Miscellaneous Exercise on Chapter 2
Maths - Exercise
Q1. The relation \(f\) is defined by $$\begin{aligned}f_{x}=\begin{cases}x^{2},0\leq x\leq 3\\ 3x,3\leq x\leq 10\end{cases}\end{aligned}$$ The relation \(g\) is defined by $$\begin{aligned}g\left( x\right) =\begin{cases}x^{2},0\leq x\leq 2\\ 3x,2\leq x\leq 10\end{cases} \end{aligned}$$ Show that \(f\) is a function and \(g\) is not a function.
Solution
The relations are defined as \[ \begin{aligned} f(x)= \begin{cases} x^{2}, & 0\le x\le 3 \\ 3x, & 3\le x\le 10 \end{cases} \\\\\ g(x)= \begin{cases} x^{2}, & 0\le x\le 2 \\ 3x, & 2\le x\le 10 \end{cases} \end{aligned} \]
To check whether \(f\) is a function, we examine the common point \(x=3\), which belongs to both intervals. Using the first rule, \[ \begin{aligned} f(3)=3^{2}=9 \end{aligned} \] and using the second rule, \[ \begin{aligned} f(3)=3\times 3=9 \end{aligned} \] Since both definitions give the same value at \(x=3\), each input has a unique output. Hence, \(f\) is a function
Now, consider the relation \(g\). The point \(x=2\) lies in both intervals. Using the first rule, \[ \begin{aligned} g(2)=2^{2}=4 \end{aligned} \] and using the second rule, \[ \begin{aligned} g(2)=3\times 2=6 \end{aligned} \]
Thus, for the same input \(x=2\), the relation \(g\) gives two different outputs. Therefore, \(g\) is not a function
Q2. If \(f (x) = x^2\) , find \(\dfrac{f(1.1)-f(1)}{(1.1-1)}\)
Solution
We are given \[ \begin{aligned} f(x)=x^{2} \end{aligned} \] and we are required to find \[ \begin{aligned} \dfrac{f(1.1)-f(1)}{1.1-1} \end{aligned} \]
Substituting the given values in the function, we get \[ \begin{aligned} \dfrac{f(1.1)-f(1)}{1.1-1} &=\dfrac{(1.1)^{2}-(1)^{2}}{1.1-1} \\ &=\dfrac{1.21-1}{0.1} \\ &=\dfrac{0.21}{0.1} \\ &=2.1 \end{aligned} \]
Q3. Find the domain of the function \(f (x)=\dfrac{x^2+2x+1}{x^2-8x+12}\)
Solution
The given function is \[ \begin{aligned} f(x)=\dfrac{x^{2}+2x+1}{x^{2}-8x+12} \end{aligned} \]
To find the domain, we first determine the values of \(x\) for which the denominator becomes zero \[ \begin{aligned} x^{2}-8x+12&=0 \\ x^{2}-6x-2x+12&=0 \\ x(x-6)-2(x-6)&=0 \\ (x-6)(x-2)&=0 \\ x&=2 \\ x&=6 \end{aligned} \]
At \(x=2\) and \(x=6\), the denominator is zero and the function is not defined. Hence, these values are excluded from the domain
Therefore, the domain of the function is \[ \begin{aligned} D=(-\infty,2)\cup(2,6)\cup(6,\infty) \end{aligned} \]
Q4. Find the domain and the range of the real function \(f\) defined by \(f (x)=\sqrt{(x-1)} \)
Solution
The given real function is \[ \begin{aligned} f(x)=\sqrt{x-1} \end{aligned} \]
For the square root to be defined, the expression inside the radical must be non-negative \[ \begin{aligned} x-1\ge 0 \\ x\ge 1 \end{aligned} \] Hence, the domain of the function is \[ \begin{aligned} D=[1,\infty) \end{aligned} \]
Since the square root of a non-negative number is always non-negative and the minimum value of \(\sqrt{x-1}\) occurs at \(x=1\), we obtain \[ \begin{aligned} R=[0,\infty) \end{aligned} \]
Q5. Find the domain and the range of the real function \(f\) defined by \(f (x) = \mid x–1\mid\)
Solution
The given real function is \[ \begin{aligned} f(x)=|x-1| \end{aligned} \]
The absolute value function is defined for all real values of \(x\). Hence, there is no restriction on \(x\). Therefore, the domain of the function is \[ \begin{aligned} D=\mathbb{R} \end{aligned} \]
Since the absolute value of any real number is always non-negative, we have \[ \begin{aligned} |x-1|\ge 0 \end{aligned} \] The minimum value of the function occurs at \(x=1\), where \(f(1)=0\). Thus, the range of the function is \[ \begin{aligned} R=[0,\infty) \end{aligned} \]
Q6. Let \(f=\left\{ \left( x,\dfrac{x^{2}}{1+x^{2}}\right) :x\in \mathbb{R} \right\}\) be a function from \(\mathbb{R}\) into \(\mathbb{R}\). Determine the range of \(f\).
Solution
The function is given by \[ \begin{aligned} f=\left\{\left(x,\dfrac{x^{2}}{1+x^{2}}\right): x\in \mathbb{R}\right\} \end{aligned} \]
For all real values of \(x\), we have \(1+x^{2}>0\). Hence, the function is well defined for every \(x\in\mathbb{R}\)
Since \(x^{2}\ge 0\) for all real \(x\), it follows that \[ \begin{aligned} \dfrac{x^{2}}{1+x^{2}} \ge 0 \end{aligned} \] Also, because \(x^{2}\lt 1+x^{2}\) for every real \(x\), \[ \begin{aligned} \dfrac{x^{2}}{1+x^{2}} < 1 \end{aligned} \]
Thus, the values of the function lie between \(0\) and \(1\). The value \(0\) is attained at \(x=0\), while the value \(1\) is never attained for any real \(x\). Therefore, the range of the function is \[ \begin{aligned} R=[0,1) \end{aligned} \]
Q7. Let \(f\), \(g\) : \(\mathbb{R}\rightarrow\mathbb{R}\) be defined, respectively by \(f(x) = x + 1, g(x) = 2x – 3\). Find \(f + g,\; f – g\) and \(\dfrac{f}{g}\)
Solution
The given functions are \[ \begin{aligned} f(x)=x+1 \\ g(x)=2x-3 \end{aligned} \]
To find \(f+g\), \[ \begin{aligned} f+g &= f(x)+g(x) \\ &=(x+1)+(2x-3) \\ &=3x-2 \end{aligned} \]
To find \(f-g\), \[ \begin{aligned} f-g &= f(x)-g(x) \\ &=(x+1)-(2x-3) \\ &=x+1-2x+3 \\ &=4-x \end{aligned} \]
To find \(\dfrac{f}{g}\), \[ \begin{aligned} \dfrac{f}{g} &= \dfrac{f(x)}{g(x)} \\ &=\dfrac{x+1}{2x-3},\ x\ne \dfrac{3}{2} \end{aligned} \]
Q8. Let \(f = \{(1,1), (2,3), (0,–1), (–1, –3)\}\) be a function from \(\mathbb{Z}\) to \(\mathbb{Z}\) defined by \(f(x) = ax + b\), for some integers \(a,\ b\). Determine \(a, b\).
Solution
The given function is \[ \begin{aligned} f=\{(1,1),(2,3),(0,-1),(-1,-3)\} \end{aligned} \] and it is defined by \[ \begin{aligned} f(x)=ax+b \end{aligned} \] where \(a\) and \(b\) are integers
Using the ordered pair \((1,1)\), \[ \begin{aligned} a+b&=1 \\ a&=1-b \end{aligned} \]
Now, using the ordered pair \((2,3)\), \[ \begin{aligned} 2a+b&=3 \\ 2(1-b)+b&=3 \\ 2-2b+b&=3 \\ -b&=1 \\ b&=-1 \end{aligned} \]
Substituting \(b=-1\) in \(a+b=1\), \[ \begin{aligned} a-1&=1 \\ a&=2 \end{aligned} \]
Hence, the required values are \[ \begin{aligned} (a,b)=(2,-1) \end{aligned} \]
Q9. Let \(R\) be a relation from \(\mathbb{N}\) to \(\mathbb{N}\) defined by \(R = {(a, b) : a, b \in
\mathbb{N}\text{ and } a = b^2\}\). Are
the following true?
(i) \((a,a) \in \mathbb{R}, \text{ for all }a \in \mathbb{N}\)
(ii) \((a,b) \in \mathbb{R}, \text{ implies } (b,a) \in \mathbb{R}\)
(iii) \((a,b) \in \mathbb{R}, (b,c) \in \mathbb{R} implies (a,c) \in \mathbb{R}\).
Justify your answer in
each case.
Solution
The given relation is \[ \begin{aligned} R=\{(a,b): a,b\in \mathbb{N}\text{ and } a=b^{2}\} \end{aligned} \]
(i) To check whether \((a,a)\in R\) for all \(a\in\mathbb{N}\), we substitute \(b=a\) in the defining condition \[ \begin{aligned} a=b^{2} \\ a=a^{2} \end{aligned} \] This equality holds only for \(a=0\) or \(a=1\). Since it is not true for all natural numbers, the statement is false
(ii) Suppose \((a,b)\in R\). Then, by definition, \[ \begin{aligned} a=b^{2} \end{aligned} \] For \((b,a)\) to belong to \(R\), we must have \[ \begin{aligned} b=a^{2} \end{aligned} \] In general, \(b=a^{2}\) does not follow from \(a=b^{2}\). Hence, the statement is false
(iii) Suppose \((a,b)\in R\) and \((b,c)\in R\). Then, \[ \begin{aligned} a=b^{2} \\ b=c^{2} \end{aligned} \] Substituting the value of \(b\), we get \[ \begin{aligned} a=(c^{2})^{2}=c^{4} \end{aligned} \] For \((a,c)\) to be in \(R\), we must have \(a=c^{2}\), which is not true in general. Hence, this statement is also false
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.
Solution
We are given the sets \[ \begin{aligned} A&=\{1,2,3,4\} \\ B&=\{1,5,9,11,15,16\} \end{aligned} \] and the relation \[ \begin{aligned} f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\} \end{aligned} \]
(i) To check whether \(f\) is a relation from \(A\) to \(B\), we observe that in every ordered pair \((a,b)\in f\), the first element \(a\) belongs to set \(A\) and the second element \(b\) belongs to set \(B\). Hence, \[ \begin{aligned} f \subseteq A\times B \end{aligned} \] Therefore, \(f\) is a relation from \(A\) to \(B\)
(ii) To check whether \(f\) is a function from \(A\) to \(B\), we examine the images of elements of \(A\). The element \(2\in A\) is associated with two different elements \(9\) and \(11\) in \(B\). Thus, \[ \begin{aligned} (2,9)\in f \text{ and } (2,11)\in f \end{aligned} \]
Since one element of the domain has more than one image, the relation \(f\) does not satisfy the definition of a function. Hence, \(f\) is not a function from \(A\) to \(B\)
Q11. Let \(f\) be the subset of \(\mathbb{Z}\) × \(\mathbb{Z}\) defined by \(f = \{(ab, a + b) : a, b \in \mathbb{Z}\}\). Is \(f\) a function from \(\mathbb{Z}\) to \(\mathbb{Z}\)? Justify your answer.
Solution
The given subset is \[ \begin{aligned} f=\{(ab,a+b): a,b\in\mathbb{Z}\} \end{aligned} \] where each ordered pair is formed using integers \(a\) and \(b\)
For \(f\) to be a function from \(\mathbb{Z}\) to \(\mathbb{Z}\), every first element in the ordered pairs must be associated with exactly one second element. Consider the first component \(ab=2\). This can be obtained in different ways, such as \[ \begin{aligned} 2&=1\times 2 \\ 2&=(-1)\times(-2) \end{aligned} \]
For \(a=1\) and \(b=2\), the corresponding ordered pair is \[ \begin{aligned} (2,1+2)=(2,3) \end{aligned} \] For \(a=-1\) and \(b=-2\), the corresponding ordered pair is \[ \begin{aligned} (2,-1+(-2))=(2,-3) \end{aligned} \]
Thus, the same first element \(2\) is associated with two different second elements \(3\) and \(-3\). Hence, a unique image does not exist for every element of the domain
Therefore, the given relation \(f\) does not satisfy the definition of a function. Hence, \(f\) is not a function from \(\mathbb{Z}\) to \(\mathbb{Z}\)
Q12. Let \(A = \{9,10,11,12,13\}\) and let \(f\) : A→N be defined by \(f (n) =\) the highest prime factor of \(n\). Find the range of \(f\).
Solution
We are given the set \[ \begin{aligned} A=\{9,10,11,12,13\} \end{aligned} \] and the function \(f:A\rightarrow\mathbb{N}\) defined by \(f(n)\) as the highest prime factor of \(n\)
Evaluating the function for each element of \(A\), \[ \begin{aligned} 9&=3\times 3 \\ 10&=2\times 5 \\ 11&=11 \\ 12&=2\times 2\times 3 \\ 13&=13 \end{aligned} \]
The highest prime factors corresponding to the elements of set \(A\) are \(3,5,11\) and \(13\). Hence, the range of the function is \[ \begin{aligned} R=\{3,5,11,13\} \end{aligned} \]
