Exercise 1.3 - Chapter 1 Relations & Functions 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 1.3$
Question $1 .$
Let $f=\{(x, y) \mid x, y \in N$ and $y=2 x\}$ be a relation on $N$. Find the domain, co-domain and range. Is this relation a function?
Solution:
$\begin{aligned}
&F=\{(x, y) \mid x, y \in N \text { and } y=2 x\} \\
&x=\{1,2,3, \ldots\} \\
&y=\{1 \times 2,2 \times 2,3 \times 2,4 \times 2,5 \times 2 \ldots\} \\
&R=\{(1,2),(2,4),(3,6),(4,8),(5,10), \ldots\}
\end{aligned}$
Domain of $\mathrm{R}=\{1,2,3,4, \ldots\}$,
Co-domain $=\{1,2,3 \ldots \ldots\}$
Range of $R=\{2,4,6,8,10, \ldots\}$
Yes, this relation is a function.
 

Question 2 .
Let $\mathrm{X}=\{3,4,6,8\}$. Determine whether the relation $\mathrm{R}=\left\{(\mathrm{x}, \mathrm{f}(\mathrm{x})) \mid \mathrm{x} \in \mathrm{X}, \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+1\right\}$ is a
function from $X$ to $N$ ?
Solution:
$\begin{aligned}
&x=\{3,4,6,8\} \\
&R=\left((x, f(x)) \mid x \in X, f(x)=X^{2}+1\right\} \\
&f(x)=x^{2}+1 \\
&f(3)=3^{2}+1=10 \\
&f(4)=4^{2}+1=17 \\
&f(6)=6^{2}+1=37 \\
&f(8)=8^{2}+1=65
\end{aligned}$

$\mathrm{R}=\{(3,10),(4,17),(6,37),(8,65)\}$
$R=\{(3,10),(4,17),(6,37),(8,65)\}$
Yes, $R$ is a function from $X$ to $N$.

Question 3.
Given the function
$f: x \rightarrow x^{2}-5 x+6$, evaluate
(i) $\mathrm{f}(-1)$
(ii) $f(2 a)$
(iii) $f(2)$
(iv) $f(x-1)$
Answer:
$f(x)=x^{2}-5 x+6$
(i) $f(-1)=(-1)^{2}-5(-1)+6=1+5+6=12$
(ii) $f(2 a)=(2 a)^{2}-5(2 a)+6=4 a^{2}-10 a+6$
(iii) $f(2)=2^{2}-5(2)+6=4-10+6=0$
(iv) $f(x-1)=(x-1)^{2}-5(x-1)+6$
$\begin{aligned}
&=x^{2}-2 x+1-5 x+5+6 \\
&=x^{2}-7 x+12
\end{aligned}$

Question $4 .$
A graph representing the function $f(x)$ is given in figure it is clear that $f(9)=2$.

(i) Find the following values of the function
(a) $f(0)$
(b) $f(7)$
(c) $f(2)$
(d) $\mathrm{f}(10)$
(ii) For what value of $x$ is $f(x)=1$ ?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under $f$ ?
Solution:
From the graph
(a) $f(0)=9$
(b) $f(7)=6$
(c) $f(2)=6$
(d) $f(10)=0$
(ii) At $x=9.5, \mathrm{f}(\mathrm{x})=1$
(iii) Domain $=\{0,1,2,3,4,5,6,7,8,9,10\}$ $=\{\mathrm{x} \mid 0<\mathrm{x}<10, \mathrm{x} \in \mathrm{R}\}$
Range $=\{x \mid 0 $=\{0,1,2,3,4,5,6,7,8,9\}$
(iv) The image of 6 under $f$ is 5 .

Question $5 .$
Let $f(x)=2 x+5$. If $x \neq 0$ then find $\frac{f(x+2)-f(2)}{x}$
Solution:

Given $f(x)=2 x+5, x \neq 0$.
$\begin{aligned}
\frac{f(x+2)-f(2)}{x} & \\
f(x) &=2 x+5 \\
& \Rightarrow f(x+2) \\
&=2(x+2)+5 \\
&=2 x+4+5=2 x+9 \\
\Rightarrow f(2) &=2(2)+5=4+5=9 \\
\therefore \frac{f(x+2)-f(2)}{x} &=\frac{2 x+9-9}{x}=\frac{2 x}{x}=2
\end{aligned}$

Question $6 .$
A function fis defined by $f(x)=2 x-3$
(i) find $\frac{f(0)+f(1)}{2}$
(ii) find $x$ such that $f(x)=0$.
(iii) find $x$ such that $f(x)=x$.
(iv) find $x$ such that $f(x)=f(1-x)$.
Solution:
Given $f(x)=2 x-3$
(i) find $\frac{f(0)+f(1)}{2}$
$f(0)=2(0)-3=-3$
$f(1)=2(1)-3=-1$
$\therefore \frac{f(0)+f(1)}{2}=\frac{-3-1}{2}=\frac{-4}{2}=-2$
(ii) $f(x)=0$
$\Rightarrow 2 \mathrm{x}-3=0$
$2 \mathrm{x}=3$
$x=\frac{3}{2}$
(iii) $f(x)=x$
$\Rightarrow 2 \mathrm{x}-3=\mathrm{x} \Rightarrow 2 \mathrm{x}-\mathrm{x}=3$
$x=3$
(iv) $f(x)=f(1-x)$
$2 x-3=2(1-x)-3$
$2 \mathrm{x}-3=2 \mathrm{x}-2 \mathrm{x}-3$
$2 \mathrm{x}+2 \mathrm{x}=2-3+3$
$4 x=2$

$\begin{aligned}
&x=\frac{2}{4} \\
&x=\frac{1}{2}
\end{aligned}$

Question $7 .$
An open box is to be made from a square piece of material, $24 \mathrm{~cm}$ on a side, by cutting equal squares from the corners and tuming up the sides as shown in figure. Express the volume $V$ of the box as a function of $x$.

Solution:
Volume of the box $=$ Volume of the cuboid
$=1 \times b \times h c u$. units
Here $1=24-2 \mathrm{x}$
$b=24-2 x$
$\mathrm{h}=\mathrm{x}$
$\therefore \mathrm{V}=(24-2 \mathrm{x})(24-2 \mathrm{x}) \times \mathrm{x}$
$=\left(576-48 x-48 x+4 x^{2}\right) x$
$\mathrm{V}=4 \mathrm{x}^{3}-96 \mathrm{x}^{2}+576 \mathrm{x}$

Question 8.
A function $f$ is defined bv $f(x)=3-2 x$. Find $x$ such that $f(x 2)=(f(x)) 2$.
Solution:
$f(x)=3-2 x$
$f\left(x^{2}\right)=3-2 x^{2}$
$(f(x))^{2}=(3-2 x)^{2}=9-12 x+4 x^{2}$
$f\left(x^{2}\right)=(f(x))^{2} \Rightarrow 3-2 x^{2}=9-12 x+4 x^{2}$
$6 x^{2}-12 x+6=0[\div 6]$
$x^{2}-2 x+1=0$
$(x-1)(x-1)=0$
$-1-1$
$x=1,1$

Question $9 .$
A plane is flying at a speed of $500 \mathrm{~km}$ per hour. Express the distance $\mathrm{d}$ travelled by the plane as function of time $r$ in hours.
Answer:
Speed of the plane $=500 \mathrm{~km} / \mathrm{hr}$
Distance travelled in " $t$ " hours
$=500 \times \mathrm{t}$ (distance $=$ speed $\times$ time $)$
$=500 \mathrm{t}$

Question $10 .$
The data in the adjacent table depicts the length of a woman's forehand and her corresponding height. Based on this data, a student finds a relationship between the height ( $y$ ) and the forehand length $(x)$ as $y=a x+b$, where $a, b$ are constants.

(i) Check if this relation is a function.
(ii) Find $a$ and $b$.
(iii) Find the height of a woman whose forehand length is $40 \mathrm{~cm}$.
(iv) Find the length of forehand of a woman if her height is $53.3$ inches.
Solution:
(i) Given $y=a x+b$
(1)
The ordered pairs are $\mathrm{R}=\{(35,56)(45,65)(50,69.5)(55,74)\}$
$\therefore$ Hence this relation is a function.
(ii) Consider any two ordered pairs $(35,56)$

Substituting $a=0.9$ in (2) we get
$\Rightarrow 65=45(.9)+b$
$\Rightarrow 65=40.5+b$
$\Rightarrow \mathrm{b}=65-40.5$
$\Rightarrow b=24.5$
$\therefore \mathrm{a}=0.9, \mathrm{~b}=24.5$
$\therefore \mathrm{y}=0.9 \mathrm{x}+24.5$
$\begin{aligned}
&\text { (iii) Given } x=40, y=? \\
&\therefore(4) \rightarrow y=0.9(40)+24.5 \\
&\Rightarrow y=36+24.5 \\
&\Rightarrow y=60.5 \text { inches } \\
&\text { (iv) Given } y=53.3 \text { inches, } x=? \\
&(4) \rightarrow 53.3=0.9 x+24.5 \\
&\Rightarrow 53.3-24.5=0.9 x \\
&\Rightarrow 28.8=0.9 x \\
&\Rightarrow x=\frac{28.8}{0.9}=32 \mathrm{~cm} \\
&\therefore \text { When } y=53.3 \text { inches, } x=32 \mathrm{~cm}
\end{aligned}$