RELATIONS AND FUNCTIONS-Exercise 2.3
Maths - Exercise
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)}.
Solution
Among the given relations, (i) and (ii) are functions, whereas (iii) is not a function. A relation is a function if every element of the domain is associated with exactly one element of the range
For (i), \[ \begin{aligned} R&=\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\} \\\\ \text{Domain}&=\{2,5,8,11,14,17\} \\\\ \text{Range}&=\{1\} \end{aligned} \] Each element of the domain has exactly one image in the range. Hence, relation (i) is a function
For (ii), \[ \begin{aligned} R&=\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\} \\\\ \text{Domain}&=\{2,4,6,8,10,12,14\} \\\\ \text{Range}&=\{1,2,3,4,5,6,7\} \end{aligned} \] Each element of the domain corresponds to exactly one element of the range. Therefore, relation (ii) is also a function
For (iii), the relation \(\{(1,3),(1,5),(2,5)\}\) is not a function because the element \(1\) in the domain is associated with two different elements \(3\) and \(5\) in the range. Hence, relation (iii) fails to satisfy the definition of a function
Q2. Find the domain and range of the following real functions:
(i) \(f(x) = –|x|\)
(ii) \(f(x)=\sqrt{9-x^2}\)
Solution
(i) We are given \[ \begin{aligned} f(x)=-|x| \end{aligned} \] The absolute value function \(|x|\) is defined for all real values of \(x\). Hence, the domain of the given function is the set of all real numbers \[ \begin{aligned} \text{Domain}=\mathbb{R} \end{aligned} \] Since \(|x|\ge 0\) for all real \(x\), we have \(-|x|\le 0\). Therefore, the range of the function is \[ \begin{aligned} \text{Range}=(-\infty,0] \end{aligned} \]
(ii) We are given \[ \begin{aligned} f(x)=\sqrt{9-x^{2}} \end{aligned} \] For the square root to be defined, the expression under the radical must be non-negative \[ \begin{aligned} 9-x^{2}\ge 0 \\ x^{2}\le 9 \\ -3\le x\le 3 \end{aligned} \] Hence, the domain of the function is \[ \begin{aligned} \text{Domain}=[-3,3] \end{aligned} \]
Since the square root function always gives non-negative values and the maximum value of \(\sqrt{9-x^{2}}\) occurs at \(x=0\), we obtain \[ \begin{aligned} 0\le f(x)\le 3 \end{aligned} \] Therefore, the range of the function is \[ \begin{aligned} \text{Range}=[0,3] \end{aligned} \]
Q3. A function \(f\) is defined by \(f(x) = 2x –5\). Write down the values of
(i) \(f(0)\)
(ii) \(f (7)\),
(iii) \(f(–3)\).
Solution
The given function is \[ \begin{aligned} f(x)=2x-5 \end{aligned} \]
(i) Substituting \(x=0\), we get \[ \begin{aligned} f(0)&=2\times 0-5 \\ &=-5 \end{aligned} \]
(ii) Substituting \(x=7\), we obtain \[ \begin{aligned} f(7)&=2\times 7-5 \\ &=14-5 \\ &=9 \end{aligned} \]
(iii) Substituting \(x=-3\), we have \[ \begin{aligned} f(-3)&=2\times(-3)-5 \\ &=-6-5 \\ &=-11 \end{aligned} \]
Q4. The function \(‘t’\) which maps temperature in degree Celsius into temperature in
degree Fahrenheit is defined by \(t(C) = \dfrac{9C}{5} + 32\).
Find (i) \(t(0)\)
(ii) \(t(28)\)
(iii) \(t(–10)\)
(iv) The value of \(C\), when \(t(C) =
212\).
Solution
The given function which converts temperature from degree Celsius to degree Fahrenheit is \[ \begin{aligned} t(C)=\dfrac{9C}{5}+32 \end{aligned} \]
(i) Substituting \(C=0\), we get \[ \begin{aligned} t(0)&=\dfrac{9\times 0}{5}+32 \\ &=32 \end{aligned} \]
(ii) Substituting \(C=28\), we obtain \[ \begin{aligned} t(28)&=\dfrac{9\times 28}{5}+32 \\ &=\dfrac{252}{5}+32 \\ &=50.4+32 \\ &=82.4 \end{aligned} \]
(iii) Substituting \(C=-10\), we have \[ \begin{aligned} t(-10)&=\dfrac{9\times (-10)}{5}+32 \\ &=-18+32 \\ &=14 \end{aligned} \]
(iv) When \(t(C)=212\), \[ \begin{aligned} 212&=\dfrac{9C}{5}+32 \\ 212-32&=\dfrac{9C}{5} \\ 180&=\dfrac{9C}{5} \\ 9C&=900 \\ C&=100 \end{aligned} \]
Q5. Find the range of each of the following functions.
(i) \(f (x) = 2 – 3x, x ∈ R, x > 0\).
(ii) \(f (x) = x2 + 2, x\) is a real number.
(iii) \(f (x) = x, x\) is a real number.
Solution
(i) We are given \[ \begin{aligned} f(x)=2-3x,\ x\in \mathbb{R},\ x>0 \end{aligned} \] Since \(x>0\), the term \(3x\) is positive and increases as \(x\) increases. Hence, \(2-3x\) decreases continuously. As \(x\to 0^{+}\), \(f(x)\to 2\), but the value \(2\) is not attained. As \(x\to \infty\), \(f(x)\to -\infty\). Therefore, \[ \begin{aligned} \text{Range}=(-\infty,2) \end{aligned} \]
(ii) We are given \[ \begin{aligned} f(x)=x^{2}+2,\ x\in \mathbb{R} \end{aligned} \] Since \(x^{2}\ge 0\) for all real \(x\), the minimum value of \(x^{2}+2\) is obtained at \(x=0\) and equals \(2\). Hence, \[ \begin{aligned} \text{Range}=[2,\infty) \end{aligned} \]
(iii) We are given \[ \begin{aligned} f(x)=x,\ x\in \mathbb{R} \end{aligned} \] As \(x\) takes all real values, the function also takes all real values. Therefore, \[ \begin{aligned} \text{Range}=\mathbb{R} \end{aligned} \]
