# Miscellaneous Exercise Relations and Function Solution - NCERT Class 11

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## Chapter 2 Ex.2.ME Question 1

The relation $$f$$ is defined by f\left( x \right) = \left\{ \begin{align}&{x^2},0 \le x \le 3\\&3x,3 \le x \le 10\end{align} \right.

The relation $$g$$ is defined by g\left( x \right) = \left\{ \begin{align}&{x^2},0 \le x \le 2\\&3x,2 \le x \le 10\end{align} \right.

Show that $$f$$ is a function and $$g$$ is not a function.

### Solution

The relation $$f$$ is defined as f\left( x \right) = \left\{ \begin{align}&{x^2},0 \le x \le 3\\&3x,3 \le x \le 10\end{align} \right.

It can be observed that for

$$0 \le x < {\rm{3}},{\rm{ }}f\left( x \right) = {x^{\rm{2}}}$$ and

$$3 < {\rm{ }}x \le 10,{\rm{ }}f\left( x \right) = 3x$$

Also, at $$x = {\rm{3}}$$

\begin{align}f\left( x \right) &= {{\rm{3}}^{\rm{2}}}\\&= {\rm{9}}\end{align}

and

\begin{align}f\left( x \right) &= {\rm{3}} \times {\rm{3}}\\& = {\rm{9}}\end{align}

i.e., $${\text{at }}x = {\rm{3}},\,f\left( x \right) = {\rm{9}}$$

Therefore, for $$0 \le x \le {\rm{1}}0$$, the images of $$f\left( x \right)$$ are unique.

Thus, the given relation is a function.

The relation $$g$$ is defined as g\left( x \right) = \left\{ \begin{align}{x^2},0 \le x \le 2\\3x,2 \le x \le 10\end{align} \right.

It can be observed that for

$$0 \le x \le 2,g\left( x \right) = {x^2}$$ and

$$2 \le x \le 10,g\left( x \right) = 3x$$

Also, at $$x = 2$$

\begin{align}g\left( x \right) &= {2^2}\\&= 4\end{align}

and

\begin{align}g\left( x \right) &= 3 \times 2\\&= 6\end{align}

Hence, element $$2$$ of the domain of the relation $$g$$ corresponds to two different images i.e., $$4$$ and $$6.$$

Hence, this relation is not a function.

Thus, $$f$$ is a function and $$g$$ is not a function.

## Chapter 2 Ex.2.ME Question 2

If $$f\left( x \right) = {x^2}$$, find $$\frac{{f\left( {1.1} \right) - f\left( 1 \right)}}{{\left( {1.1 - 1} \right)}}$$.

### Solution

It is given that $$f\left( x \right) = {x^2}$$

Therefore,

\begin{align}\frac{{f\left( {1.1} \right) - f\left( 1 \right)}}{{\left( {1.1 - 1} \right)}} &= \frac{{{{\left( {1.1} \right)}^2} - {{\left( 1 \right)}^2}}}{{\left( {1.1 - 1} \right)}}\\&= \frac{{1.21 - 1}}{{0.1}}\\&= \frac{{0.21}}{{0.1}}\\&= 2.1\end{align}

## Chapter 2 Ex.2.ME Question 3

Find the domain of the function $$f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}$$

### Solution

The given function is $$f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}$$.

\begin{align}f\left( x \right) &= \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}\\&= \frac{{{x^2} + 2x + 1}}{{\left( {x - 6} \right)\left( {x - 2} \right)}}\end{align}

It can be seen that, the function $$f$$ is defined for all real numbers except at $$x = {\rm{6}}$$ and $$x = {\rm{2}}$$.

Hence, the domain of the function $$f$$ is set of real numbers except $$6$$ and $$2$$. i.e., $${\bf{R}} - \left\{ {2,6} \right\}$$.

## Chapter 2 Ex.2.ME Question 4

Find the domain and the range of the real function $$f$$ defined by $$f\left( x \right) = \sqrt {\left( {x - 1} \right)}$$.

### Solution

The given real function is$$f\left( x \right) = \sqrt {\left( {x - 1} \right)}$$.

It can be seen that $$\sqrt {\left( {x - 1} \right)}$$ is defined for $$x \ge 1$$.

Therefore, the domain of $$f$$ is the set of all real numbers greater than or equal to $$1$$ i.e., the domain of $$f = \left[ {1,\infty } \right)$$.

As

\begin{align}&\Rightarrow \;x \ge 1\\&\Rightarrow \; \left( {x - 1} \right) \ge 0\\&\Rightarrow \; \sqrt {\left( {x - 1} \right)} \ge 0\end{align}

Therefore, the range of $$f$$ is the set of all real numbers greater than or equal to $$0$$ i.e., the range of $$f = \left[ {0,\infty } \right)$$.

## Chapter 2 Ex.2.ME Question 5

Find the domain and the range of the real function $$f$$ defined by $$f\left( x \right) = \left| {x - 1} \right|.$$

### Solution

The given real function is $$f\left( x \right) = \left| {x - 1} \right|$$

It is clear that $$\left| {x - {\rm{1}}} \right|$$ is defined for all real numbers.

Therefore, Domain of $$f = {\bf{R}}$$

Also, for $$x \in {\bf{R}},\left| {x - 1} \right|{\rm{ }}$$assumes all real numbers.

Hence, the range of $$f$$ is the set of all non-negative real numbers.

## Chapter 2 Ex.2.ME Question 6

Let $$f = \left\{ {\left( {x,\frac{{{x^2}}}{{1 + {x^2}}}} \right):x \in R} \right\}$$ be a function from $$R$$ into $$R.$$ Determine the range of $$f$$.

### Solution

It is given that $$f = \left\{ {\left( {x,\frac{{{x^2}}}{{1 + {x^2}}}} \right):x \in R} \right\}$$

Therefore,$$\left\{ {(0,0),\left( { \pm 0.5,\frac{1}{5}} \right),\left( { \pm 1,\frac{1}{2}} \right),\left( { \pm 1.5,\frac{9}{{13}}} \right),\left( { \pm 2,\frac{4}{5}} \right),\left( { \pm 3,\frac{9}{{10}}} \right),\left( { \pm 4,\frac{{16}}{{17}}} \right), \ldots } \right\}$$

The range of $$f$$ is the set of all second elements. It can be observed that all these elements are greater than or equal to $$0$$ but less than $$1$$ (denominator is greater than numerator).

Thus, range of $$f = \left[ {0,{\rm{ }}1} \right)$$

## Chapter 2 Ex.2.ME Question 7

Let $$f,g:{\bf{R}} \to {\bf{R}}$$ be defined, respectively by $$f\left( x \right) = x + 1,\;g\left( x \right) = 2x - 3$$. Find $$f + g,{\rm{ }}f - g$$ and $$\frac{f}{g}$$.

### Solution

Since $$f,g:{\bf{R}} \to {\bf{R}}$$ is defined as $$f\left( x \right) = x + {\rm{1}},\;g\left( x \right) = {\rm{2}}x - {\rm{3}}$$

Hence,

\begin{align}\left( {f + g} \right)\left( x \right) &= f\left( x \right) + g\left( x \right)\\&= \left( {x + {\rm{1}}} \right) + \left( {{\rm{2}}x - {\rm{3}}} \right)\\&= {\rm{3}}x - {\rm{2}}.\end{align}

Therefore, $$\left( {f + g} \right)\left( x \right) = {\rm{3}}x - {\rm{2}}$$

Now,

\begin{align}\left( {f - g} \right)\left( x \right) &= f\left( x \right) - g\left( x \right)\\&= \left( {x + 1} \right) - \left( {2x - 3} \right)\\&= x + 1 - 2x + 3\\&= - x + 4\end{align}

Therefore, $$\left( {f - g} \right)\left( x \right) = - x + {\rm{4}}$$

Now,

\begin{align}\left( {\frac{f}{g}} \right)\left( x \right) &= \frac{{f\left( x \right)}}{{g\left( x \right)}},g\left( x \right) \ne 0,x \in R\\&= \frac{{x + 1}}{{2x - 3}},2x - 3,\;x \ne \frac{3}{2}\end{align}

Therefore,$$\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{x + 1}}{{2x - 3}},\;x \ne \frac{3}{2}$$

## Chapter 2 Ex.2.ME Question 8

Let  $$f = \left\{ {\left( {1,1} \right),\left( {2,3} \right),\left( {0, - 1} \right),\left( { - 1, - 3} \right)} \right\}$$ be a function from $$Z$$ to $$Z$$ defined by $$f\left( x \right) = ax + b,$$ for some integers $$a, \,b.$$ Determine $$a,\, b.$$

### Solution

It is given that $$f = \left\{ {\left( {1,1} \right),\left( {2,3} \right),\left( {0, - 1} \right),\left( { - 1, - 3} \right)} \right\}$$ and $$f\left( x \right) = ax + b$$

\begin{align}&\left( {{\rm{1}},{\rm{ 1}}} \right) \in {\rm{f}} \Rightarrow f\left( {\rm{1}} \right) = {\rm{1}} \Rightarrow {\rm{a}} \times {\rm{1 + b = 1}}\\&\Rightarrow \;{\rm{a + b = 1}}\\&\left( {0, - 1} \right) \in f \Rightarrow f\left( 0 \right) = - 1 \Rightarrow a \times 0 + b = - 1\\&\Rightarrow\; b = - 1\end{align}

On substituting $$b = - {\rm{1}}$$ in $${\rm{ }}a + b = {\rm{1}}$$, we obtain

\begin{align}a + \left( {-{\rm{1}}} \right) &= {\rm{1}}\\a &= {\rm{1}} + {\rm{1}}\\&= {\rm{2}}\end{align}

Thus, the respective values of $$a$$ and $$b$$ are $$2$$ and $$–1.$$

## Chapter 2 Ex.2.ME Question 9

Let $$R$$ be a relation from $$N$$ to $$N$$ defined by $$R = \left\{ {\left( {a,b} \right):a,b \in {\bf{N}}{\text{ and }}a = {b^2}} \right\}$$. Are the following true?

(i) $$\left( {a,a} \right) \in R,$$ for all $$a \in {\bf{N}}$$

(ii) $$\left( {a,{\rm{ }}b} \right) \in R,$$ implies $$\left( {b,a} \right) \in R$$

(iii) $$\left( {a,b} \right) \in R,\;\left( {b,c} \right) \in R$$ implies $$\left( {a,c} \right) \in R$$

### Solution

It is given that $$R = \left\{ {\left( {a,b} \right):a,b \in {\bf{N}}{\text{ and }}a = {b^2}} \right\}$$

(i)  It can be seen that $$2 \in {\bf{N}}$$; however, $${\rm{ 2}} \ne {{\rm{2}}^2} = 4$$

Therefore, the statement “$$\left( {a,a} \right) \in R,$$ for all $$a \in {\bf{N}}$$” is not true.

(ii) It can be seen that $$\left( {9,3} \right) \in {\bf{N}}$$ because $${\rm{9}},{\rm{3}} \in {\bf{N}}$$and $$9 = {3^2}.$$

Now, $${\rm{3}} \ne {{\rm{9}}^2} = {\rm{81}};$$ therefore, $$\left( {3,9} \right) \notin {\bf{N}}$$

Therefore, the statement “$$\left( {a,{\rm{ }}b} \right) \in R,$$ implies $$\left( {b,a} \right) \in R$$” is not true.

(iii) It can be seen that $$(9,3) \in R,\left( {16,{\rm{ }}4} \right) \in R$$ because $$9,3,16,4 \in {\bf{N}}$$ and $$9 = {3^2},\;{\rm{16}} = {{\rm{4}}^2}$$

Now, $$9 \ne {4^2} = 16;$$ therefore, $$\left( {9,4} \right) \notin {\bf{N}}$$.

Therefore, the statement “$$\left( {a,b} \right) \in R,\;\left( {b,c} \right) \in R$$ implies $$\left( {a,c} \right) \in R$$” is not true.

## Chapter 2 Ex.2.ME Question 10

Let $$A = \left\{ {1,2,3,4} \right\},\;B = \left\{ {1,5,9,11,15,16} \right\}$$ and $$f = \left\{ {\left( {1,5} \right),\left( {2,9} \right),\left( {3,1} \right),\left( {4,5} \right),\left( {2,11} \right)} \right\}.$$ Are the following are true?

(i) $$f$$ is a relation from $$A$$ to $$B$$

(ii) $$f$$ is a function from $$A$$ to $$B.$$

### Solution

It is given that $$A = \left\{ {1,2,3,4} \right\},\;B = \left\{ {1,5,9,11,15,16} \right\}$$ and $$f = \left\{ {\left( {1,5} \right),\left( {2,9} \right),\left( {3,1} \right),\left( {4,5} \right),\left( {2,11} \right)} \right\}.$$

Therefore,

A \times B = \left\{ \begin{align}&\left( {1,1} \right),\left( {1,5} \right),\left( {1,9} \right),\left( {1,11} \right),\left( {1,15} \right),\left( {1,16} \right),\left( {2,1} \right),\left( {2,5} \right),\left( {2,9} \right),\\&\left( {2,11} \right),\left( {2,15} \right),\left( {2,16} \right),\left( {3,1} \right),\left( {3,5} \right),\left( {3,9} \right),\left( {3,11} \right),\left( {3,15} \right),\\&\left( {3,16} \right),\left( {4,1} \right),\left( {4,{\rm{ }}5} \right),\left( {4,9} \right),\left( {4,11} \right),\left( {4,15} \right),\left( {4,16} \right)\end{align} \right\}

(i) A relation from a non-empty set $$A$$ to a non-empty set $$B$$ is a subset of the Cartesian product $$A \times B$$.

It is observed that $$f$$ is a subset of $$A \times B$$.

Thus, $$f$$ is a relation from $$A$$ to $$B.$$

(ii) Since the same first element i.e., $$2$$ corresponds to two different images i.e., $$9$$ and $$11.$$

Thus, $$f$$ is not a function from $$A$$ to $$B.$$

## Chapter 2 Ex.2.ME Question 11

Let $$f$$ be the subset of $${\bf{Z}} \times {\bf{Z}}$$ defined by $$f = \left\{ {\left( {ab,{\rm{ }}a + b} \right):a,b \in {\bf{Z}}} \right\}.$$Is $$f$$ a function from $$Z$$ to $$Z$$? Justify your answer.

### Solution

The relation $$f$$ is defined as $$f = \left\{ {\left( {ab,{\rm{ }}a + b} \right):a,b \in {\bf{Z}}} \right\}.$$

We know that a relation $$f$$ from a set $$A$$ to a set $$B$$ is said to be a function if every element of set $$A$$ has unique images in set $$B.$$

Since, $$2,6,-2,-6 \in {\bf{Z}},\left( {2 \times 6,2 + 6} \right),\left( {-2{\rm{ }} \times -6,-2{\rm{ }} + \left( {-6} \right)} \right) \in f$$

i.e., $$\left( {12,8} \right),\left( {12, - 8} \right) \in f$$

It can be seen that the same first element i.e., $$12$$ corresponds to two different images i.e., $$8$$ and $$–8.$$

Thus, relation $$f$$ is not a function.

## Chapter 2 Ex.2.ME Question 12

Let $$A = \left\{ {9,10,11,12,13} \right\}$$ and let $$f:A \to {\bf{N}}$$ be defined by $$f\left( n \right) =$$ the highest prime factor of $$n$$. Find the range of $$f$$.

### Solution

It is given that $$A = \left\{ {9,10,11,12,13} \right\}$$ and $$f:A \to {\bf{N}}$$ is defined by

$$f\left( n \right) =$$ the highest prime factor of $$n$$

Hence,

Prime factor of $$9 = 3$$

Prime factor of $$10 = 2,5$$

Prime factor of $$11 = 11$$

Prime factor of $$12 = 2,3$$

Prime factor of $$13 = 13$$

Therefore,

$$f\left( 9 \right) =$$ the highest prime factor of $$9 = 3$$

$$~f\left( 10 \right)=$$ the highest prime factor of $$10 = 5$$

$$f\left( {11} \right) =$$ the highest prime factor of $$11 = 11$$

$$~f\left( 12 \right)=$$ the highest prime factor of $$12 = 3$$

$$f\left( {13} \right) =$$ the highest prime factor of $$13 = 13$$

The range of $$f$$ is the set of all $$f\left( n \right),$$ where $$n \in A.$$

Therefore, Range of $$f = \left\{ {3,5,11,13} \right\}$$

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