CBSE Class 11 Mathematics Relations and Functions Assignment Set 05

Relations and Functions

Extra/HOT Questions

Question 1. Find x and y if \((x^2 - 3x, y^2 - 5y) = (-2, -6)\).
Answer: Given that the two ordered pairs are equal, we can equate their corresponding components:
\(x^2 - 3x = -2\)
\(y^2 - 5y = -6\)
Solving the first equation for \(x\):
\(x^2 - 3x + 2 = 0\)
\((x - 1)(x - 2) = 0\)
\(x = 1\) or \(x = 2\)
Solving the second equation for \(y\):
\(y^2 - 5y + 6 = 0\)
\((y - 2)(y - 3) = 0\)
\(y = 2\) or \(y = 3\)
Thus, the possible ordered pairs for \((x, y)\) are \((1, 2), (1, 3), (2, 2), (2, 3)\).
In simple words: Equate the first parts to find x and the second parts to find y.

Exam Tip: Two ordered pairs \((a, b)\) and \((c, d)\) are equal if and only if \(a = c\) and \(b = d\).

Question 2. Draw the graph of the following functions: a) Modulus function in \([-4, 4]\), b) Signum function in \([-6, 6]\), c) Greatest integer function in \([-3, 4]\).
Answer:
a) Modulus function \(f(x) = |x|\) for \(x \in [-4, 4]\):
The graph consists of two straight line segments starting from the origin \((0,0)\) and extending to \((-4, 4)\) and \((4, 4)\), forming a V-shape.
b) Signum function \(f(x) = \text{sgn}(x)\) for \(x \in [-6, 6]\):
The graph is a set of three horizontal lines: \(y = -1\) for \(x \in [-6, 0)\), \(y = 0\) at \(x = 0\), and \(y = 1\) for \(x \in (0, 6]\), with open circles at \((0, -1)\) and \((0, 1)\) and a solid dot at \((0, 0)\).
c) Greatest integer function \(f(x) = [x]\) for \(x \in [-3, 4]\):
The graph consists of a series of horizontal step-like line segments (staircase pattern) for each unit interval in \([-3, 4]\).
In simple words: Understand the basic shapes of standard functions like V-shape for modulus, steps for greatest integer, and horizontal lines for signum.

Exam Tip: Pay close attention to closed and open intervals (solid dots vs. hollow circles) when graphing step and signum functions.

Question 3. Find the domain of the following functions: a) \(f(x) = \frac{x^2 - 1}{x - 1}\), b) \(f(x) = \frac{3x + 1}{x^2 - 5x + 6}\), c) \(f(x) = \frac{2x - 3}{(x - 1)(x + 2)}\).
Answer:
a) For \(f(x) = \frac{x^2 - 1}{x - 1}\), the denominator must not be zero:
\(x - 1 \neq 0 \implies x \neq 1\)
Domain = \(\mathbb{R} - \{1\}\)
b) For \(f(x) = \frac{3x + 1}{x^2 - 5x + 6}\), the denominator must not be zero:
\(x^2 - 5x + 6 \neq 0\)
\((x - 2)(x - 3) \neq 0 \implies x \neq 2\) and \(x \neq 3\)
Domain = \(\mathbb{R} - \{2, 3\}\)
c) For \(f(x) = \frac{2x - 3}{(x - 1)(x + 2)}\), the denominator must not be zero:
\((x - 1)(x + 2) \neq 0 \implies x \neq 1\) and \(x \neq -2\)
Domain = \(\mathbb{R} - \{-2, 1\}\)
In simple words: The domain of a rational function is all real numbers except those that make the denominator zero.

Exam Tip: Always factorize quadratic denominators completely to find all restricted values of x.

Question 4. Find the domain and range of the following functions: a) \(f(x) = \frac{1}{9 - x^2}\), b) \(f(x) = \sqrt{x^2 - 1}\), c) \(f(x) = \frac{1}{x^2 + 4}\), d) \(f(x) = \frac{|x|}{x}\).
Answer:
a) \(f(x) = \frac{1}{9 - x^2}\):
Denominator \(9 - x^2 \neq 0 \implies x \neq \pm 3\). Domain = \(\mathbb{R} - \{-3, 3\}\).
For range, let \(y = \frac{1}{9 - x^2} \implies 9 - x^2 = \frac{1}{y} \implies x^2 = 9 - \frac{1}{y} \geq 0 \implies \frac{9y - 1}{y} \geq 0\).
Thus, range is \((-\infty, 0) \cup [\frac{1}{9}, \infty)\).
b) \(f(x) = \sqrt{x^2 - 1}\):
For domain, \(x^2 - 1 \geq 0 \implies x \in (-\infty, -1] \cup [1, \infty)\).
Range = \([0, \infty)\) since square root always yields non-negative values.
c) \(f(x) = \frac{1}{x^2 + 4}\):
Denominator \(x^2 + 4 \neq 0\) for all real \(x\). Domain = \(\mathbb{R}\).
Since \(x^2 \geq 0\), \(x^2 + 4 \geq 4\), so \(0 < \frac{1}{x^2 + 4} \leq \frac{1}{4}\). Range = \((0, \frac{1}{4}]\).
d) \(f(x) = \frac{|x|}{x}\):
Denominator \(x \neq 0\). Domain = \(\mathbb{R} - \{0\}\).
For \(x > 0\), \(f(x) = 1\). For \(x < 0\), \(f(x) = -1\). Range = \(\{-1, 1\}\).
In simple words: Domain is valid input values of x, and range is all resulting output values of y.

Exam Tip: For square root functions, set the expression inside the radical greater than or equal to zero.

Question 5. If \(f(x) = x^2 + \frac{1}{x^2}\) then show that \(f(a) = f(\frac{1}{a})\) and also evaluate \(f(\frac{3}{2}) - f(\frac{2}{3})\).
Answer:
Given \(f(x) = x^2 + \frac{1}{x^2}\).
\(f(a) = a^2 + \frac{1}{a^2}\)
\(f(\frac{1}{a}) = (\frac{1}{a})^2 + \frac{1}{(\frac{1}{a})^2} = \frac{1}{a^2} + a^2 = a^2 + \frac{1}{a^2}\)
Hence, \(f(a) = f(\frac{1}{a})\).
Now, evaluating \(f(\frac{3}{2})\):
\(f(\frac{3}{2}) = (\frac{3}{2})^2 + \frac{1}{(\frac{3}{2})^2} = \frac{9}{4} + \frac{4}{9} = \frac{81 + 16}{36} = \frac{97}{36}\)
Evaluating \(f(\frac{2}{3})\):
Since \(f(a) = f(\frac{1}{a})\), \(f(\frac{2}{3}) = f(\frac{3}{2}) = \frac{97}{36}\).
Therefore, \(f(\frac{3}{2}) - f(\frac{2}{3}) = \frac{97}{36} - \frac{97}{36} = 0\).
In simple words: Substitute the given values directly into the function and simplify.

Exam Tip: Use symmetry properties like \(f(a) = f(1/a)\) to save time during calculations.

Question 6. Let \(R = \{(x, y) / x, y \in \mathbb{N}, x + 2y = 13\}\) then write R as an ordered pair and also find the domain and range.
Answer:
Given \(x + 2y = 13\) where \(x, y \in \mathbb{N}\).
\(x = 13 - 2y\)
Since \(x \in \mathbb{N}\), \(13 - 2y > 0 \implies 2y < 13 \implies y \leq 6\).
Testing values for \(y \in \{1, 2, 3, 4, 5, 6\}\):
If \(y = 1 \implies x = 13 - 2(1) = 11\) (Ordered pair: \((11, 1)\))
If \(y = 2 \implies x = 13 - 2(2) = 9\) (Ordered pair: \((9, 2)\))
If \(y = 3 \implies x = 13 - 2(3) = 7\) (Ordered pair: \((7, 3)\))
If \(y = 4 \implies x = 13 - 2(4) = 5\) (Ordered pair: \((5, 4)\))
If \(y = 5 \implies x = 13 - 2(5) = 3\) (Ordered pair: \((3, 5)\))
If \(y = 6 \implies x = 13 - 2(6) = 1\) (Ordered pair: \((1, 6)\))
Therefore, \(R = \{(11, 1), (9, 2), (7, 3), (5, 4), (3, 5), (1, 6)\}\).
Domain = Set of first elements = \(\{1, 3, 5, 7, 9, 11\}\).
Range = Set of second elements = \(\{1, 2, 3, 4, 5, 6\}\).
In simple words: Find all natural number pairs that satisfy the given equation and list them.

Exam Tip: Always verify that the generated coordinates satisfy the given natural number condition.

Question 7. Let \(A = \{x / x \text{ is a natural number } < 12\}\) and R be a relation in A defined by \((x, y) \in R\) if \(x + y = 12\), then write R.
Answer:
Set \(A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}\).
The relation \(R\) is defined as \(R = \{(x, y) \in A \times A : x + y = 12\}\).
Finding all pairs from set A whose sum is 12:
\(R = \{(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3), (10, 2), (11, 1)\}\).
In simple words: List pairs of numbers from 1 to 11 that add up to 12.

Exam Tip: Make sure both elements of every ordered pair belong strictly to set A.

Question 8. A function f is defined on the set of natural numbers as \(f(x) = \begin{cases} x^2 & \text{if } 1 \leq x < 5 \\ x + 3 & \text{if } 5 < x \leq 8 \\ \frac{x - 3}{2} & \text{if } 8 < x \leq 11 \end{cases}\). Write the function in roster form and also find the domain and range of the function.
Answer:
Domain consists of all integer values of x specified in the intervals: \(x \in \{1, 2, 3, 4, 6, 7, 8, 9, 10, 11\}\) (Note: 5 is excluded).
Calculating function values for each element in the domain:
For \(x = 1 \implies f(1) = 1^2 = 1\)
For \(x = 2 \implies f(2) = 2^2 = 4\)
For \(x = 3 \implies f(3) = 3^2 = 9\)
For \(x = 4 \implies f(4) = 4^2 = 16\)
For \(x = 6 \implies f(6) = 6 + 3 = 9\)
For \(x = 7 \implies f(7) = 7 + 3 = 10\)
For \(x = 8 \implies f(8) = 8 + 3 = 11\)
For \(x = 9 \implies f(9) = \frac{9 - 3}{2} = 3\)
For \(x = 10 \implies f(10) = \frac{10 - 3}{2} = 3.5\) (not a natural number, but function is defined on natural numbers; assuming real outputs: 3.5)
For \(x = 11 \implies f(11) = \frac{11 - 3}{2} = 4\)
Roster form: \(f = \{(1, 1), (2, 4), (3, 9), (4, 16), (6, 9), (7, 10), (8, 11), (9, 3), (10, 3.5), (11, 4)\}\).
Domain = \(\{1, 2, 3, 4, 6, 7, 8, 9, 10, 11\}\).
Range = \(\{1, 3, 3.5, 4, 9, 10, 11, 16\}\).
In simple words: Evaluate the piecewise function for each integer in the given ranges.

Exam Tip: Be careful with strict inequalities (< and >) versus non-strict inequalities (\(\leq\) and \(\geq\)).

Question 9. Let \(A = \{1, 2, 3, 4\}\), \(B = \{-1, 0, 1\}\) and \(C = \{3, 4\}\) then verify the following: a) \(A \times (B \cup C) = (A \times B) \cup (A \times C)\), b) \(A \times (B - C) = (A \times B) - (A \times C)\), c) \(A \times (B \cap C) = (A \times B) \cap (A \times C)\).
Answer:
Given sets: \(A = \{1, 2, 3, 4\}\), \(B = \{-1, 0, 1\}\), \(C = \{3, 4\}\).
a) Verification of \(A \times (B \cup C) = (A \times B) \cup (A \times C)\):
\(B \cup C = \{-1, 0, 1, 3, 4\}\)
\(A \times (B \cup C) = \{(1,-1), (1,0), (1,1), (1,3), (1,4), (2,-1), (2,0), (2,1), (2,3), (2,4), (3,-1), (3,0), (3,1), (3,3), (3,4), (4,-1), (4,0), (4,1), (4,3), (4,4)\}\).
Both \(A \times B\) and \(A \times C\) unioned give the exact same set of ordered pairs. Hence verified.
b) Verification of \(A \times (B - C) = (A \times B) - (A \times C)\):
\(B - C = \{-1, 0, 1\} - \{3, 4\} = \{-1, 0, 1\}\)
\(A \times (B - C) = A \times B\). Since elements of C are not in B, \((A \times B) - (A \times C)\) also equals \(A \times B\). Hence verified.
c) Verification of \(A \times (B \cap C) = (A \times B) \cap (A \times C)\):
\(B \cap C = \{-1, 0, 1\} \cap \{3, 4\} = \emptyset\)
\(A \times (B \cap C) = A \times \emptyset = \emptyset\).
Intersection of \(A \times B\) and \(A \times C\) has no common ordered pairs since second coordinates come from disjoint sets. Hence verified.
In simple words: Cartesian product distributes over union, difference, and intersection of sets.

Exam Tip: Set identity proofs can be verified by evaluating LHS and RHS set operations step by step.

Question 10. If \(A = \{-3, -2, 0, 2, 3\}\) write the subset B of \(A \times A\) such that first element of B is either -3 or +3.
Answer:
Given \(A = \{-3, -2, 0, 2, 3\}\).
\(A \times A\) is the set of all ordered pairs with both coordinates in A.
Subset B consists of all ordered pairs \((x, y) \in A \times A\) such that \(x = -3\) or \(x = 3\).
Listing the pairs:semanticweb
When \(x = -3\): \((-3, -3), (-3, -2), (-3, 0), (-3, 2), (-3, 3)\)
When \(x = 3\): \((3, -3), (3, -2), (3, 0), (3, 2), (3, 3)\)
\(B = \{(-3, -3), (-3, -2), (-3, 0), (-3, 2), (-3, 3), (3, -3), (3, -2), (3, 0), (3, 2), (3, 3)\}\).
In simple words: Select all pairs from the Cartesian product where the first number is -3 or 3.

Exam Tip: Ensure all elements in the subset B have their first component strictly chosen from the specified values.