Exercise 1.3 - Chapter 1 - Applications of Matrices and Determinants - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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EX 1.3
Question 1.
Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and $B$ denote the set of the sections. Define a relation from $A$ to $B$ as " $x$ related toy if the student $x$ belongs to the section $\mathrm{y}$ ". Is this relation a function? What can you say about the inverse relation? Explain your answer.
Solution:
(i) $\mathrm{A}=\left\{\right.$ set of students in $11^{\text {th }}$ standard $\}$
$B=\{$ set of sections in 11 sup $>$ th standard $\}$
$\mathrm{R}: \mathrm{A} \rightarrow \mathrm{B} \Rightarrow \mathrm{x}$ related to $\mathrm{y}$
$\Rightarrow$ Every students in eleventh Standard must in one section of the eleventh standard.
$\Rightarrow$ It is a function.
Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.
Question 2.
Write the values of $f$ at $-4,1,-2,7,0$ if
$f(x)= \begin{cases}-x+4 & \text { if }-\infty Solution:
$
\begin{aligned}
& f(-4)=-(-4)+4=8 \\
& f(1)=1-1^2=0 \\
& f(-2)=(-2)^2-(-2)=4+2=6 \\
& f(7)=0 \\
& f(0)=0
\end{aligned}
$
Question 3.
Write the values of $f$ at $-3,5,2,-1,0$ if

$
f(x)= \begin{cases}x^2+x-5 & \text { if } x \in(-\infty, 0) \\ x^2+3 x-2 & \text { if } x \in(3, \infty) \\ x^2 & \text { if } x \in(0,2) \\ x^2-3 & \text { otherwise }\end{cases}
$
Solution:
$
\begin{aligned}
& f(-3)=(-3)^2-3-5=9-8=1 \\
& f(5)=(5)^2+3(5)-2=25+15-2=38 \\
& f(2)=4-3=1 \\
& f(-1)=(-1)^2+(-1)-5=1-6=-5 \\
& f(0)=0-3=-3
\end{aligned}
$
Question 4.
State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?
(i) If $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}]$ and $/=\{(\mathrm{a}, \mathrm{c}),(\mathrm{b}, \mathrm{c}),(\mathrm{c}, \mathrm{b})\}$;(f: $\mathrm{A} \rightarrow \mathrm{A})$.
(ii) If $\mathrm{X}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}$ and $/=\{(\mathrm{x}, \mathrm{y}),(\mathrm{x}, \mathrm{z}),(\mathrm{z}, \mathrm{x})\}$; (f: $\mathrm{X} \rightarrow \mathrm{X})$.
Solution:
(i) $f: A \rightarrow A$

It is a function but it is not $1-1$ and not onto function.
(ii) $\mathrm{f}: \mathrm{X} \rightarrow \mathrm{X}$

$\mathrm{x} \in \mathrm{X}$ (Domain) has two $i$ not a function.
Question 5.
Let $A=\{1,2,3,4\}$ and $B=\{a, b, c, d\}$. Give a function from $A \rightarrow B$ for each of the following:
(i) neither one-to-one nor onto.
(ii) not one-to-one but onto.
(iii) one-to-one but not onto.
(iv) one-to-one and onto.
Solution:
$
\begin{aligned}
& \mathrm{A}=\{1,2,3,4\} \\
& \mathrm{B}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}
\end{aligned}
$

(i)

(ii)

$R=\{(1, \mathrm{~b})(2, \mathrm{~b})(3, \mathrm{c})(4, \mathrm{~d})\}$ is not $1-1$ and not onto
(iii) Not possible

(iv)

Question 6.
Find the domain of $\frac{1}{1-2 \sin x}$
Solution:
The Denominator $\neq 0 \Rightarrow 1-2 \sin x \neq 0$
$
\Rightarrow \sin x \neq \frac{1}{2} \text { (i.e.,) } x \neq \frac{\pi}{6}
$
So domain $=\mathrm{R}-\left\{n \pi+(-1)^n \frac{\pi}{6}\right\}, n \in \mathrm{Z}$
Question 7.
Find the largest possible domain of the real valued function $f(x)=\frac{\sqrt{4-x^2}}{\sqrt{x^2-9}}$
Solution:

$
f(x)=\frac{\sqrt{4-x^2}}{\sqrt{x^2-9}}

$\therefore$ No largest possible domain The domain is null set
Question 8.
Find the range of the function $\frac{1}{2 \cos x-1}$
Solution:
The range of $\cos x$ is -1 to 1
$
-1 \leq \cos x \leq 1
$
$(\times$ by 2$)-2 \leq 2 \cos x \leq 2$
adding -1 throughout
$
-2-1 \leq 2 \cos x-1 \leq 2-1
$
(i.e.) $-3 \leq 2 \cos x-1 \leq 1$
so $1 \leq \frac{1}{2 \cos x-1} \leq \frac{-1}{3}$
The range is outside $\frac{-1}{3}$ and 1
i.e., range is $\left(-\infty, \frac{-1}{3}\right] \cup[1, \infty)$

Question 9.
Show that the relation $x y=-2$ is a function for a suitable domain. Find the domain and the range of the function.
Solution:
$
\mathrm{xy}=-2 \Rightarrow \mathrm{y}=-2 / \mathrm{x}
$
which is a function
The domain is $(-\infty, 0) \cup(0, \infty)$ and range is $\mathrm{R}-\{0\}$
Question 10.
If $f, g: R \rightarrow R$ are defined by $f(x)=|x|+x$ and $g(x)=|x|-x$, find gof and fog.
Solution:
$
f(x)=\left\{\begin{array}{cc}
0 & x<0 \\
2 x & x>0
\end{array} \text { and } g(x)=\left\{\begin{array}{cc}
-2 x & x<0 \\
0 & x>0
\end{array}\right.\right.
$
Now $(f o g)(x)=0$ and $g o f(x)=0$

Question 11.
If $f, g, h$ are real valued functions defined on $R$, then prove that $(f+g) o h=f o h+g o h$. What can you say about $f o(g+h)$ ? Justify your answer.
Solution:
Let $\mathrm{f}+\mathrm{g}=\mathrm{k}$
Now $k \odot h=k(h(x))$
$=(\mathrm{f}+\mathrm{g}((\mathrm{h}(\mathrm{x}))$
$=\mathrm{f}[\mathrm{h}(\mathrm{x})]+\mathrm{g}[\mathrm{h}(\mathrm{x})]$
$=$ foh + goh
(i.e.,) $(f+g)(o) h=$ foh + goh
$f o(g+h)$ is also a function
Question 12.
If $f: R \rightarrow R$ is defined by $f(x)=3 x-5$, prove that $f$ is a bijection and find its inverse.
Solution:
$
\mathrm{P}(\mathrm{x})=3 \mathrm{x}-5
$
Let $y=3 x-5 \Rightarrow 3 x=y+5$
$
x=\frac{y+5}{3}
$
Let $g(y)=\frac{y+5}{3}$
Now $g \circ f(x)=g[(f(x)]=g(3 x-5)$
$=\frac{3 x-5+5}{3}=x$
also $f o g(y)=f[g(y)]=f\left[\frac{y+5}{3}\right]$
$
=3\left(\frac{y+5}{3}\right)-5=y+5-5=y
$
Thus $g \circ f=\mathrm{I}_x$ and $f o g=\mathrm{I}_y$
$f$ and $g$ are bijections and inverse to each other. Hence $f$ is a bijection and $f^{-1}(y)=\frac{y+5}{3}$
Replacing $y$ by $x$ we get $f^{-1}(x)=\frac{x+5}{3}$
Question 13.
The weight of the muscles of a man is a function of his body weight $\mathrm{x}$ and can be expressed as $W(x)=$ $0.35 \mathrm{x}$. Determine the domain of this function.
Solution:
$
\mathrm{W}(\mathrm{x})=0.35 \mathrm{x}
$
Since body weight $\mathrm{x}$ is positive and if it increases then $\mathrm{W}(\mathrm{x})$ also increase.
Domain is $(0, \infty)$ i.e., $\mathrm{x}>0$

Question 14.
The distance of an object falling is a function of time $t$ and can be expressed as $s(t)=-16 t^2$. Graph the function and determine if it is one-to-one.
Solution:
$s(t)=-16 t^2$
Suppose $S\left(t_1\right)=S\left(t_2\right)$
since time cannot be negative, we to take $t_1=t_2$
Hence it is one-one.

Question 15.
The total cost of airfare on a given route is comprised of the base cost $\mathrm{C}$ and the fuel surcharge $\mathrm{S}$ in rupee. Both $\mathrm{C}$ and $\mathrm{S}$ are functions of the mileage $\mathrm{m} ; \mathrm{C}(\mathrm{m})=0.4 \mathrm{~m}+50$ and $\mathrm{S}(\mathrm{m})=0.03 \mathrm{~m}$. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Solution:
$\mathrm{C}$ - base cost
$\mathrm{S}=$ fuel surcharge,
$\mathrm{m}=$ mileage
$\mathrm{C}(\mathrm{m})=0.4 \mathrm{~m}+50$
$\mathrm{S}(\mathrm{m})=0.03 \mathrm{~m}$
Total cost $=\mathrm{C}(\mathrm{m})+\mathrm{S}(\mathrm{m})$
$=0.4 \mathrm{~m}+50+0.03 \mathrm{~m}$
$=0.43 \mathrm{~m}+50$
for 1600 miles
$
\mathrm{T}(\mathrm{c})=0.43(1600)+50=688+50=₹ 738
$
Question 16.
A salesperson whose annual earnings can be represented by the function $\mathrm{A}(\mathrm{x})=30,000+0.04 \mathrm{x}$, where $\mathrm{x}$ is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function $S(x)=25,000+0.05 x$. Find $(A+S)(x)$ and determine the total family income if they each sell Rupees $1,50,00,000$ worth of merchandise.
Solution:
$A(x)=30,000+0.04 \mathrm{x}$, where $\mathrm{x}$ is merchandise rupee value
$\mathrm{S}(\mathrm{x})=25000+0.05 \mathrm{x}$
$(\mathrm{A}+\mathrm{S})(\mathrm{x})=\mathrm{A}(\mathrm{x})+\mathrm{S}(\mathrm{x})$
$=30000+0.04 \mathrm{x}+25000+0.05 \mathrm{x}$
$=55000+0.09 \mathrm{x}$
$(\mathrm{A}+\mathrm{S})(\mathrm{x})=55000+0.09 \mathrm{x}$
They each sell $\mathrm{x}=1,50,00,000$ worth of merchandise
$(\mathrm{A}+\mathrm{S}) \mathrm{x}=55000+0.09(1,50,00,000)$
$=55000+13,50,000$
$\therefore$ Total income of family $=₹ 14,05,000$

Question 17.
The function for exchanging American dollars for Singapore Dollar on a given day is $f(x)=1.23 x$, where $\mathrm{x}$ represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is $g(y)=50.50 \mathrm{y}$, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee. Solution: $\mathrm{f}(\mathrm{x})=1.23 \mathrm{x}$ where $\mathrm{x}$ is number of American dollars. $g(y)=50.50 \mathrm{y}$ where $y$ is number of Singapore dollars.
$$
\begin{aligned}
& \operatorname{gof}(\mathrm{x})=\mathrm{g}(\mathrm{f}(\mathrm{x})) \\
& =\mathrm{g}(1.23 \mathrm{x}) \\
& =50.50(1.23 \mathrm{x}) \\
& =62.115 \mathrm{x}
\end{aligned}
$
Question 18 .
The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100 . He estimates that if the menu price of the meal is $\mathrm{x}$ rupees, then the number of customers who will order that meal at that price in an evening is given by the function $D(x)=200-x$. Express his day revenue, total cost and profit on this meal as functions of $\mathrm{x}$.
Solution:
cost of one meal $=₹ 100$
Total cost $=₹ 100(200-\mathrm{x})$
Number of customers $=200-\mathrm{x}$
Day revenue $=₹(200-x) x$
Total profit $=$ day revenue - total cost
$
=(200-x) x-(100)(200-x)
$
Question 19.
The formula for converting from Fahrenheit to Celsius temperatures is $y=\frac{5 x}{9}-\frac{160}{9}$
Find the inverse of this function and determine whether the inverse is also a function.
Solution:

Question 20.
A simple cipher takes a number and codes it, using the function $f(x)=3 x-4$. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line $\mathrm{y}=\mathrm{x}$ (by drawing the lines).
Solution:
$
f(x)=3 x-4
$
Let $y=3 x-4$