NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3


Chapter 2 Ex.2.3 Question 1

Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) \(\left\{ {\left( {{\rm{2}},{\rm{1}}} \right),\left( {{\rm{5}},{\rm{1}}} \right),\left( {{\rm{8}},{\rm{1}}} \right),\left( {{\rm{11}},{\rm{1}}} \right),\left( {{\rm{14}},{\rm{1}}} \right),\left( {{\rm{17}},{\rm{1}}} \right)} \right\}\)

(ii) \(\left\{ {\left( {{\rm{2}},{\rm{1}}} \right),\left( {{\rm{4}},{\rm{2}}} \right),\left( {{\rm{6}},{\rm{3}}} \right),\left( {{\rm{8}},{\rm{4}}} \right),\left( {{\rm{1}}0,{\rm{5}}} \right),\left( {{\rm{12}},{\rm{6}}} \right),\left( {{\rm{14}},{\rm{7}}} \right)} \right\}\)

(iii) \(\left\{ {\left( {{\rm{1}},{\rm{3}}} \right),\left( {{\rm{1}},{\rm{5}}} \right),\left( {{\rm{2}},{\rm{5}}} \right)} \right\}\)

 

Solution

 

(i) \(\left\{ {\left( {{\rm{2}},{\rm{1}}} \right),\left( {{\rm{5}},{\rm{1}}} \right),\left( {{\rm{8}},{\rm{1}}} \right),\left( {{\rm{11}},{\rm{1}}} \right),\left( {{\rm{14}},{\rm{1}}} \right),\left( {{\rm{17}},{\rm{1}}} \right)} \right\}\)

Since \({\rm{2}},{\rm{5}},{\rm{8}},{\rm{11}},{\rm{14}},\) and \(17\) are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain \( = \left\{ {2,5,8,11,14,17} \right\}\) and range \( = \left\{ 1 \right\}\)

(ii) \(\left\{ {\left( {{\rm{2}},{\rm{1}}} \right),\left( {{\rm{4}},{\rm{2}}} \right),\left( {{\rm{6}},{\rm{3}}} \right),\left( {{\rm{8}},{\rm{4}}} \right),\left( {{\rm{1}}0,{\rm{5}}} \right),\left( {{\rm{12}},{\rm{6}}} \right),\left( {{\rm{14}},{\rm{7}}} \right)} \right\}\)

Since \({\rm{2}},{\rm{4}},{\rm{6}},{\rm{8}},{\rm{1}}0,12\) and \(14\) are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain \( = \left\{ {2,4,6,8,10,12,14} \right\}\) and range \( = \left\{ {1,2,3,4,5,6,7} \right\}\)

(iii) \(\left\{ {\left( {{\rm{1}},{\rm{3}}} \right),\left( {{\rm{1}},{\rm{5}}} \right),\left( {{\rm{2}},{\rm{5}}} \right)} \right\}\)

Since the same first element i.e., \(1\) corresponds to two different images i.e., \(3\) and \(5,\) this relation is not a function.

Chapter 2 Ex.2.3 Question 2

Find the domain and range of the following real function:

(i) \(f\left( x \right) = -\left| x \right|\)

(ii) \(f\left( x \right) = \sqrt {9 - {x^2}}\)

 

Solution

 

(i) \(f\left( x \right) = -\left| x \right|,x \in {\bf{R}}\)

We know that, \(\left| x \right| = \left\{ \begin{array}{l}x,{\rm{ if }}\;x \ge {\rm{0}}\\ - x,{\rm{ if }}\;x < {\rm{0}}\end{array} \right.\)

Therefore, \(f(x) = \left| { - x} \right| = \left\{ \begin{array}{l} - x,{\rm{ if }}\,x \ge {\rm{0}}\\x,{\rm{ if }}\,x < {\rm{0}}\end{array} \right.\)

Since \(f\left( x \right)\) is defined for \(x \in {\bf{R}}\), the domain of \(f\left( x \right) = {\bf{R}}\)

It can be observed that the range of \(f\left( x \right) = -\left| x \right|\) is all real numbers except positive real numbers.

Therefore, the range of \(f\left( x \right) is \; \left( {-\infty ,0} \right]\)

(ii) \(f\left( x \right) = \sqrt {9 - {x^2}} \)

Since \(\sqrt {9 - {x^2}} \) is defined for all real numbers that are greater than or equal to \(-3\) and less than or equal to \(3,\) the domain of \(f\left( x \right)\) is \(\left\{ {x:-3 \le x \le 3} \right\}\) or \(\left[ -3,\; 3 \right].\)

For any value of \(x\) such that \(-3 \le x \le 3\), the value of \(f\left( x \right)\) will lie between \(0\) and \(3.\)

Therefore, the range of \(f\left( x \right)\) is \(\left\{ {x:0 \le x \le 3} \right\}\) or  \(\left[ 0,\, 3 \right]\)

Chapter 2 Ex.2.3 Question 3

A function \(f\) is defined by \(f\left( x \right) = 2x-5\). Write down the values of

(i)  \(f\left( 0 \right)\)

(ii)  \(f\left( 7 \right)\)

(iii) \(f\left( {-3} \right)\)

 

Solution

 

The given function is \(f\left( x \right) = 2x-5\).

Therefore,

(i) \({\rm{ }}f\left( 0 \right) = {\rm{2}} \times 0 - {\rm{5}} = 0 - {\rm{5}} = - {\rm{5}}\)

(ii) \(f\left( {\rm{7}} \right) = {\rm{2}} \times 7 - {\rm{5}} = 14 - {\rm{5}} = {\rm{9}}\)

(iii) \(f\left( {-{\rm{3}}} \right) = {\rm{2}} \times \left( { - 3} \right) - 5 = - 6 - 5 = - {\rm{11}}\)

Chapter 2 Ex.2.3 Question 4

The function \('t'\) which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by \(t\left( C \right) = \frac{{9C}}{5} + 32 \).

Find

(i) \(t\left( 0 \right)\)

(ii) \(t\left( {28} \right)\)

(iii) \(t\left( {-10} \right)\)

(iv) The value of \(C,\) when \(~t\left( C \right)=212\).

 

Solution

 

The given function is \(t\left( C \right) = \frac{{9C}}{5} + 32\)

Therefore,

(i) \(\begin{align}t\left( 0 \right) = \frac{{9 \times 0}}{5} + 32 = 0 + 32 = 32 \end{align}\)

(ii) \(\begin{align}t\left( {28} \right) = \frac{{9 \times 28}}{5} + 32 = \frac{{252 + 160}}{5} = \frac{{412}}{5} = 82.4 \end{align}\)

(iii) \(\begin{align}t\left( { - 10} \right) = \frac{{9 \times ( - 10)}}{5} + 32 = 9 \times ( - 2) + 32 = - 18 + 32 = 14 \end{align}\)

(iv) It is given that

\[\begin{align}\;\;\;\;t&\left( C \right) = 212\\ \Rightarrow \;&\frac{{9C}}{5} + 32 = 212\\ \Rightarrow \;&\frac{{9C}}{5} = 212 - 32\\ \Rightarrow \;&C = \frac{{180 \times 5}}{9}\\&\quad= 100\end{align}\]

Thus, the value of \('t'\), when \(t\left( C \right) = 212\) is \(100.\)

Chapter 2 Ex.2.3 Question 5

Find the range of each of the following functions.

(i) \(f\left( x \right)~=2-3x,\,x\in R,x>0.\)

(ii) \(f\left( x \right)~={{x}^{2}}+2,\,x\) is a real number.

(iii) \(f\left( x \right)~=x,\,~x\) is a real number.

 

Solution

 

(i) \(f\left( x \right)~=2-3x,x\in \mathbf{R},x>0.\)

The values of \(f\left( x \right)\) for various values of real numbers \(x > 0\) can be written in the tabular form as

\(x\)

\(0.01 \)

 \(0.1\)

\(0.9\)

\(1\)

\(2\)

\(2.5\)

\(4\)

\( 5\)

\( \ldots \)

\(f\left( x \right)\)

\(1.97\)

\(1.7\)

\(-0.7\)

\(-1\)

\(-4\)

\(-5.5\)

\( -10\)

\(-13\)

\( \ldots \)

Thus, it can be clearly observed that the range of \(f\) is the set of all real numbers less than \(2.\) i.e., range of \(f = \left( {-\infty ,2} \right)\)

Alternative Method:

Let \(x > 0\)

\[\begin{array}{l} \Rightarrow 3x > 0\\ \Rightarrow 2 - 3x < 2\\ \Rightarrow f\left( x \right) < 2\end{array}\]

Therefore, Range of \(f = \left( {-\infty ,2} \right)\)

(ii) \(f\left( x \right)~={{x}^{2}}+2,x\) is a real number.

The values of \(f(x)\) for various values of real numbers \(x\) can be written in the tabular form as

\(x\)

0

\( \pm 0.3\)

\( \pm 0.8\)

\( \pm 1\)

\( \pm 2\)

\( \pm 3\)

\( \ldots \)

\(f\left( x \right)\)

2

2.09

2.64

3

6

11

\( \ldots \)

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than \(2.\) i.e., range of \(f = \left[ {2,\infty } \right)\)

Alternative Method:

Let \(x\) be any real number i.e., \({x^2} \ge 0\).

Accordingly,

\[\begin{align}&{x^2} \ge 0\\ \Rightarrow &\;{x^2} + {\rm{2}} \ge 0 + {\rm{2}}\\ \Rightarrow &\;{x^2} + {\rm{2}} \ge {\rm{2}}\\ \Rightarrow &\;f\left( x \right) \ge {\rm{2}}\end{align}\]

Therefore, Range of \(f = \left[ {2,\infty } \right)\)

(iii) \(f\left( x \right) = x,\,x\) is a real number

It is clear that, the range of \(f\) is the set of all real numbers.

Therefore, Range of \(f = {\bf{R}}.\)

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