NCERT Solutions Class 11 Maths Chapter 2

NCERT Solutions Class 11

Maths

Chapter 2 – Relations and Functions

Q.1 Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

Answer:

We have, A = {9, 10, 11, 12, 13}
and f(n) = the highest prime factor of n
f(9) = the highest prime factor of 9 =3,
f(10) = the highest prime factor of 10 = 5,
f(11) = the highest prime factor of 11 = 11,
f(12) = the highest prime factor of 12 = 3,
f(13) = the highest prime factor of 13 = 13

The range of function = {Values of f(n)}, n ∈ N.
The range of f(n) = {3, 5, 11, 13}.

Q.2 Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(ii) f (x) = x² + 2, x is a real number.
(iii) f (x) = x, x is a real number.

Answer:

(i) Since, x > 0
So, – 3x < 0
2 – 3x < 2
⇒ f(x) < 2
Therefore, Range of f = (–∞, 2)

$\begin{array}{l}\left(\text{ii}\right)\text{f}\left(\text{x}\right)={\text{x}}^2+2,\text{\hspace{0.33em}x is a real number.}\\ \text{Let \hspace{0.33em}y}={\text{x}}^2+2\\ \Rightarrow {\text{x}}^2=\text{y}-2\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}x}=\sqrt{\text{y}-2}\\ \mathrm{Since},\text{x is a real number.}\\ \text{So, y}-\text{2}\ge 0\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\ge 2\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[2,0\right)\\ \mathrm{Therefore},\text{the range of function}\mathrm{is}\left[2,0\right).\\ \left(\mathrm{iii}\right)\text{f}\left(\text{x}\right)=\text{x},\text{x is a real number.}\\ \text{x is real number}\Rightarrow \text{f}\left(\text{x}\right)\text{is also a real number.}\\ \text{So, range of function}=\text{R}.\end{array}$

Q.3 Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.

Answer:

Given, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) B ∩ C = {}
= ∅
A × (B ∩ C)
= A × ∅
= ∅
A × B = {(1, 1), (1, 2),(1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
(A × B) ∩ (A × C)
= ∅
L.H.S.= R.H.S. Hence proved.

(ii) Here,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5),
(2, 6), (2, 7), (2, 8), (3, 5), (3, 6),
(3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Since,
(A × C) ∩ (B × D) = {(1, 5), (1, 6), (2, 5), (2, 6)}
= (A × C)
So, (A × C) is a subset of (B × D).

Q.4 State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ ∅) = ∅ .

Answer:

(i) False,
Since, P = {m, n} and Q = {n, m}
P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True.

Q.5 Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

Answer:

$\begin{array}{l}\text{f}=\left\{\left(\mathrm{ab},\text{a}+\text{b}\right):\text{a},\text{b}\in \text{Z}\right\}\\ \mathrm{This}\text{can be written as:}\\ \text{f}\left(\mathrm{ab}\right)=\text{a}+\text{b},\forall \text{a,b}\in \text{Z}\\ \mathrm{Substituting}\text{a}=2,-2\text{and b}=3,-3\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(2\times 3\right)=2+3\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(6\right)=5\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(-2\times -3\right)=\left(-2\right)+\left(-3\right)\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(6\right)=-5\\ \mathrm{Since},\text{element 6 has two different images 5 and}-\text{5.}\\ \text{So, relation f is not a function because elements have no}\\ \text{unique images.}\end{array}$

Q.6 Let A = {1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?

(i) f is a relation from A to B
(ii) f is a function from A to B.
Justify your answer in each case.

Answer:

Since, A x B ={(1, 1), (1, 5), (1, 9), (1, 11),
(1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11),
(2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11),
(3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11),
(4, 15), (4, 16)}

(i) True.
Since, f is a subset of A x B, so, f is a relation.

(ii) False.
Since, image of 2 are 9 and 11 i.e., each element has not unique image. So, this relation is not a function.

Q.7

$\begin{array}{l}\text{Let R be a relation from N to N defined by}\\ \text{R = {(a, b): a, b}\in {\text{N and a = b}}^{\text{2}}\text{}.}\\ \text{Are the following true?}\\ \text{(i) (a,a)}\in \text{R, for all a}\in \text{N}\\ \text{(ii) (a,b)}\in \text{R, implies (b,a)}\in \text{R}\\ \text{(iii) (a,b)}\in \text{R, (b,c)}\in \text{R implies (a,c)}\in \text{R.}\end{array}$

Answer:

$\begin{array}{l}\begin{array}{l}\mathrm{Since},\text{\hspace{0.33em}R}=\left\{\left(\text{a},\text{b}\right):\text{a},\text{b}\in \mathrm{Nanda}={\text{b}}^2\right\}\\ \left(\text{i}\right)\mathrm{Since},3\in \text{N}\Rightarrow 3\ne 3^2=9\\ \mathrm{So},\left(\text{a},\text{a}\right)\in \text{R},\mathrm{foralla}\in \text{N is not true.}\\ \left(\mathrm{ii}\right)\mathrm{No},\left(\text{a},\text{b}\right)\in \text{R},\mathrm{implies}\left(\text{b},\text{a}\right)\in \text{R not true.}\end{array}\\ \begin{array}{l}\text{Because, \hspace{0.33em}if\hspace{0.33em}}\left(4,2\right)\in \text{R}\Rightarrow 4=2^2\\ \mathrm{and}\left(2,4\right)\in \text{R}\Rightarrow 2\ne 4^2=16\\ \mathrm{So},\left(\text{a},\text{b}\right)\in \text{R},\mathrm{implies}\left(\text{b},\text{a}\right)\in \text{R not true.}\\ \left(\mathrm{iii}\right)\mathrm{No},\left(\text{a},\text{b}\right)\in \text{R},\left(\text{b},\text{c}\right)\in \mathrm{Rimplies}\left(\text{a},\text{c}\right)\in \text{R is not true}.\end{array}\\ \begin{array}{l}\mathrm{Because},\\ \left(25,5\right)\in \text{R},\left(36,6\right)\in \text{R as}25,5,36,6\in \text{N}\end{array}\\ \begin{array}{l}\text{and\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}25}=5^2,36=6^2.\\ 25\ne 36\Rightarrow \left(25,36\right)\notin \text{R}\\ \mathrm{So},\left(\text{a},\text{b}\right)\in \text{R},\left(\text{b},\text{c}\right)\in \mathrm{Rimplies}\left(\text{a},\text{c}\right)\in \text{R is not true}.\end{array}\end{array}$

Q.8

$\begin{array}{l}\text{Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from}\\ \text{Z to Z defined by f(x) = ax + b, for some integers a, b.}\\ \text{Determine a, b.}\end{array}$

Answer:

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}\\ \mathrm{and}\mathrm{f}=\left\{\right(1,1),(2,3),(0,-1),(-1,-3\left)\right\}\\ \mathrm{Then},\\ \mathrm{f}\left(1\right)=1\Rightarrow \mathrm{a}\left(1\right)+\mathrm{b}=1\\ \Rightarrow \mathrm{a}+\mathrm{b}=1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{f}\left(0\right)=-1\Rightarrow \mathrm{a}\left(0\right)+\mathrm{b}=-1\\ \Rightarrow \mathrm{b}=-1\\ \mathrm{Substituting}\text{value of b in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{a}+(-1)=1\\ \mathrm{a}=1+1\\ \mathrm{a}=2\\ \mathrm{Thus},\text{a}=\text{2 and}\mathrm{b}=-1.\end{array}$

Q.9

$\begin{array}{l}\text{Let f, g : R}\mathrm{\to }\text{R be defined, respectively by f(x) = x + 1,}\\ \text{g(x) = 2x – 3. Find f + g, f – g and}\frac{\text{f}}{\text{g}}\text{.}\end{array}$

Answer:

(f + g)(x) = f(x) + g(x)
= x + 1 + 2x – 3
= 3x – 2
(f – g)(x) = f(x) – g(x)
= (x + 1) – (2x – 3)
= – x + 4

$\begin{array}{l}\left(\frac{\text{f}}{\text{g}}\right)\left(\text{x}\right)=\frac{\text{f}\left(\text{x}\right)}{\text{g}\left(\text{x}\right)}\\ =\frac{\text{x}+1}{2\text{x}-3},\text{where x}\ne \frac{3}{2}\end{array}$

Q.10

$\begin{array}{l}\text{Let f =}\left\{\left(\text{x,\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}}{{\text{1+x}}^{\text{2}}}\right)\text{: \hspace{0.17em}x}\in \text{R}\right\}\text{be a function from R to R.}\\ \text{Determine the range of f.}\end{array}$

Answer:

$\begin{array}{l}\begin{array}{l}\text{f}=\left\{\left(\text{x},\frac{{\text{x}}^2}{1+{\text{x}}^2}\right):\text{\hspace{0.33em}x}\in \text{R}\right\}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{from}\mathrm{R}\mathrm{to}\mathrm{R}\\ \Rightarrow \text{f}:\text{R}\to \text{R and f}\left(\text{x}\right)=\frac{{\text{x}}^2}{1+{\text{x}}^2},1+{\text{x}}^2\ne 0\\ \mathrm{Range}\text{of f:}\\ \text{Let\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}=\frac{{\text{x}}^2}{1+{\text{x}}^2}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\left(1+{\text{x}}^2\right)={\text{x}}^2\end{array}\\ \begin{array}{l}\Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}+{\mathrm{yx}}^2={\text{x}}^2\\ \Rightarrow \left(\text{y}-1\right){\text{x}}^2+\text{y}=0\end{array}\\ \begin{array}{l}\Rightarrow {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}}^2=\frac{\text{y}}{1-\text{y}}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}=\pm \sqrt{\frac{\text{y}}{1-\text{y}}}\\ \mathrm{Since},\text{x is a real number. So,}\end{array}\\ \begin{array}{l}\frac{\text{y}}{1-\text{y}}\ge 0\\ \Rightarrow \frac{\text{y}}{\text{y}-1}\le 0\\ \Rightarrow \frac{\text{y}-1}{\text{y}-1}\le 0\\ \Rightarrow 0\le \text{y}<1\\ \Rightarrow \text{y}\in \left[0,1\right)\\ \mathrm{Thus},\text{range of function is [0,1).}\end{array}\end{array}$

Q.11

$\text{Find the domain and the range of the real function f defined by f (x) =}\left|\text{x –\hspace{0.17em}1}\right|\text{.}$

Answer:

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{f}\left(\mathrm{x}\right)=|\mathrm{x}-1|\\ \mathrm{Here},\text{f}\left(\mathrm{x}\right)\text{is well defined for x}\in \text{R.}\\ \text{So, Domain of function is R.}\\ \mathrm{Now},|\mathrm{x}-1|\ge 0\text{for all x}\in \mathrm{R}\\ \Rightarrow 0\le |\mathrm{x}-1|\le \mathrm{\infty }\text{\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ \Rightarrow 0\le \text{\hspace{0.17em}f}\left(\mathrm{x}\right)\le \mathrm{\infty }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)\in [0,\mathrm{\infty })\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ \mathrm{Hence},\text{range of function}=[0,\mathrm{\infty }).\end{array}$

Q.12

$\text{Find the domain and the range of the real function f defined by (x) =}\sqrt{\left(\text{x – 1}\right)}\text{.}$

Answer:

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{(\mathrm{x}-1)}\\ \mathrm{Since},\text{f}\left(\mathrm{x}\right)\text{is a real number.}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}-1\ge 0\\ \Rightarrow \mathrm{x}\ge 1\\ \mathrm{So},\text{domain of function is [1,}\mathrm{\infty }\text{).}\\ \text{Since,}\sqrt{(\mathrm{x}-1)}\ge 0\text{for x}\ge 1.\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)\ge 0\text{for x}\ge 1\\ \mathrm{Thus},\text{the range of function is [0,}\mathrm{\infty }\text{).}\end{array}$

Q.13

$\text{Find the domain of the function f(x) =}\frac{{\text{x}}^{\text{2}}\text{+2x+1}}{{\text{x}}^{\text{2}}\text{– 8x+12}}\text{.}$

Answer:

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{f}\left(\text{x}\right)=\frac{{\text{x}}^2+2\text{x}+1}{{\text{x}}^2-8\text{x}+12}\\ \mathrm{Here},{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}}^2-8\text{x}+12=\left(\text{x}-2\right)\left(\text{x}-6\right)\\ \mathrm{So},\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=\frac{{\text{x}}^2+2\text{x}+1}{\left(\text{x}-2\right)\left(\text{x}-6\right)}\\ \mathrm{Here},\text{the function f}\left(\text{x}\right)\text{is defined for all the real values}\\ \text{except 2 and 6. So, the domain of f}\left(\text{x}\right)\text{is R}-\left\{2,6\right\}.\end{array}$

Q.14

$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2,\mathrm{find}\frac{\mathrm{f}\left(1.1\right)–\mathrm{f}\left(1\right)}{\left(1.1–1\right)}.$

Answer:

$\begin{array}{l}\mathrm{Since},\text{f}\left(\text{x}\right)={\text{x}}^2\\ \frac{\text{f}\left(1.1\right)-\text{f}\left(1\right)}{\left(1.1-1\right)}=\frac{{\left(1.1\right)}^2-{\left(1\right)}^2}{0.1}\\ =\frac{\left(1.1-1\right)\left(1.1+1\right)}{0.1}\\ =\frac{0.1\times 2.1}{0.1}\\ =2.1\\ \therefore \frac{\text{f}\left(1.1\right)-\text{f}\left(1\right)}{\left(1.1-1\right)}=2.1\end{array}$

Q.15

$\begin{array}{l}\text{The function ‘t’ which maps temperature in degree Celsius\hspace{0.33em}}\mathrm{into}\\ \text{temperature in degree Fahrenheit is defined by t}\left(\text{C}\right)=\frac{\text{9C}}{\text{5}}\text{+32.}\\ \text{Find}\\ \text{(i) \hspace{0.33em}t(0)}\\ \text{(ii)\hspace{0.33em}t(28)}\\ \text{(iii)\hspace{0.33em}t(–10)}\\ \text{(iv)\hspace{0.33em}The value of C, when t(C) = 212.}\end{array}$

Answer:

$\begin{array}{l}\begin{array}{l}\mathrm{The}\text{given function is:}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(\text{C}\right)=\frac{9\text{C}}{5}+32\\ \left(\text{i}\right)\text{\hspace{0.33em}Putting C}=0\text{in t}\left(\text{C}\right),\text{we get}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(0\right)=\frac{9\left(0\right)}{5}+32\\ =32\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}Putting C}=28\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(28\right)=\frac{9\left(28\right)}{5}+32\\ =82.4\\ \left(\mathrm{iii}\right)\text{\hspace{0.33em}Putting C}=-10\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(-10\right)=\frac{9\left(-10\right)}{5}+32\\ \begin{array}{l}=14\\ \left(\mathrm{iv}\right)\text{\hspace{0.33em}t}\left(\text{C}\right)=212\\ \frac{9\text{C}}{5}+32=212\\ \frac{9\text{C}}{5}=212-32\end{array}\\ \begin{array}{l}=180\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}C}=180\times \frac{5}{9}\\ =100\\ \mathrm{Thus},\text{the value of t is 100 when t}\left(\text{C}\right)=212.\end{array}\end{array}$

Q.16 A function f is defined by f(x) = 2x –5. Write down the values of

(i) f (0),
(ii) f (7),
(iii) f (–3).

Answer:

f(x) = 2x –5

(i) Substituting x = 0 in f(x), we get
f(0) = 2(0) – 5
= – 5

(ii) Substituting x = 7 in f(x), we get
f(7) = 2(7) – 5
= 9

(iii) Substituting x = – 3 in f(x), we get
f(– 3) = 2(– 3) – 5
= – 11

Q.17

$\begin{array}{l}\text{Find the domain and range of the following real functions:}\\ \begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}f}\left(\text{x}\right)\text{= –}\left|\text{x}\right|\\ \left(\text{ii}\right)\text{\hspace{0.33em}f}\left(\text{x}\right)\text{=}\sqrt{\text{9}-{\text{x}}^{\text{2}}}\end{array}\end{array}$

Answer:

$\begin{array}{l}\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=-\left|\text{x}\right|,\forall \text{x}\in \text{R}\\ \mathrm{Since},\left|\text{x}\right|=\{\begin{array}{l}\text{x},\text{\hspace{0.33em}x}\ge 0\\ -\text{x},\text{\hspace{0.33em}x}<0\end{array}\\ \mathrm{So},\text{\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=-\left|\text{x}\right|=\{\begin{array}{l}-\text{x},\text{\hspace{0.33em}x}\ge 0\\ \text{\hspace{0.33em}x},\text{\hspace{0.33em}x}<0\end{array}\\ \mathrm{Since},\text{x}\in \text{R},\text{so domain of f}\left(\text{x}\right)\text{is R.}\\ \mathrm{Here},-\left|\text{x}\right|\text{is negative number for all real number, R.}\end{array}\\ \text{Therefore, Range of f}\left(\text{x}\right)\text{is (}-\infty ,\text{0].}\\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{We}\text{have,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=\sqrt{9-{\text{x}}^2}\\ \text{f}\left(\text{x}\right)\text{is defined for all x satisfying}\end{array}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}9}-{\text{x}}^{\text{2}}\ge 0\\ \Rightarrow {\text{x}}^2-9\le 0\\ \Rightarrow \left(\text{x}-3\right)\left(\text{x}+3\right)\le 0\end{array}\\ \begin{array}{l}\Rightarrow -3\le \text{x}\le 3\\ \Rightarrow \text{\hspace{0.33em}x}\in \left[-3,3\right]\\ \mathrm{Hence},\text{Domain of f}\left(\text{x}\right)=\left[-3,3\right]\end{array}\\ \mathrm{Let}\text{y}=\text{f}\left(\text{x}\right).\mathrm{Then},\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}=\sqrt{9-{\text{x}}^2}\\ \Rightarrow {\text{\hspace{0.33em}y}}^2=9-{\text{x}}^2\\ \Rightarrow {\text{\hspace{0.33em}x}}^2=9-{\text{y}}^2\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}x}=\sqrt{9-{\text{y}}^2}\end{array}\\ \mathrm{It}\text{is clear that x will take real values, if}\\ \begin{array}{l}9-{\text{y}}^2\ge 0\\ \Rightarrow {\text{\hspace{0.33em}y}}^2-9\le 0\\ \Rightarrow \left(\text{y}-3\right)\left(\text{y}+3\right)\le 0\end{array}\\ \Rightarrow -3\le \text{y}\le 3\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[-3,3\right]\\ \mathrm{Since},\text{y}=\sqrt{9-{\text{x}}^2}\ge 0\text{for all x}\in \left[-3,3\right]\\ \therefore \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[0,3\right]\text{for all x}\in \left[-3,3\right]\\ \mathrm{Hence},\text{range of f}\left(\text{x}\right)=\left[0,3\right]\end{array}$

Q.18 Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1, 3), (1, 5), (2, 5)}.

Answer:

(i) Given relation is a function, because there is no two different images of same elements.
Domain = {2, 5, 8, 11, 14, 17} and Range = {1}

(ii) Given relation is a function, because each element has unique image.
Domain = {2, 4, 6, 8, 10, 12, 14} and
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) The given relation is not a function, because here one element has two different images,
i.e., 2 has two images 3 and 5.

Q.19 Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R

Answer:

R = {(a, b): a, b ∈ Z, a – b is an integer}
Since, a and b both are integer Z and their difference is also an integer.
Therefore, domain of R = Z
and Range of R = Z.

Q.20 Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer:

Since, A = {x, y, z} and B = {1, 2} So, A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
n(A × B) = 6
The number of subsets of A × B is 2⁶. Therefore, number of relations from A into B will be 2⁶.

Q.21 Write the relation R = {(x, x³): x is a prime number less than 10} in roster form.

Answer:

R = {(x, x³): x is a prime number less than 10}
= {(x, x³): x = 2, 3, 5, 7}
= {(2, 8), (3, 27), (5, 125), (7, 343)}

Q.22 Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

Answer:

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}} = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Q.23 Let A = {1, 2, 3, 4, 6}. Let R be the relation on
A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.

(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.

Answer:

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3) , (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4 6}
(iii) Range of R = {1, 2, 3, 4, 6}

Q.24 The Fig 2.7 shows a relationship between the sets P and Q. Write this relation.
(i) in set-builder form (ii) roster form.
What is its domain and range?

Answer:

Here, P = {5, 6, 7}, Q = {3, 4, 5}
(i) The set-builder form is R = {(x, y): y = x – 2; for x ∈P}
(ii) In roster form:
R = {(5, 3), (6, 4), (7, 5)}
Domain = {5, 6, 7}, Range = {3, 4, 5}

Q.25 A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Answer:

Here, A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
= {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Which is a roster form.

Q.26 Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

Answer:

R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈N} = {(1, 6), (2, 7), (3, 8)}
Domain of R = {1, 2, 3}
Range of R = {6, 7, 8}

Q.27 Let A = {1, 2, 3,…, 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

Answer:

We have:
A = {1, 2, 3,…, 14}
Relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain of Relation R = {1, 2, 3, 4}
Codomain = A
Range of relation R = {3, 6, 9, 12}

Q.28 The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1).
Find the set A and the remaining elements of A × A.

Answer:

Number of elements in the cartesian product A × A = 9
Some given elements of A × A are (–1, 0) and (0, 1).
Let number of elements in set A, n(A) = 3
and number of elements in set A × A = 9
i.e., n(A × A) = n(A) × n(A)
9 = { n(A)}²
n(A) = 3
So, A = {–1, 0, 1}
Then, A × A = {(–1, –1), (–1, 0), (–1, 1), (0, –1,), (0, 0), (0,1), (1, –1,), (1, 0), (1, 1)}
Remaining elements of A × A are: (–1, –1), (–1, 1), (0, –1,), (0, 0), (1, –1,), (1, 0), (1, 1).

Q.29 Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

Answer:

Number of elements in set A = 3
Number of elements in set B = 2,
Some elements of A × B are (x, 1), (y, 2), (z, 1).
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B} So, A = {x, y, z} and B = {1, 2}.

Q.30 If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Answer:

Given, A × B = {(a, x),(a , y), (b, x), (b, y)}
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}.
Then, A = {a, b}, B = {x, y}.

Q.31 If A = {–1, 1}, find A × A × A.

Answer:

Since, A = {–1, 1}
A × A = {(–1, –1), (–1, 1,), (1,–1), (1, 1)}
A × A × A = {–1, 1} × {(–1, –1), (–1, 1,), (1,–1), (1, 1)}
= {(1,–1, –1), (–1,–1, 1,), (–1,1,–1), (–1,1, 1), (1, –1, –1), (1,–1, 1,), (1, 1,–1), (1, 1, 1)}

Q.32 If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Answer:

$\begin{array}{l}\text{Given:\hspace{0.33em}G}=\left\{\text{7},\text{8}\right\}\text{and H}=\left\{\text{5},\text{4},\text{2}\right\},\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}G}\times \text{H}=\left\{\left(7,5\right),\left(7,4\right),\left(7,2\right),\left(8,5\right),\left(8,4\right),\left(8,2\right)\right\}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}H}\times \text{G}=\left\{\left(5,7\right),\left(5,8\right),\left(4,7\right),\left(4,8\right),\left(2,7\right),\left(2,8\right)\right\}\end{array}$

Q.33 If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).

Answer:

$\begin{array}{l}\mathrm{Number}\text{of elements in set A}=3\\ \text{i}.\text{e}.,\text{\hspace{0.33em}n}\left(\text{A}\right)=3\\ \mathrm{Elements}\text{in set B}=\left\{3,4,5\right\}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}\left(\text{B}\right)=3\\ \mathrm{So},\\ \text{Number of elements in A}\times \text{B}=\text{n}\left(\text{A}\right)\times \text{n}\left(\text{B}\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}\left(\text{A}\times \text{B}\right)=3\times 3\\ =9\\ \mathrm{Thus},\text{\hspace{0.33em}number\hspace{0.33em}of\hspace{0.33em}elements\hspace{0.33em}in\hspace{0.33em}A}\times \text{B\hspace{0.33em}is\hspace{0.33em}9.}\end{array}$

Q.34

$\text{If\hspace{0.33em}}\left(\frac{\text{x}}{\text{3}}\text{+1, y-}\frac{\text{2}}{\text{3}}\right)\text{=}\left(\frac{\text{5}}{\text{3}}\text{,}\frac{\text{1}}{\text{3}}\right)\text{, find the values of x and y.}$

Answer:

$\begin{array}{l}\mathrm{Given}:\\ (\frac{\mathrm{x}}{3}+1,\mathrm{y}-\frac{2}{3})=(\frac{5}{3},\frac{1}{3})\\ \mathrm{Comparing}\text{both sides, we get}\\ \frac{\mathrm{x}}{3}+1=\frac{5}{3}\Rightarrow \frac{\mathrm{x}}{3}=\frac{5}{3}-1\\ \frac{\mathrm{x}}{3}=\frac{2}{3}\Rightarrow \mathrm{x}=2\\ \mathrm{and}\mathrm{y}-\frac{2}{3}=\frac{1}{3}\Rightarrow \mathrm{y}=\frac{1}{3}+\frac{2}{3}\\ \mathrm{y}=\frac{3}{3}=1\\ \mathrm{Thus},\text{x}=2\text{and y}=\text{1.}\end{array}$