = x_2 \). Yes.
• Onto: For any \( y \in \mathbf{R} \), let \( y = 3 - 4x \). Then \( x = \frac{3-y}{4} \). Since \( y \in \mathbf{R} \), \( x \) is also real. Yes.
• Therefore, f is bijective.

(ii) \( f(x) = 1 + x^2 \)

• One-one: \( f(1) = 2, f(-1) = 2 \). No.
• Onto: Minimum value is 1 (since \( x^2 \geq 0 \)). Any value less than 1 (e.g., 0) in the codomain has no pre-image. No.
• Therefore, f is neither one-one nor onto.

Solution:

• One-one:
  Let \( f(a_1, b_1) = f(a_2, b_2) \).
  \( \Rightarrow (b_1, a_1) = (b_2, a_2) \).
  \( \Rightarrow b_1 = b_2 \) and \( a_1 = a_2 \).
  \( \Rightarrow (a_1, b_1) = (a_2, b_2) \).
  So, f is one-one.
• Onto:
  For any arbitrary element \( (b, a) \in B \times A \), there exists \( (a, b) \in A \times B \) such that \( f(a, b) = (b, a) \).
  So, f is onto.

Therefore, f is a bijective function.

Solution:

• One-one check:
  \( f(1) = \frac{1+1}{2} = 1 \)
  \( f(2) = \frac{2}{2} = 1 \)
  Since \( f(1) = f(2) \) but \( 1 \neq 2 \), the function is not one-one.
• Onto check:
  For any \( y \in \mathbf{N} \), consider \( n = 2y \) (which is even and in \( \mathbf{N} \)).
  Then \( f(2y) = \frac{2y}{2} = y \).
  So every natural number has a pre-image. It is onto.

Since it is not one-one, it is not bijective.

Solution:

• One-one:
  Let \( f(x_1) = f(x_2) \).
  \( \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
  \( (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \)
  \( x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \)
  \( -3x_1 - 2x_2 = -3x_2 - 2x_1 \)
  \( -x_1 = -x_2 \Rightarrow x_1 = x_2 \)
  Yes, it is one-one.
• Onto:
  Let \( y = \frac{x-2}{x-3} \).
  \( y(x-3) = x-2 \)
  \( xy - 3y = x - 2 \)
  \( xy - x = 3y - 2 \)
  \( x(y-1) = 3y - 2 \)
  \( x = \frac{3y-2}{y-1} \)
  For any \( y \in B \) (so \( y \neq 1 \)), \( x \) is defined.
  We must check if \( x = 3 \). If \( \frac{3y-2}{y-1} = 3 \), then \( 3y-2 = 3y-3 \Rightarrow -2 = -3 \) (False). So \( x \neq 3 \), meaning \( x \in A \).
  Yes, it is onto.

Answer: (D)

Solution:

• \( f(1) = 1, f(-1) = 1 \). Not one-one (Many-one).
• Range is \( [0, \infty) \). Negative numbers in codomain \( \mathbf{R} \) have no pre-image. Not onto.

Therefore, f is neither one-one nor onto.

Answer: (A)

Solution:

• \( 3x_1 = 3x_2 \Rightarrow x_1 = x_2 \). One-one.
• For any \( y \in \mathbf{R} \), \( x = y/3 \in \mathbf{R} \). Onto.

Therefore, f is one-one onto.