Exercise 1.5 - Chapter 1 Sets Relations and Functions 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

EX 1.5
Choose the correct or the most suitable answer.
Question 1.
If $A=\left\{(x, y): y=e^x ; x \in R\right\}$ and $B=\left\{(x, y): y=e^{-x}, x \in R\right\}$ then $n(A \cap B)$
(a) Infinity
(b) 0
(c) 1
(d) 2
Solution:
(c) 1
Hint.
$\mathrm{A} \cap \mathrm{B}=(0,1)$
$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=1$

Question 2.
IfA $\{(x, y): y=\sin x, x \in R)$ and $8=(x, y): y=\cos x, x \in R)$ then $A \cap B$ contains
(a) no element
(b) infinitely many elements
(c) only one element
(d) cannot be determined.
Solution:
(b) infinitely many elements
Question 3.
The relation $\mathrm{R}$ defined on a set $\mathrm{A}=\{0,-1,1,2\}$ by $\mathrm{xRy}$ if $\left|\mathrm{x}^2+\mathrm{y}^2\right| \leq 2$, then which one of the following is true?

(a) $\mathrm{R}=\{(0,0),(0,-1),(0,1),(-1,0),(-1,1),(1,2),(1,0)\}$
(b) $R=\{(0,0),(0,-1),(0,1),(-1,0),(1,0)$
(c) Domain of $R$ is $\{0,-1,1,2\}$
Solution:
(a) Range of $\mathrm{R}$ is $\{0,-1,1\}$
Hint.
$
\begin{aligned}
& A=\{0,-1,1,2\} \\
& \left|x^2+y^2\right| \leq 2
\end{aligned}
$
The values of $x$ and $y$ can be $0,-1$ or 1
So range $=\{0,-1,1\}$
Question 4 .
If $\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+|\mathrm{x}+2|, \mathrm{x} \in \mathrm{R}$, then

(a) $f(x)=\left\{\begin{array}{lll}-2 x & \text { if } & x \in(-\infty,-2] \\ 4 & \text { if } & x \in(-2,2] \\ 2 x & \text { if } & x \in(2, \infty)\end{array}\right.$
(b) $f(x)=\left\{\begin{array}{lll}2 x & \text { if } & x \in(-\infty,-2] \\ 4 x & \text { if } & x \in(-2,2] \\ -2 x & \text { if } & x \in(2, \infty)\end{array}\right.$
(c) $f(x)=\left\{\begin{array}{lll}-2 x & \text { if } & x \in(-\infty,-2] \\ -4 x & \text { if } & x \in(-2,2] \\ 2 x & \text { if } & x \in(2, \infty)\end{array}\right.$
(d) $f(x)=\left\{\begin{array}{lll}-2 x & \text { if } & x \in(-\infty,-2] \\ 2 x & \text { if } & x \in(-2,2] \\ 2 x & \text { if } & x \in(2, \infty)\end{array}\right.$
Solution:
(a) $f(x)=\left\{\begin{array}{lll}-2 x & \text { if } & x \in(-\infty,-2] \\ 4 & \text { if } & x \in(-2,2] \\ 2 x & \text { if } & x \in(2, \infty)\end{array}\right.$
Hint.
$
\begin{gathered}
f(x)=|\mathrm{x}-2|+|\mathrm{x}+2| \\
f(x)=\left\{\begin{array}{c}
-(x-2)-(x+2)=-2 x, x \in(-\infty,-2] \\
-(x-2)+(x+2)=4, x \in(-2,2] \\
x-2+x+2=2 x, x \in(2, \infty]
\end{array}\right. \\
\therefore f(x)=\quad\left\{\begin{array}{c}
-2 x, x \in(-\infty,-2] \\
4, x \in(-2,2] \\
2 x, x \in(2, \infty]
\end{array}\right.
\end{gathered}
$

Question 5.
Let $R$ be the set of all real numbers. Consider the following subsets of the plane $R x R: S=\{(x, y): y=x+1$ and $0<\mathrm{x}<2\}$ and $\mathrm{T}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}-\mathrm{y}$ is an integer $\}$ Then which of the following is true?
(a) $\mathrm{T}$ is an equivalence relation but $\mathrm{S}$ is not an equivalence relation.
(b) Neither $\mathrm{S}$ nor $\mathrm{T}$ is an equivalence relation
(c) Both $\mathrm{S}$ and $\mathrm{T}$ are equivalence relation
(d) $\mathrm{S}$ is an equivalence relation but $\mathrm{T}$ is not an equivalence relation.
Solution:
(a) $\mathrm{T}$ is an equivalence relation but $\mathrm{S}$ is not an equivalence relation.
Hint.
$(0,1),(1,2)$ it is not an equivalence relation
$\mathrm{T}$ is an equivalence relation
Question 6.
Let $A$ and $B$ be subsets of the universal set $N$, the set of natural numbers. Then $A^{\prime} \cup\left[(A \cap B) \cup B^{\prime}\right]$ is .........
(a) $\mathrm{A}$
(b) A'
(c) $\mathrm{B}$
(d) $\mathrm{N}$
Solution:
(d) $\mathrm{N}$
Hint.

Question 7.
The number of students who take both the subjects Mathematics and Chemistry is 70 . This represents $10 \%$ of the enrollment in Mathematics and $14 \%$ of the enrollment in Chemistry. How many students take at least one of these two subjects?
(a) 1120
(b) 1130
(c) 1100
(d) insufficient data
Solution:
(b) 1130
Hint.
$
\begin{aligned}
& \mathrm{n}(M \cup C)=n(M)+n(C)-n(M \cap C) \\
& =700+500-70=1130
\end{aligned}
$
Question 8 .
If $n[(A \times B) \cap(A \times C)]=8$ and $n(B \cap C)=2$, then $n(A)$ is
(a) 6
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
Question 9.
If $n(A)=2$ and $n(B \cup C)=3$, then $n[(A \times B) \cup(A \times C)]$ is ......
(a) $2^3$
(b) $3^2$
(c) 6
(d) 5
Solution:
(c) 6
Hint.

$
n[(A \times B) \cup(A \times C)]=n[A \times(B \cup C)]=n(A) \times n(B \cup C)=2 \times 3=6
$
Question 10.
If two sets $\mathrm{A}$ and $\mathrm{B}$ have 17 elements in common, then the number of elements common to the set $\mathrm{A} \times \mathrm{B}$ and $\mathrm{B} \times \mathrm{A}$ is
(a) $2^{17}$
(b) $17^2$
(c) 34
(d) insufficient data
Solution:
(b) $17^2$
Hint.
$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=17$
So $n[(A \times B) \cap(B \times A)]$
$=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=17 \times 17=17^2$
Question 11.
For non-empty sets $A$ and $B$, if $A \subset B$ then $(A \times B) \cap(B \times A)$ is equal to
(a) $A \cap B$
(b) $A \times A$
(c) $B \times B$
(d) None of these

Solution:
(b) $\mathrm{A} \times \mathrm{A}$
Hint.
When $\mathrm{A} \subset \mathrm{B},(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})=\mathrm{A} \times \mathrm{A}$
Question 12 .
The number of relations on a set containing 3 elements is
(a) 9
(b) 81
(c) 512
(d) 1024
Solution:
(c) 512
Hint.
Number of relations $=2^{n^2}=2^{3^2}=2^9=512$
Question 13.
Let $\mathrm{R}$ be the universal relation on a set $\mathrm{X}$ with more than one element. Then $\mathrm{R}$ is
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Solution:
(c) Transitive
Question 14.
Let $\mathrm{X}=\{1,2,3,4\}$ and $\mathrm{R}=\{(1,1),(1,2),(1,3),(2,2),(3,3),(2,1),(3,1),(1,4),(4,1)\}$. Then $\mathrm{R}$ is $\ldots \ldots \ldots$
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(b) Symmetric
Hint.
$(4,4\} \notin \mathrm{R} \Rightarrow \mathrm{R}$ is not reflexive
$(1,4),(4,1) \in R \Rightarrow R$ is symmetric
$(1,4),(4,1) \in \mathrm{R}$ but $(4,4) \notin \mathrm{R}$
So $\mathrm{R}$ is not transitive

Question 15.
The range of the function $\frac{1}{1-2 \sin x}$ is
(a) $(-\infty,-1) \cup\left(\frac{1}{3}, \infty\right)$
(b) $\left(-1, \frac{1}{3}\right)$
(c) $\left[-1, \frac{1}{3}\right]$
t) $(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)$
Solution:
$
\begin{aligned}
& -1 \leq \sin x \leq 1 \\
& \text { So }-2 \leq 2 \sin x \leq 2 \\
& \Rightarrow 2 \geq-2 \sin x \geq-2 \\
& 2+1 \geq-2 \sin x+1 \geq-2+1 \\
& \text { (i.e.) } 3 \geq 1-2 \sin x \geq-1 \\
& \Rightarrow \frac{1}{3} \leq \frac{1}{1-2 \sin x} \leq-1 \\
& \Rightarrow \text { The range is }(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)
\end{aligned}
$
Question 16.
The range of the function $f(x)=|\lfloor x\rfloor-x|, x \in \mathrm{R}$ is
Solution:
(c) $[0,1)$
Hint. $f(x)=|\lfloor x\rfloor-x| f(x)=\lfloor x\rfloor-x$
$f(0)=0-0=0$
$f(6.5)=6-6.5=-.5$
$f(-7.2)=8-7.2=.8$
$\therefore$ Range is $(0,1)$
Question 17.
The rule $f(x)=\mathrm{x}^2$ is a bijection if the domain and the co-domain are given by .....
(a) R,R
(b) R, $(0, \infty)$
(c) $(0, \infty), R$
(d) $[0, \infty),[0, \infty)$
Solution:
(d) $[0, \infty),[0, \infty)$

Question 18.
The number of constant functions from a set containing $\mathrm{m}$ elements to a set containing $\mathrm{n}$ elements is
(a) $\mathrm{mn}$
(b) $\mathrm{m}$
(c) $n$
(d) $m+n$
Solution:
(c) $\mathrm{n}$

Question 19.
The function $f:[0,2 \pi] \rightarrow[-1,1]$ defined by $f(x)=\sin x$ is
(a) One to one
(b) Onto
(c) Bijection
(d) Cannot be defined
Solution:
(b) Onto
For $x=\pi / 4, x=3 \pi / 4, f(x)=\frac{1}{\sqrt{2}}$
So it is not one-to-one
So it is an onto function
Question 20.
If the function $f:[-3,3] \rightarrow S$ defined by $f(x)=x^2$ is onto, then $S$ is (a) $[-9,9]$
(b) $\mathrm{R}$
(c) $[-3,3]$
(d) $[0,9]$
Solution:
(d) $[0,9]$
Question 21.
Let $X=\{1,2,3,4\}, Y=\{a, b, c, d)$ and $f=\{(1, a),(4, b),(2, c),(3, d)(2, d)\}$. Then $f$ is
(a) An one-to-one function
(b) An onto function
(c) A function which is not one-to-one
(d) Not a function
Solution:
(d) Not a function
Hint.

Since the element 2 has two images, it is not a function
Question 22.
The inverse of $f(x)=\left\{\begin{array}{ccc}x & ; & x<1 \\ x^2 & ; & 1 \leq x \leq 4 \\ 8 \sqrt{x} & ; & x>4\end{array}\right.$ is
(a) $f^{-1}(x)=\left\{\begin{array}{ccc}x & ; & x<1 \\ \sqrt{x} & ; & 1 \leq x \leq 16 \\ \frac{x^2}{64} ; & x>16\end{array}\right.$
(b) $f^{-1}(x)=\left\{\begin{array}{ccc}-x & ; & x<1 \\ \sqrt{x} & ; & 1 \leq x \leq 16 \\ \frac{x^2}{64} & ; & x>16\end{array}\right.$
(c) $f^{-1}(x)=\left\{\begin{array}{ccc}x^2 & ; & x<1 \\ \sqrt{x} & ; & 1 \leq x \leq 16 \\ \frac{x^2}{64} & ; & x>16\end{array}\right.$
(d) $f^{-1}(x)=\left\{\begin{array}{ccc}2 x & ; & x<1 \\ \sqrt{x} & ; & 1 \leq x \leq 16 \\ \frac{x^2}{8} & ; & x>16\end{array}\right.$
Solution:
(a)

$
f^{-1}(x)=\left\{\begin{array}{ccc}
x & ; & x<1 \\
\sqrt{x} & ; & 1 \leq x \leq 16 \\
\frac{x^2}{64} & ; & x>16
\end{array}\right.
$

$\begin{aligned}
& \text { Hint. } f(x)=\left\{\begin{array}{cc}
x & x<1 \\
x^2 & 1 \leq x \leq 4 \\
8 \sqrt{x} & x>4
\end{array}\right. \\
& y=x \text { then } x=y \Rightarrow f^{-1}(x)=x \\
& \text { When } y=x^2, x=\sqrt{y} \text {. So } f^{-1}(x)=\sqrt{x} \\
& \text { When } y=8 \sqrt{x} \\
& \sqrt{x}=\frac{y}{8}(\text { or }) x=\frac{y^2}{64} \\
& \therefore f^{-1}(x)=\frac{x^2}{64} \\
& \therefore f^{-1}(x)=\left\{\begin{array}{cc}
x & x<1 \\
\sqrt{x} & 1 \leq x \leq 16 \\
\frac{x^2}{64} & x>16
\end{array}\right. \\
&
\end{aligned}$

Question 23.
Let $f: R \rightarrow R$ be defined by $f(x)=1-|x|$. Then the range of $f$ is
(a) $\mathrm{R}$
(b) $(1, \infty)$
(c) $(-1, \infty)$
(d) $(-\infty, 1]$
Solution:
(d) $(-\infty, 1]$
Hint.
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=1-|x|$
For example,
$
\begin{aligned}
& f(1)=1-1=0 \\
& f(8)=1-8=-1 \\
& f(-9)=1-9=-8 \\
& f(-0.2)=1-0.2=0.8 \\
& \text { so range }=(-\infty, 1]
\end{aligned}
$
Question 24.
The function $f: R \rightarrow R$ is defined by $f(x)=\sin x+\cos x$ is .....
(a) An odd function
(b) Neither an odd function nor an even function

(c) An even function
(d) Both odd function and even function
Solution:
(b) Neither an odd function nor an even function
Question 25.
The function $f: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $f(x)=\frac{\left(x^2+\cos x\right)\left(1+x^4\right)}{(x-\sin x)\left(2 x-x^3\right)}+e^{-|x|}$ is
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function arid even function
Solution:
(c) An even function
$
\begin{aligned}
f(x) & =\frac{\left(x^2+\cos x\right)\left(1+x^4\right)}{(x-\sin x)\left(2 x-x^3\right)}+e^{-|x|} \\
f(-x) & =\frac{x^2+\cos (-x)\left(1+(-x)^4\right)}{[-x-\sin (-x)]\left[-2 x-(-x)^3\right]}+e^{-|-x|} \\
& =\frac{\left(x^2+\cos x\right)\left(1+x^4\right)}{(-x+\sin x)\left(-2 x+x^3\right)}+e^{-|x|} \\
& =\frac{\left(x^2+\cos x\right)\left(1+x^4\right)}{-(x-\sin x)(-1)\left(2 x-x^3\right)}+e^{-|x|} \\
& =\frac{\left(x^2+\cos x\right)\left(1+x^4\right)}{(x-\sin x)\left(2 x-x^3\right)}+e^{-|x|}=f(x)
\end{aligned}
$
$\Rightarrow f(x)$ is an even function