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 \(\left( {a,c} \right) \in R\)

Justify your answer in each case.

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 \(B.\)

Justify your answer in each case.

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 \(f = \left\{ {3,5,11,13} \right\}\)