Exercise 1.4 - Chapter 1 Relations & Functions 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 1.4
Question $1 .$
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

(i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.
 

Question 2.
Let $f: A \rightarrow B$ be a function defined by $f(x)=\frac{x}{2}-1$, Where $A=\{2,4,6,10,12\}$,
$\mathrm{B}=\{0,1,2,4,5,9\}$. Represent $\mathrm{f}$ by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution:
f: $\mathrm{A} \rightarrow \mathrm{B}$
$A=\{2,4,6,10,12\}, B=\{0,1,2,4,5,9\}$
$f(x)=\frac{x}{2}-1$
$f(2)=\frac{2}{2}-1=0$
$f(4)=\frac{4}{2}-1=1$
$f(6)=\frac{6}{2}-1=2$
$f(10)=\frac{10}{2}-1=4$

$f(12)=\frac{12}{2}-1=5$
(i) Set of ordered pairs $=\{(2,0),(4,1),(6,2),(10,4),(12,5)\}$

(ii) a table

Question $3 .$
Represent the function $f=\{(1,2),(2,2),(3,2),(4,3),(5,4)\}$ through
(i) an arrow diagram
(ii) a table form
(iii) a graph
Solution:
$f=\{(1,2),(2,2),(3,2),(4,3),(5,4)\}$

Question $4 .$
Show that the function $f: N \rightarrow N$ defined by $f(x)=2 x-1$ is one -one but not onto.
Solution:
f: $\mathrm{N} \rightarrow \mathrm{N}$
$f(x)=2 x-1$
$N=\{1,2,3,4,5, \ldots\}$
$f(1)=2(1)-1=1$
$f(2)=2(2)-1=3$
$f(3)=2(3)-1=5$
$f(4)=2(4)-1=7$
$f(5)=2(5)-1=9$

In the figure, for different elements in $x$, there are different images in $f(x)$.
Hence $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ is a one-one function.
A function $f: N \rightarrow N$ is said to be onto function if the range of $f$ is equal to the co-domain of $f$
Range $=\{1,3,5,7,9, \ldots\}$
Co-domain $=\{1,2,3, . .\}$
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.
 

Question $5 .$
Show that the function $f: N \rightarrow N$ defined by $f(m)=m^{2}+m+3$ is one - one function.
Solution:
f: $\mathrm{N} \rightarrow \mathrm{N}$
$f(m)=m^{2}+m+3$
$\mathrm{N}=\{1,2,3,4,5 \ldots \ldots\} \mathrm{m} \in \mathrm{N}$
$f(m)=m^{2}+m+3$
$f(1)=1^{2}+1+3=5$
$f(2)=2^{2}+2+3=9$
$f(3)=3^{2}+3+3=15$
$f(4)=4^{2}+4+3=23$

In the figure, for different elements in the $(\mathrm{X})$ domain, there are different images in $f(x)$. Hence $f: N$ $\rightarrow \mathrm{N}$ is a one to one but not onto function as the range of $\mathrm{f}$ is not equal to co-domain.
Hence it is proved.

Question $6 .$
Let $\mathrm{A}=\{1,2,3,4\}$ and $\mathrm{B}=\mathrm{N}$.
Let $f: A \rightarrow B$ be defined by $f(x)=x^{3}$ then,
(i) find the range of $f$
(ii) identify the type of function
Answer:
$\begin{aligned}
&A=\{1,2,3,4\} \\
&B=\{1,2,3,4,5, \ldots\} \\
&f(x)=x^{3} \\
&f(1)=1^{3}=1 \\
&f(2)=2^{3}=8 \\
&f(3)=3^{3}=27 \\
&f(4)=4^{3}=64
\end{aligned}$
(i) Range $=\{1,8,27,64\}$
(ii) one -one and into function.
 

Question $7 .$
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+1$
(ii) $f: R \rightarrow R$ defined by $f(x)=3-4 x^{2}$
Solution:
(i) $f: R \rightarrow R$
$f(x)=2 x+1$
$f(1)=2(1)+1=3$
$f(2)=2(2)+1=5$

$\begin{aligned}
&f(-1)=2(-1)+1=-1 \\
&f(0)=2(0)+1=1
\end{aligned}$
It is a bijective function. Distinct elements of $A$ have distinct images in $B$ and every element in $B$ has a pre-image in $\mathrm{A}$.
$\begin{aligned}
&\text { (ii) } f: R \rightarrow R ; f(x)=3-4 x^{2} \\
&f(1)=3-4\left(1^{2}\right)=3-4=-1 \\
&f(2)=3-4\left(2^{2}\right)=3-16=-13 \\
&f(-1)=3-4(-1)^{2}=3-4=-1
\end{aligned}$
It is not bijective function since it is not one-one
 

Question 8.
Let $A=\{-1,1\}$ and $B=\{0,2\}$. If the function $f: A \rightarrow B$ defined by $f(x)=a x+b$ is an onto function? Find $a$ and $b$.
Solution:
$\mathrm{A}=\{-1,1\}, \mathrm{B}=\{0,2\}$
f. $\mathrm{A} \rightarrow \mathrm{B}, \mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$
$f(-1)=a(-1)+b=-a+b$
$f(1)=a(1)+b=a+b$
Since $f(x)$ is onto, $f(-1)=0$
$\Rightarrow-\mathrm{a}+\mathrm{b}=0$... (1)
$\& f(1)=2$
$\Rightarrow a+b=2$... (2)
$-a+b=0$
$a+b=2$ $2 b=2$
$b=1$
$\therefore(2) \Rightarrow a+1=2$
$a=2-1$
$a=1$
$\therefore \quad a=1, b=1$

Question $9 .$
If the function $\mathrm{f}$ is defined by
$f(x)=\left[\begin{array}{cll}x+2 & \text { if } & x>1 \\ 2 & \text { if } & -1 \leq x \leq 1 \\ x-1 & \text { if } & -3 (i) $\mathrm{f}(3)$
(ii) $f(0)$
(iii) $f(-1.5)$
(iv) $f(2)+f(-2)$
Solution:
(i) $f(3) \Rightarrow f(x)=x+2 \Rightarrow 3+2=5$
(ii) $f(0) \Rightarrow 2$
(iii) $f(-1.5)=x-1$
$=-1.5-1=-2.5$
(iv) $f(2)+f(-2)$
$f(2)=2+2=4 \quad[\because f(x)=x+2]$
$f(-2)=-2-1=-3 \quad[\because f(x)=x-1]$
$f(2)+f(-2)=4-3=1$
 

Question $10 .$
A function $\mathrm{f}:[-5,9] \rightarrow \mathrm{R}$ is defined as follows:
$f(x)=\left[\begin{array}{l}6 x+1 \text { if }-5 \leq x<2 \\ 5 x^{2}-1 \text { if } 2 \leq x<6 \\ 3 x-4 \text { if } 6 \leq x \leq 9\end{array}\right.$
Find (i) $f(-3)+f(2)$
(ii) $\quad f(7)-f(1)$
(iii) $2 f(4)+f(8)$
(iv) $\frac{2 f(-2)-f(6)}{f(4)+f(-2)}$

Solution:
f : $[-5,9] \rightarrow \mathrm{R}$
(i) $f(-3)+f(2)$
$f(-3)=6 x+1=6(-3)+1=-17$
$f(2)=5 \times 2-1=5\left(2^{2}\right)-1=19$
$\therefore f(-3)+f(2)=-17+19=2$
(ii) $f(7)-f(1)$
$f(7)=3 x-4=3(7)-4=17$
$f(1)=6 x+1=6(1)+1=7$
$f(7)-f(1)=17-7=10$
(iii) $2 f(4)+f(8)$
$f(4)=5 x^{2}-1=5 \times 4^{2}-1=79$
$f(8)=3 x-4=3 \times 8-4=20$
$\therefore 2 \mathrm{f}(4)+\mathrm{f}(8)=2 \times 79+20=178$
(iv) $\frac{2 f(-2)-f(6)}{f(4)+f(-2)}$
$\begin{aligned}
f(-2)=& 6 x+1=6(-2)+1=-11 \\
f(6) &=3 x-4=3(6)-4=14 \\
f(4)=& 5 x^{2}-1=5\left(4^{2}\right)-1=79 \\
f(-2)=& 6 x+1=6(-2)+1=-11 \\
\frac{2 f(-2)-f(6)}{f(4)+f(-2)}=& \frac{2(-11)-14}{79+(-11)}=\frac{-22-14}{68} \\
&=\frac{-36}{68}=\frac{-9}{17}
\end{aligned}$

Question 11
The distance $S$ an object travels under the influence of gravity in time $t$ seconds is 12 given by $S(t)$ $=\frac{1}{3} \mathrm{gt}^{2}+\mathrm{at}+\mathrm{b}$, where, ( $\mathrm{g}$ is the acceleration due to gravity), $\mathrm{a}, \mathrm{b}$ are constants. Check if the function $S(t)$ is one-one.
Answer.
$S(t)=\frac{1}{2} g t^{2}+a t+b$
Let the time be $1,2,3 \ldots \mathrm{n}$ seconds
$\begin{aligned}
&S(1)=\frac{1}{2} g(1)^{2}+a(1)+b \\
&=\frac{g}{2}+a+b \\
&S(2)=\frac{1}{2} g(2)^{2}+a(2)+b \\
&=\frac{4 g}{2}+2 a+b \\
&=2 g+2 a+b \\
&S(3)=\frac{1}{2} g(3)^{2}+a(3)+6 \\
&=\frac{9}{2} g+3 a+b
\end{aligned}$
For every different value of t, there will be different distance.
$\therefore$ It is a one-one function.
 

Question $12 .$
The function ' $t$ ' which maps temperature in Celsius $(\mathrm{C})$ into temperature in Fahrenheit (F) is defined by $t(C)=F$ where $F=\frac{9}{5} C+32$. Find,
(i) $\mathrm{t}(0)$
(ii) $\mathrm{t}(28)$
(iii) t(-10)
(iv) the value of $\mathrm{C}$ when $\mathrm{t}(\mathrm{C})=212$
(v) the temperature when the Celsius value is equal to the Farenheit value.
Solution:
(1)
$\begin{aligned}
t(0) &=\mathrm{F} \\
\mathrm{F} &=\frac{9}{5}(\mathrm{C})+32=\frac{9}{5}(0)+32=32^{\circ} \mathrm{F}
\end{aligned}$
(ii)
$t(28)=\mathrm{F}=\frac{9}{5}(28)+32=\frac{252}{5}+32$

$=50.4+32=82.4^{\circ} \mathrm{F}$
(iii) $t(-10)=\mathrm{F}=\frac{9}{5}(-10)+32=14^{\circ} \mathrm{F}$
(iv) $t(\mathrm{C})=212$
$\text { i.e } \begin{aligned}
\frac{9}{5}(\mathrm{C})+32 &=212 \Rightarrow \frac{9}{5} \mathrm{C}=212-32=180 \\
\frac{9}{5} \mathrm{C} &=180 \Rightarrow \mathrm{C}=\frac{180 \times 5}{9}=100^{\circ} \mathrm{C} \\
\mathrm{C} &=100^{\circ} \mathrm{C} .
\end{aligned}$
(v) when $\mathrm{C}=\mathrm{F}$
$\begin{aligned}
\frac{9}{5} \mathrm{C}+32 &=\mathrm{C} \\
32 &=\mathrm{C}-\frac{9}{5} \mathrm{C} \\
32 &=\mathrm{C}\left(1-\frac{9}{5}\right) \\
32 &=\mathrm{C}\left(\frac{5-9}{5}\right) \\
32 &=\mathrm{C}\left(\frac{-4}{5}\right) \\
\mathrm{C} &=\frac{8}{32 \times \frac{-5}{4}} \\
\mathrm{C} &=-40^{\circ}
\end{aligned}$