CBSE Class 12 Mathematics Relations And Functions Assignment Set 13

Question 1. Define one-one and onto function. Give example and also draw diagram.
Answer:
1. One-one function (Injective function): A function \(f: A \to B\) is defined to be one-one if distinct elements of \(A\) have distinct images in \(B\). That is, for all \(x_1, x_2 \in A\),
\(f(x_1) = f(x_2) \implies x_1 = x_2\)
Alternatively, \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\).
2. Onto function (Surjective function): A function \(f: A \to B\) is said to be onto if every element of \(B\) is the image of some element of \(A\). That is, for every \(y \in B\), there exists an element \(x \in A\) such that \(f(x) = y\).

Example: Let \(A = \{1, 2, 3\}\) and \(B = \{4, 5, 6\}\). Define \(f: A \to B\) by \(f(1) = 4, f(2) = 5, f(3) = 6\). This function is both one-one and onto.
In simple words: A one-one function assigns different outputs to different inputs, while an onto function covers every single possible output in the target set.

Exam Tip: To prove a function is one-one, always start by assuming \(f(x_1) = f(x_2)\) and show that \(x_1 = x_2\). To prove onto, express \(x\) in terms of \(y\) from \(y = f(x)\) and check if all elements in the codomain have a pre-image.

Question 2. What is the range of the function \(f(x) = \frac{|x-1|}{x-1}\)?
Answer:
Given function: \(f(x) = \frac{|x-1|}{x-1}\), where \(x \neq 1\).
Case 1: When \(x > 1\), \(|x-1| = x-1\), so
\(f(x) = \frac{x-1}{x-1} = 1\)
Case 2: When \(x < 1\), \(|x-1| = -(x-1)\), so
\(f(x) = \frac{-(x-1)}{x-1} = -1\)
Thus, the range of the function is the set \(\{ -1, 1 \}\).
In simple words: The function only ever outputs either positive 1 or negative 1 depending on whether x is greater than or less than 1.

Exam Tip: Always remember to check the domain restriction (\(x \neq 1\)) when dealing with modulus functions in denominators.

Question 3. Let \(A = \{1, 2, 3\}\), \(B = \{4, 5, 6, 7\}\) and let \(f = \{(1, 4), (2, 5), (3, 6)\}\) be a function from \(A\) to \(B\). State whether \(f\) is one-one or not.
Answer:
Given \(A = \{1, 2, 3\}\) and \(B = \{4, 5, 6, 7\}\).
The function is given by \(f = \{(1, 4), (2, 5), (3, 6)\}\).
Here, distinct elements of \(A\) have distinct images in \(B\):
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)
Since no two different elements in domain \(A\) share the same image in \(B\), the function \(f\) is one-one.
In simple words: Every input in A points to a completely separate output in B, so it is one-one.

Exam Tip: For roster-form functions, check if any second coordinate (y-value) repeats. If y-values repeat, it is not one-one.

Question 4. Let \(f: N \to N\) be defined as
\(f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}\)
\(\forall n \in N\). Find whether the function is bijective.
Answer:
Let us check injectivity (one-one):
Take \(n_1 = 1\) (odd) and \(n_2 = 2\) (even).
\(f(1) = \frac{1+1}{2} = 1\)
\(f(2) = \frac{2}{2} = 1\)
Here, \(f(1) = f(2) = 1\), but \(1 \neq 2\).
Therefore, \(f\) is not one-one.

Let us check surjectivity (onto):
For any natural number \(y \in N\), we can choose an even number \(n = 2y \in N\).
Then \(f(2y) = \frac{2y}{2} = y\).
Thus, every element \(y \ln N\) has a pre-image in the domain \(N\).
Therefore, \(f\) is onto.

Since the function is not one-one, it is not bijective.
In simple words: Even though every number in the target set has a match, two different numbers give the exact same output, so it fails to be one-one and thus is not bijective.

Exam Tip: To show a function is not one-one, finding a single counterexample where two different inputs yield the same output is sufficient.

Question 5. Let \(A = R - \{3\}\) and \(B = R - \{1\}\). The function \(f: A \to B\) defined by \(f(x) = \frac{x-2}{x-3}\). Is \(f\) one-one and onto? Justify your answer.
Answer:
Injectivity (One-one):
Let \(x_1, x_2 \in A\) such that \(f(x_1) = f(x_2)\).
\(\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}\)
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6\)
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\)
\(x_1 = x_2\)
Therefore, \(f\) is one-one.

Surjectivity (Onto):
Let \(y \in B\) (where \(y \neq 1\)). Let \(y = f(x)\).
\(y = \frac{x-2}{x-3}\)
\(y(x - 3) = x - 2\)
\(yx - 3y = x - 2\)
\(yx - x = 3y - 2\)
\(x(y - 1) = 3y - 2\)
\(x = \frac{3y - 2}{y - 1}\)
Since \(y \neq 1\), \(x\) is a real number, and since \(y \neq 1\), we can check if \(x = 3\):
If \(\frac{3y - 2}{y - 1} = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3\) (which is impossible).
Thus, \(x \in A\) for all \(y \in B\).
Hence, for every \(y \in B\), there exists \(x = \frac{3y-2}{y-1} \in A\) such that \(f(x) = y\).
Therefore, \(f\) is onto.
In simple words: Algebraic cross-multiplication proves every output comes from a unique input, and every target value has a valid pre-image, making the function both one-one and onto.

Exam Tip: When proving onto for rational functions, always express \(x\) in terms of \(y\) and ensure the denominator for \(x\) cannot be zero for any value in the codomain.

Question 6. \(f: R \to R\) defined by \(f(x) = 3-4x\). Prove \(f\) is a bijective function i.e., one-one and onto.
Answer:
One-one:
Let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\).
\(3 - 4x_1 = 3 - 4x_2\)
\(-4x_1 = -4x_2\)
\(x_1 = x_2\)
Thus, \(f\) is one-one.

Onto:
Let \(y \in R\) (codomain). Let \(y = f(x)\).
\(y = 3 - 4x\)
\(4x = 3 - y\)
\(x = \frac{3-y}{4}\)
For any real number \(y \in R\), \(x = \frac{3-y}{4}\) is also a real number (\(x \in R\)).
Also, \(f\left(\frac{3-y}{4}\right) = 3 - 4\left(\frac{3-y}{4}\right) = 3 - (3 - y) = y\).
Thus, every element in the codomain has a pre-image in the domain. Therefore, \(f\) is onto.

Since \(f\) is both one-one and onto, it is a bijective function.
In simple words: Linear functions with non-zero slopes are always one-to-one and cover all real numbers, making them bijective.

Exam Tip: Linear functions of the form \(f(x) = ax + b\) (where \(a \neq 0\)) are always bijective from \(R \to R\).

Question 7. Is the following function \(f(x) = 1+x^2\), \(f: R \to R\) is one-one, onto?
Answer:
One-one check:
Let \(x_1 = 1\) and \(x_2 = -1\).
\(f(1) = 1 + (1)^2 = 2\)
\(f(-1) = 1 + (-1)^2 = 2\)
Here, \(f(1) = f(-1) = 2\), but \(1 \neq -1\).
Therefore, \(f\) is not one-one.

Onto check:
Notice that for all \(x \in R\), \(x^2 \geq 0\), which implies \(1 + x^2 \geq 1\).
Thus, the range of \(f\) is \([1, \infty)\), which is a proper subset of the codomain \(R\).
Negative real numbers in the codomain (e.g., \(-1\)) do not have any pre-image in \(R\).
Therefore, \(f\) is not onto.
In simple words: Because squaring negative numbers gives positive results and the outputs can never be less than 1, the function fails both tests.

Exam Tip: Even powers like \(x^2\) on real domains \(R \to R\) are never one-one because \(f(x) = f(-x)\).

Question 8. Test the injectivity and surjectivity of \(f: N \to N\) : \(f(x) = x^2\).
Answer:
Injectivity (One-one):
Let \(x_1, x_2 \in N\) such that \(f(x_1) = f(x_2)\).
\(x_1^2 = x_2^2\)
Since \(x_1, x_2 \in N\) (positive integers), we take the positive square root:
\(x_1 = x_2\)
Therefore, \(f\) is one-one.

Surjectivity (Onto):
Let \(y \in N\) be an element in the codomain. Let \(y = f(x) = x^2\), which means \(x = \sqrt{y}\).
For \(y = 2 \in N\), \(x = \sqrt{2}\), but \(\sqrt{2} \notin N\) (not a natural number).
Thus, elements like \(2, 3, 5, 6\) have no pre-image in domain \(N\).
Therefore, \(f\) is not onto.
In simple words: While different natural numbers give different squares, not every natural number is a perfect square (like 2 or 3), so it is not onto.

Exam Tip: Always pay close attention to the domain and codomain sets (\(N\) vs \(R\)) as they completely change whether a function is onto or not.

Question 9. Show that the function \(f: R \to R\) given by \(f(x) = x^3\) is injective.
Answer:
Let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\).
\(x_1^3 = x_3^3\)
Taking the cube root on both sides:
\(x_1 = x_2\)
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\) for all \(x_1, x_2 \in R\), the function \(f\) is injective (one-one).
In simple words: Odd powers preserve signs, ensuring that every input has a unique output, proving the function is one-one.

Exam Tip: Odd degree polynomial functions \(f(x) = x^{2n+1}\) on real numbers are always strictly increasing and hence one-one.

Question 10. If \(f: R \to R\) such that \(f(x) = x^2 - 3x + 2\). Find \(f(f(x))\).
Answer:
Given \(f(x) = x^2 - 3x + 2\).
We need to find \(f(f(x))\):
\(f(f(x)) = (f(x))^2 - 3(f(x)) + 2\)
Substitute \(f(x) = x^2 - 3x + 2\):
\(f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Now expand each term:
1. \((x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(-3x)(2) + 2(x^2)(2)\)
\(= x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2\)
\(= x^4 - 6x^3 + 13x^2 - 12x + 4\)

2. \(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\)

3. Add the constant term \(+ 2\).

Combine all terms together:
\(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\)
\(= x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\)
\(= x^4 - 6x^3 + 10x^2 - 3x\)
In simple words: Substitute the entire function expression in place of every x inside the original function, then expand and simplify carefully.

Exam Tip: Group like powers of \(x\) carefully during polynomial expansion to avoid calculation errors.