# NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3

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## 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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