This MCQ module is based on: Exercises and Summary – Relations and Functions
TOPIC 7 OF 45
Exercises and Summary – Relations and Functions
🎓 Class 11
Mathematics
CBSE
Theory
Ch 2 — Relations and Functions
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Exercises and Summary – Relations and Functions
Targeting Class 11 level in Functions, with Advanced difficulty.
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2.4.2 Algebra of Real Functions
In this section, we shall learn how to add two real functions, subtract a real function from another, multiply a real function by a scalar (by a scalar we mean a real number), multiply two real functions and divide one real function by another.
(i) Addition of Two Real Functions
Let \(f: X \rightarrow \mathbf{R}\) and \(g: X \rightarrow \mathbf{R}\) be any two real functions, where \(X \subset \mathbf{R}\). Then, we define \((f + g): X \rightarrow \mathbf{R}\) by
\[(f + g)(x) = f(x) + g(x), \text{ for all } x \in X.\]
(ii) Subtraction of a Real Function from Another
Let \(f: X \rightarrow \mathbf{R}\) and \(g: X \rightarrow \mathbf{R}\) be any two real functions, where \(X \subset \mathbf{R}\). Then, we define \((f - g): X \rightarrow \mathbf{R}\) by
\[(f - g)(x) = f(x) - g(x), \text{ for all } x \in X.\]
(iii) Multiplication by a Scalar
Let \(f: X \rightarrow \mathbf{R}\) be a real valued function and \(\alpha\) be a scalar. Then the product \(\alpha f\) is a function from X to \(\mathbf{R}\) defined by
\[(\alpha f)(x) = \alpha \cdot f(x), \text{ for all } x \in X.\]
(iv) Multiplication of Two Real Functions
The product (or multiplication) of two real functions \(f: X \rightarrow \mathbf{R}\) and \(g: X \rightarrow \mathbf{R}\) is a function \(fg: X \rightarrow \mathbf{R}\) defined by
\[(fg)(x) = f(x) \cdot g(x), \text{ for all } x \in X.\]
This is also called pointwise multiplication.
(v) Quotient of Two Real Functions
Let \(f\) and \(g\) be two real functions defined from \(X \rightarrow \mathbf{R}\), where \(X \subset \mathbf{R}\). The quotient of \(f\) by \(g\) denoted by \(\frac{f}{g}\) is a function defined by
\[\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, \text{ provided } g(x) \neq 0, \; x \in X.\]
Worked Examples -- Algebra of Functions
Example 16
Let \(f(x) = x^2\) and \(g(x) = 2x + 1\) be two real functions. Find \((f + g)(x)\), \((f - g)(x)\), \((fg)(x)\), and \(\left(\frac{f}{g}\right)(x)\).
Solution
We have:
\((f + g)(x) = f(x) + g(x) = x^2 + 2x + 1 = (x + 1)^2\)
\((f - g)(x) = f(x) - g(x) = x^2 - 2x - 1\)
\((fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2\)
\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1},\quad x \neq -\frac{1}{2}\)
\((f + g)(x) = f(x) + g(x) = x^2 + 2x + 1 = (x + 1)^2\)
\((f - g)(x) = f(x) - g(x) = x^2 - 2x - 1\)
\((fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2\)
\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1},\quad x \neq -\frac{1}{2}\)
Example 17
Let \(f(x) = \sqrt{x}\) and \(g(x) = x\) be two functions defined over the set of non-negative real numbers. Find \((f + g)(x)\), \((f - g)(x)\), \((fg)(x)\) and \(\left(\frac{f}{g}\right)(x)\).
Solution
We have:
\((f + g)(x) = \sqrt{x} + x\)
\((f - g)(x) = \sqrt{x} - x\)
\((fg)(x) = \sqrt{x} \cdot x = x^{3/2}\)
\(\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} = x^{-1/2},\quad x \neq 0\)
\((f + g)(x) = \sqrt{x} + x\)
\((f - g)(x) = \sqrt{x} - x\)
\((fg)(x) = \sqrt{x} \cdot x = x^{3/2}\)
\(\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} = x^{-1/2},\quad x \neq 0\)
Example 18
Let \(\mathbf{R}\) be the set of real numbers. Define the real function \(f: \mathbf{R} \rightarrow \mathbf{R}\) by \(f(x) = x + 10\) and sketch the graph of this function.
Solution
Here \(f(0) = 10\), \(f(1) = 11\), \(f(2) = 12\), ..., \(f(10) = 20\), etc., and \(f(-1) = 9\), \(f(-2) = 8\), ..., \(f(-10) = 0\), and so on.
Therefore, the shape of the graph of \(f(x) = x + 10\) is a straight line (Fig 2.16).
Remark: The function \(f\) defined by \(f(x) = mx + c\), \(x \in \mathbf{R}\), is called a linear function, where \(m\) and \(c\) are constants. The above function is an example of a linear function.
Therefore, the shape of the graph of \(f(x) = x + 10\) is a straight line (Fig 2.16).
Remark: The function \(f\) defined by \(f(x) = mx + c\), \(x \in \mathbf{R}\), is called a linear function, where \(m\) and \(c\) are constants. The above function is an example of a linear function.
Fig 2.16 -- Graph of the linear function f(x) = x + 10
Miscellaneous Examples
Example 19
Let R be a relation from \(\mathbf{Q}\) to \(\mathbf{Q}\) defined by \(R = \{(a, b) : a, b \in \mathbf{Q}\) and \(a - b \in \mathbf{Z}\}\). Show that:
(i) \((a, a) \in R\) for all \(a \in \mathbf{Q}\) (ii) \((a, b) \in R\) implies \((b, a) \in R\) (iii) \((a, b) \in R\) and \((b, c) \in R\) implies \((a, c) \in R\).
Solution
(i) Since \(a - a = 0 \in \mathbf{Z}\), it follows that \((a, a) \in R\).
(ii) \((a, b) \in R\) implies that \(a - b \in \mathbf{Z}\). So, \(b - a = -(a - b) \in \mathbf{Z}\). Therefore, \((b, a) \in R\).
(iii) \((a, b) \in R\) and \((b, c) \in R\) implies that \(a - b \in \mathbf{Z}\) and \(b - c \in \mathbf{Z}\). So, \(a - c = (a - b) + (b - c) \in \mathbf{Z}\). Therefore, \((a, c) \in R\).
(ii) \((a, b) \in R\) implies that \(a - b \in \mathbf{Z}\). So, \(b - a = -(a - b) \in \mathbf{Z}\). Therefore, \((b, a) \in R\).
(iii) \((a, b) \in R\) and \((b, c) \in R\) implies that \(a - b \in \mathbf{Z}\) and \(b - c \in \mathbf{Z}\). So, \(a - c = (a - b) + (b - c) \in \mathbf{Z}\). Therefore, \((a, c) \in R\).
Example 20
Let \(f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}\) be a linear function from \(\mathbf{Z}\) into \(\mathbf{Z}\). Find \(f(x)\).
Solution
Since \(f\) is a linear function, \(f(x) = mx + c\). Also, since \((1, 1) \in f\) and \((0, -1) \in f\),
\(f(1) = m + c = 1\) and \(f(0) = c = -1\).
This gives \(m = 2\) and \(c = -1\).
Therefore \(f(x) = 2x - 1\).
\(f(1) = m + c = 1\) and \(f(0) = c = -1\).
This gives \(m = 2\) and \(c = -1\).
Therefore \(f(x) = 2x - 1\).
Example 21
Find the domain of the function \(f(x) = \dfrac{x^2 + 3x + 5}{x^2 - 5x + 4}\).
Solution
Since \(x^2 - 5x + 4 = (x - 4)(x - 1)\), the function \(f\) is defined for all real numbers except at \(x = 4\) and \(x = 1\). Hence the domain of \(f\) is \(\mathbf{R} - \{1, 4\}\).
Example 22
The function \(f\) is defined by:
\[f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}\]Draw the graph of \(f(x)\).
Solution
Here, for \(x < 0\): \(f(-4) = 1 - (-4) = 5\), \(f(-3) = 4\), \(f(-2) = 3\), \(f(-1) = 2\); etc.
At \(x = 0\): \(f(0) = 1\).
For \(x > 0\): \(f(1) = 2\), \(f(2) = 3\), \(f(3) = 4\), \(f(4) = 5\); and so on.
Thus the graph of \(f\) is shown in Fig 2.17.
At \(x = 0\): \(f(0) = 1\).
For \(x > 0\): \(f(1) = 2\), \(f(2) = 3\), \(f(3) = 4\), \(f(4) = 5\); and so on.
Thus the graph of \(f\) is shown in Fig 2.17.
Fig 2.17 -- Graph of the piecewise function
Exercise 2.3
Q1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) \(\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}\)
(ii) \(\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}\)
(iii) \(\{(1,3),(1,5),(2,5)\}\)
(i) \(\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}\)
(ii) \(\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}\)
(iii) \(\{(1,3),(1,5),(2,5)\}\)
(i) It is a function because every first element is distinct. Domain = \(\{2, 5, 8, 11, 14, 17\}\), Range = \(\{1\}\).
(ii) It is a function because every first element is distinct. Domain = \(\{2, 4, 6, 8, 10, 12, 14\}\), Range = \(\{1, 2, 3, 4, 5, 6, 7\}\).
(iii) It is not a function because element 1 maps to both 3 and 5.
(ii) It is a function because every first element is distinct. Domain = \(\{2, 4, 6, 8, 10, 12, 14\}\), Range = \(\{1, 2, 3, 4, 5, 6, 7\}\).
(iii) It is not a function because element 1 maps to both 3 and 5.
Q2. Find the domain and range of the following real functions:
(i) \(f(x) = -|x|\) (ii) \(f(x) = \sqrt{9 - x^2}\)
(i) \(f(x) = -|x|\) (ii) \(f(x) = \sqrt{9 - x^2}\)
(i) \(f(x) = -|x|\) is defined for all \(x \in \mathbf{R}\).
Domain = \(\mathbf{R}\). Since \(|x| \geq 0\), we have \(-|x| \leq 0\). Range = \((-\infty, 0]\).
(ii) \(f(x) = \sqrt{9 - x^2}\). For \(f\) to be defined, \(9 - x^2 \geq 0\), i.e., \(x^2 \leq 9\), i.e., \(-3 \leq x \leq 3\).
Domain = \([-3, 3]\). The maximum value of \(\sqrt{9 - x^2}\) is \(\sqrt{9} = 3\) (at \(x = 0\)) and the minimum is 0 (at \(x = \pm 3\)).
Range = \([0, 3]\).
Domain = \(\mathbf{R}\). Since \(|x| \geq 0\), we have \(-|x| \leq 0\). Range = \((-\infty, 0]\).
(ii) \(f(x) = \sqrt{9 - x^2}\). For \(f\) to be defined, \(9 - x^2 \geq 0\), i.e., \(x^2 \leq 9\), i.e., \(-3 \leq x \leq 3\).
Domain = \([-3, 3]\). The maximum value of \(\sqrt{9 - x^2}\) is \(\sqrt{9} = 3\) (at \(x = 0\)) and the minimum is 0 (at \(x = \pm 3\)).
Range = \([0, 3]\).
Q3. A function \(f\) is defined by \(f(x) = 2x - 5\). Write down the values of:
(i) \(f(0)\) (ii) \(f(7)\) (iii) \(f(-3)\)
(i) \(f(0)\) (ii) \(f(7)\) (iii) \(f(-3)\)
(i) \(f(0) = 2(0) - 5 = -5\)
(ii) \(f(7) = 2(7) - 5 = 14 - 5 = 9\)
(iii) \(f(-3) = 2(-3) - 5 = -6 - 5 = -11\)
(ii) \(f(7) = 2(7) - 5 = 14 - 5 = 9\)
(iii) \(f(-3) = 2(-3) - 5 = -6 - 5 = -11\)
Q4. The function \(t\) which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by \(t(C) = \frac{9C}{5} + 32\).
Find (i) \(t(0)\) (ii) \(t(28)\) (iii) \(t(-10)\) (iv) The value of C, when \(t(C) = 212\).
Find (i) \(t(0)\) (ii) \(t(28)\) (iii) \(t(-10)\) (iv) The value of C, when \(t(C) = 212\).
(i) \(t(0) = \frac{9(0)}{5} + 32 = 32\)
(ii) \(t(28) = \frac{9(28)}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4\)
(iii) \(t(-10) = \frac{9(-10)}{5} + 32 = -18 + 32 = 14\)
(iv) \(212 = \frac{9C}{5} + 32\) gives \(\frac{9C}{5} = 180\), so \(C = \frac{180 \times 5}{9} = 100\).
(ii) \(t(28) = \frac{9(28)}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4\)
(iii) \(t(-10) = \frac{9(-10)}{5} + 32 = -18 + 32 = 14\)
(iv) \(212 = \frac{9C}{5} + 32\) gives \(\frac{9C}{5} = 180\), so \(C = \frac{180 \times 5}{9} = 100\).
Q5. Find the range of each of the following functions:
(i) \(f(x) = 2 - 3x,\; x \in \mathbf{R},\; x > 0\)
(ii) \(f(x) = x^2 + 2,\; x\) is a real number
(iii) \(f(x) = x,\; x\) is a real number
(i) \(f(x) = 2 - 3x,\; x \in \mathbf{R},\; x > 0\)
(ii) \(f(x) = x^2 + 2,\; x\) is a real number
(iii) \(f(x) = x,\; x\) is a real number
(i) Since \(x > 0\), we have \(3x > 0\), so \(-3x < 0\), thus \(2 - 3x < 2\). Range = \((-\infty, 2)\).
(ii) Since \(x^2 \geq 0\) for all real \(x\), we have \(x^2 + 2 \geq 2\). Range = \([2, \infty)\).
(iii) \(f(x) = x\) is the identity function. Range = \(\mathbf{R}\).
(ii) Since \(x^2 \geq 0\) for all real \(x\), we have \(x^2 + 2 \geq 2\). Range = \([2, \infty)\).
(iii) \(f(x) = x\) is the identity function. Range = \(\mathbf{R}\).
Miscellaneous Exercise on Chapter 2
Q1. The relation \(f\) is defined by \(f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases}\).
The relation \(g\) is defined by \(g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases}\).
Show that \(f\) is a function and \(g\) is not a function.
The relation \(g\) is defined by \(g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases}\).
Show that \(f\) is a function and \(g\) is not a function.
For \(f\): At \(x = 3\), \(f(3) = 3^2 = 9\) and also \(f(3) = 3 \times 3 = 9\). Both rules give the same value, so \(f\) is well-defined. Hence \(f\) is a function.
For \(g\): At \(x = 2\), \(g(2) = 2^2 = 4\) and also \(g(2) = 3 \times 2 = 6\). Both rules give different values (4 and 6). Since one element maps to two different values, \(g\) is not a function.
For \(g\): At \(x = 2\), \(g(2) = 2^2 = 4\) and also \(g(2) = 3 \times 2 = 6\). Both rules give different values (4 and 6). Since one element maps to two different values, \(g\) is not a function.
Q2. If \(f(x) = x^2\), find \(\frac{f(1.1) - f(1)}{(1.1 - 1)}\).
\(\frac{f(1.1) - f(1)}{1.1 - 1} = \frac{(1.1)^2 - 1^2}{0.1} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1\).
Q3. Find the domain of the function \(f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}\).
\(x^2 - 8x + 12 = (x - 2)(x - 6)\). The function is undefined when denominator = 0, i.e., \(x = 2\) or \(x = 6\).
Domain = \(\mathbf{R} - \{2, 6\}\).
Domain = \(\mathbf{R} - \{2, 6\}\).
Q4. Find the domain and the range of the real function \(f\) defined by \(f(x) = \sqrt{(x - 1)}\).
For \(f\) to be defined, \(x - 1 \geq 0\), i.e., \(x \geq 1\). Domain = \([1, \infty)\).
Since \(\sqrt{x - 1} \geq 0\) for all \(x\) in the domain, Range = \([0, \infty)\).
Since \(\sqrt{x - 1} \geq 0\) for all \(x\) in the domain, Range = \([0, \infty)\).
Q5. Find the domain and the range of the real function \(f\) defined by \(f(x) = |x - 1|\).
\(f(x) = |x - 1|\) is defined for all \(x \in \mathbf{R}\). Domain = \(\mathbf{R}\).
Since \(|x - 1| \geq 0\) for all real \(x\), Range = \([0, \infty)\).
Since \(|x - 1| \geq 0\) for all real \(x\), Range = \([0, \infty)\).
Q6. Let \(f = \{(x, \frac{x^2}{1 + x^2}) : x \in \mathbf{R}\}\) be a function from \(\mathbf{R}\) into \(\mathbf{R}\). Determine the range of \(f\).
Let \(y = \frac{x^2}{1 + x^2}\). We note that:
- When \(x = 0\), \(y = 0\).
- For all \(x \neq 0\), \(x^2 > 0\) and \(1 + x^2 > x^2\), so \(0 < \frac{x^2}{1 + x^2} < 1\).
As \(x \to \infty\), \(y \to 1\) but never reaches 1.
Range = \([0, 1)\).
- When \(x = 0\), \(y = 0\).
- For all \(x \neq 0\), \(x^2 > 0\) and \(1 + x^2 > x^2\), so \(0 < \frac{x^2}{1 + x^2} < 1\).
As \(x \to \infty\), \(y \to 1\) but never reaches 1.
Range = \([0, 1)\).
Q7. Let \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) be defined, respectively by \(f(x) = x + 1\), \(g(x) = 2x - 3\). Find \(f + g\), \(f - g\) and \(\frac{f}{g}\).
\((f + g)(x) = (x + 1) + (2x - 3) = 3x - 2\)
\((f - g)(x) = (x + 1) - (2x - 3) = -x + 4\)
\(\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3},\quad x \neq \frac{3}{2}\)
\((f - g)(x) = (x + 1) - (2x - 3) = -x + 4\)
\(\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3},\quad x \neq \frac{3}{2}\)
Q8. Let \(f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}\) be a function from \(\mathbf{Z}\) to \(\mathbf{Z}\) defined by \(f(x) = ax + b\), for some integers \(a, b\). Determine \(a, b\).
From \(f(0) = -1\): \(a(0) + b = -1\), so \(b = -1\).
From \(f(1) = 1\): \(a(1) + (-1) = 1\), so \(a = 2\).
Verification: \(f(2) = 2(2) - 1 = 3\) and \(f(-1) = 2(-1) - 1 = -3\). Both check out.
Therefore \(a = 2\) and \(b = -1\).
From \(f(1) = 1\): \(a(1) + (-1) = 1\), so \(a = 2\).
Verification: \(f(2) = 2(2) - 1 = 3\) and \(f(-1) = 2(-1) - 1 = -3\). Both check out.
Therefore \(a = 2\) and \(b = -1\).
Q9. Let R be a relation from \(\mathbf{N}\) to \(\mathbf{N}\) defined by \(R = \{(a, b) : a, b \in \mathbf{N}\) and \(a = b^2\}\). Are the following true?
(i) \((a, a) \in R\), for all \(a \in \mathbf{N}\) (ii) \((a, b) \in R\), implies \((b, a) \in R\) (iii) \((a, b) \in R\), \((b, c) \in R\) implies \((a, c) \in R\).
(i) \((a, a) \in R\), for all \(a \in \mathbf{N}\) (ii) \((a, b) \in R\), implies \((b, a) \in R\) (iii) \((a, b) \in R\), \((b, c) \in R\) implies \((a, c) \in R\).
(i) False. For \((a, a) \in R\), we need \(a = a^2\), which is true only for \(a = 1\). For example, \((2, 2) \notin R\) since \(2 \neq 2^2 = 4\).
(ii) False. \((4, 2) \in R\) since \(4 = 2^2\), but \((2, 4) \notin R\) since \(2 \neq 4^2 = 16\).
(iii) False. \((16, 4) \in R\) and \((4, 2) \in R\), but \((16, 2) \notin R\) since \(16 \neq 2^2 = 4\).
(ii) False. \((4, 2) \in R\) since \(4 = 2^2\), but \((2, 4) \notin R\) since \(2 \neq 4^2 = 16\).
(iii) False. \((16, 4) \in R\) and \((4, 2) \in R\), but \((16, 2) \notin R\) since \(16 \neq 2^2 = 4\).
Q10. Let \(A = \{1, 2, 3, 4\}\), \(B = \{1, 5, 9, 11, 15, 16\}\) and \(f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}\). Are the following true?
(i) \(f\) is a relation from A to B (ii) \(f\) is a function from A to B
Justify your answer in each case.
(i) \(f\) is a relation from A to B (ii) \(f\) is a function from A to B
Justify your answer in each case.
(i) True. \(f\) is a subset of \(A \times B\), so it is a relation from A to B.
(ii) False. Element 2 has two images (9 and 11), i.e., \((2, 9) \in f\) and \((2, 11) \in f\). Since 2 maps to two different elements, \(f\) is not a function.
(ii) False. Element 2 has two images (9 and 11), i.e., \((2, 9) \in f\) and \((2, 11) \in f\). Since 2 maps to two different elements, \(f\) is not a function.
Q11. Let \(f\) be the subset of \(\mathbf{Z} \times \mathbf{Z}\) defined by \(f = \{(ab, a + b) : a, b \in \mathbf{Z}\}\). Is \(f\) a function from \(\mathbf{Z}\) to \(\mathbf{Z}\)? Justify your answer.
No, \(f\) is not a function. Consider the element \(ab = 2\):
When \(a = 1, b = 2\): \(ab = 2\) and \(a + b = 3\), giving \((2, 3)\).
When \(a = -1, b = -2\): \(ab = 2\) and \(a + b = -3\), giving \((2, -3)\).
Since the element 2 maps to both 3 and \(-3\), \(f\) is not a function.
When \(a = 1, b = 2\): \(ab = 2\) and \(a + b = 3\), giving \((2, 3)\).
When \(a = -1, b = -2\): \(ab = 2\) and \(a + b = -3\), giving \((2, -3)\).
Since the element 2 maps to both 3 and \(-3\), \(f\) is not a function.
Q12. Let \(A = \{9, 10, 11, 12, 13\}\) and let \(f: A \rightarrow \mathbf{N}\) be defined by \(f(n) =\) the highest prime factor of \(n\). Find the range of \(f\).
\(f(9) = 3\) (since \(9 = 3^2\)), \(f(10) = 5\) (since \(10 = 2 \times 5\)), \(f(11) = 11\) (prime), \(f(12) = 3\) (since \(12 = 2^2 \times 3\)), \(f(13) = 13\) (prime).
Range of \(f = \{3, 5, 11, 13\}\).
Range of \(f = \{3, 5, 11, 13\}\).
Interactive: Algebra of Functions
Enter two function rules and a value of x to compute f+g, f-g, fg, and f/g
Click "Compute" to see results.
Activity: Function or Not?
L5 Evaluate
Materials: Paper, pencil
Challenge: For each scenario below, decide whether the described mapping is a function. Justify your reasoning using the definition.
- Mapping each student in your class to their roll number. (One student, one roll number.)
- Mapping each student to the sports they play. (A student may play multiple sports.)
- Mapping each positive integer to its factors. (For example, 6 maps to 1, 2, 3, 6.)
- Mapping each country to its capital city. (Each country has exactly one capital.)
Answers:
- (1) Function -- each student has exactly one roll number.
- (2) Not a function -- a student may play multiple sports (multiple images).
- (3) Not a function -- a number has multiple factors (multiple images).
- (4) Function -- each country has exactly one capital city.
Summary
- Ordered pair: A pair of elements grouped together in a particular order, written as \((a, b)\). Two ordered pairs are equal if and only if corresponding elements are equal.
- Cartesian product: \(A \times B = \{(a, b) : a \in A,\; b \in B\}\). In particular, \(\mathbf{R} \times \mathbf{R} = \{(x, y) : x, y \in \mathbf{R}\}\) and \(\mathbf{R} \times \mathbf{R} \times \mathbf{R} = \{(x, y, z) : x, y, z \in \mathbf{R}\}\).
- If \((a, b) = (x, y)\), then \(a = x\) and \(b = y\).
- If \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = pq\).
- In general, \(A \times B \neq B \times A\).
- Relation: A relation R from a set A to a set B is a subset of the Cartesian product \(A \times B\) obtained by describing a relationship between the first element \(x\) and the second element \(y\) of the ordered pairs, where \((x, y) \in R\).
- The image of an element \(x\) under a relation R is given by \(y\), where \((x, y) \in R\).
- The domain of R is the set of all first elements of the ordered pairs in a relation R.
- The range of the relation R is the set of all second elements of the ordered pairs in a relation R.
- Function: A function \(f\) from a set A to a set B is a specific type of relation for which every element \(x\) of set A has one and only one image \(y\) in set B. We write \(f: A \rightarrow B\), where \(f(x) = y\).
- A is the domain and B is the codomain of \(f\).
- The range of the function is the set of images.
- A real function has the set of real numbers or one of its subsets both as its domain and as its range.
- Algebra of functions: For functions \(f: X \rightarrow \mathbf{R}\) and \(g: X \rightarrow \mathbf{R}\), we have:
\((f + g)(x) = f(x) + g(x)\), \((f - g)(x) = f(x) - g(x)\)
\((fg)(x) = f(x) \cdot g(x)\), \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\) where \(g(x) \neq 0\)
\((\alpha f)(x) = \alpha \cdot f(x)\), where \(\alpha\) is a real number.
Historical Note
The word FUNCTION first appeared in a Latin manuscript by Gottfried Wilhelm Leibniz in 1673, where he used it to describe quantities related to a curve such as coordinates and slope. On July 5, 1698, Johan Bernoulli, in a letter to Leibniz, for the first time deliberately assigned a specialised use of the term "function" in the analytical sense. Leibniz implied his approval by the end of that month. The notion has evolved significantly since then, becoming one of the central concepts in all of modern mathematics.
Competency-Based Questions
Scenario: A mobile phone company charges customers based on their data usage. The monthly bill function is defined piecewise: \(B(x) = \begin{cases} 199, & 0 \leq x \leq 2 \\ 199 + 50(x - 2), & x > 2 \end{cases}\) where \(x\) is data usage in GB and \(B(x)\) is the bill in rupees.
Q1. What is the monthly bill for a customer who uses 1.5 GB of data?
L3 ApplyAnswer: (b) Rs. 199. Since \(0 \leq 1.5 \leq 2\), \(B(1.5) = 199\).
Q2. Is B(x) a function? Verify by checking whether the piecewise definition gives a unique output at the boundary point x = 2.
L4 AnalyseAnswer: At \(x = 2\): From the first piece, \(B(2) = 199\). From the second piece, \(B(2) = 199 + 50(2 - 2) = 199\). Both give the same value, so \(B\) is well-defined at the boundary. Since every input \(x \geq 0\) gives exactly one output, \(B(x)\) is a function.
Q3. A customer's bill is Rs. 449. How much data did they use? Show your working.
L3 ApplyAnswer: Since 449 > 199, the customer used more than 2 GB.
\(449 = 199 + 50(x - 2)\)
\(250 = 50(x - 2)\)
\(x - 2 = 5\)
\(x = 7\) GB.
\(449 = 199 + 50(x - 2)\)
\(250 = 50(x - 2)\)
\(x - 2 = 5\)
\(x = 7\) GB.
Q4. The company introduces a new plan where the first 3 GB are free and additional data costs Rs. 40 per GB. Write this as a function and determine its domain and range.
L6 CreateAnswer:
\(C(x) = \begin{cases} 0, & 0 \leq x \leq 3 \\ 40(x - 3), & x > 3 \end{cases}\)
Domain = \([0, \infty)\) (non-negative data usage).
Range = \([0, \infty)\) (bill can be 0 or any positive value).
Check at boundary: \(C(3) = 0\) from piece 1 and \(C(3) = 40(0) = 0\) from piece 2. Well-defined, so \(C\) is a valid function.
\(C(x) = \begin{cases} 0, & 0 \leq x \leq 3 \\ 40(x - 3), & x > 3 \end{cases}\)
Domain = \([0, \infty)\) (non-negative data usage).
Range = \([0, \infty)\) (bill can be 0 or any positive value).
Check at boundary: \(C(3) = 0\) from piece 1 and \(C(3) = 40(0) = 0\) from piece 2. Well-defined, so \(C\) is a valid function.
Assertion--Reason Questions
Assertion (A): \((f + g)(x) = (g + f)(x)\) for all real functions \(f\) and \(g\).
Reason (R): Addition of real numbers is commutative.
Reason (R): Addition of real numbers is commutative.
Answer: (a) -- \((f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x)\) by commutativity of real number addition.
Assertion (A): The domain of \(f(x) = \frac{1}{x - 2}\) is \(\mathbf{R} - \{2\}\).
Reason (R): A rational function is undefined where its denominator equals zero.
Reason (R): A rational function is undefined where its denominator equals zero.
Answer: (a) -- The denominator \(x - 2 = 0\) when \(x = 2\). So \(f\) is defined for all reals except 2. Both statements are true and R explains A.
Assertion (A): If \(f(x) = x^2\) and \(g(x) = x\), then \(\left(\frac{f}{g}\right)(x) = x\) for all \(x \in \mathbf{R}\).
Reason (R): \(\frac{f(x)}{g(x)} = \frac{x^2}{x} = x\).
Reason (R): \(\frac{f(x)}{g(x)} = \frac{x^2}{x} = x\).
Answer: (d) -- A is false because \(\frac{f}{g}\) is not defined at \(x = 0\) (since \(g(0) = 0\)). The correct domain is \(\mathbf{R} - \{0\}\). R is technically a true simplification but omits the crucial restriction \(x \neq 0\).
Frequently Asked Questions
How to solve domain and range problems in Class 11?
To find domain: identify all x-values for which f(x) is defined. To find range: express x in terms of y from y = f(x) and find valid y-values, or analyze the graph.
What exercises are in NCERT Chapter 2 Relations and Functions?
Chapter 2 contains Exercise 2.1 (Cartesian products), Exercise 2.2 (Relations), Exercise 2.3 (Functions, domain, range, graphs), and a Miscellaneous Exercise.
How to graph functions for Class 11 exercises?
Plot key points, identify the shape, mark domain and range on axes, and note asymptotes or discontinuities. Practice standard functions: identity, constant, polynomial, modulus, signum.
What are common mistakes in Relations and Functions?
Common errors include confusing codomain with range, not checking all domain elements, errors in Cartesian products, and incorrect piecewise function graphs.
How to verify if a given relation is a function?
Check that every element in the domain maps to exactly one element in the codomain. No first element should repeat with a different second element.
Frequently Asked Questions — Relations and Functions
What is Exercises and Summary - Relations and Functions in NCERT Class 11 Mathematics?
Exercises and Summary - Relations and Functions is a key concept covered in NCERT Class 11 Mathematics, Chapter 2: Relations and Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Exercises and Summary - Relations and Functions step by step?
To solve problems on Exercises and Summary - Relations and Functions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Relations and Functions?
The essential formulas of Chapter 2 (Relations and Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Exercises and Summary - Relations and Functions important for the Class 11 board exam?
Exercises and Summary - Relations and Functions is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Exercises and Summary - Relations and Functions?
Common mistakes in Exercises and Summary - Relations and Functions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Exercises and Summary - Relations and Functions?
End-of-chapter NCERT exercises for Exercises and Summary - Relations and Functions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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