CBSE Class 11 Maths Revision Notes Chapter 2

Q.1 If P = {2, 3}, form the set P × P × P.

Ans

P × P × P = {(2, 2, 2), (2, 2, 3), (2, 3, 2), (2, 3, 3), (3, 2, 2), (3, 2, 3), (3, 3, 2), (3, 3, 3)}

Q.2  In given figure write the relation in

(i) Set–builder from (ii) Roster form
Also write the domain and the range.

Ans

Q.3 a) If (x – 2, y + 3) = (4, 1), find the values of x and y.
b) If set A = {2, 3} and B = {3, 4}, find the number of elements in A × B.

Ans

a) Since ordered pairs are equal, so
x – 2 = 4 and y + 3 = 1
x = 4 + 2 = 6 and y = 1 – 3 = –2

b) n(A × B) = n(A) × n(B) = 2 × 2 = 4 elements

Q.4 Find the range:

a) f = {(x, x²): x ∈ R)} be a function from R into R.
b) A = {9, 10, 11, 13, 17, 18} and f : A → N defined by f(x) = highest prime factor of x.

Ans

a) Range : set of all positive real numbers.
b) Factors of 9 = 3 × 3 ⇒ highest prime factor = 3
Factors of 10 = 2 × 5 ⇒ highest prime factor = 5
Factors of 11 = 1 × 11 ⇒ highest prime factor = 11
Factors of 13 = 1 × 13 ⇒ highest prime factor = 13
Factors of 18 = 2 × 3 × 3 ⇒ highest prime factor = 3
∴ range = {3, 5, 11, 13}

Q.5 Let R be a relation from N to N defined by R = {(x, y): x, y ∈ N and y = x²}.
Are the following statements true? Give reason also.

(i) (x, x) ∈ R ∀ x ∈ N
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ N
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ N
(iv) Is the relation a function?

Ans

i) No, as only (1, 1) ∈ R, (2, 2), (3, 3), …, ∉ R.
ii) No, (2, 4) ∈ R but (4, 2) ∉ R.
iii) No, as (2, 4) ∈ R and (4, 16) ∈ R but (2, 16) ∉ R.
iv) Yes, as all elements in the domain have a unique image in the co-domain.

Q.6 Find the domain of the function

$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2–7\mathrm{x}+12}.$

Ans

Since, x² – 7x +12 = (x – 4)(x – 3), the function is defined for all real numbers
except x = 4 and x = 3. Hence, the domain of f is R – {3, 4}.

Q.7 A function f is defined by f(x) = x² – 5. Find the value of

(i) f(0).
(ii) f(–3).

$\left(\mathrm{iii}\right)\mathrm{f}\left(\frac{2}{3}\right).$

Ans

(i) f(x) = x² – 5 ⇒ f(0) = 0² – 5 = –5
(ii) f(–3) = (–3)² – 5 = 9 – 5 = 4

$\left(\mathrm{iii}\right)\mathrm{f}\left(\frac{2}{3}\right)={\left(\frac{2}{3}\right)}^2–5=\frac{4}{9}–5=\frac{4–45}{9}=\frac{–41}{9}$

Q.8 Let A = {2, 3, 4}, B = {4, 5}, C = {5, 6, 7}. Find (A × B) ∩ (A × C).

Ans

A × B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}
A × C = {(2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)}
(A × B) ∩ (A × C) = {(2, 5), (3, 5), (4, 5)}

Q.9 Let A = {1, 2, 6, 8} and let R be a relation on A defined by
{(a, b): a, b ∈ A, b is exactly divisible by a}

a) Write R in roster form.
b) Find the domain of R.
c) Find the range of R.

Ans

Q.10 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)}

a) Is f a relation from A to B? Why?
b) Is f a function from A to B? Why?

Ans

a) Yes, f is a relation because every element of f is a subset of A × B.
b) No, because the element 2 of A has two image in B, i.e., 9 and 11.
The mapping is not unique.

Q.11 Let f, g: R → R defined by f(x) = x – 1, g(x) = 2x – 5. Find

a) f – g.
b) f/g.

Ans

a) (f – g)x = f(x) – g(x)
= x – 1 – (2x – 5)
= x – 1 –2x + 5
= 4 – x

b) (f/g)x = f(x)/g(x) = (x – 1)/(2x – 5), where x ≠ 5/2.

Q.12

$\text{If\hspace{0.33em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2,\text{\hspace{0.33em}find\hspace{0.33em}}\frac{\mathrm{f}\left(2.1\right)–\mathrm{f}\left(1\right)}{2.1–1}$

Ans

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2 \\ \therefore \mathrm{f}\left(2,1\right)={\left(2.1\right)}^2=4.41 \\ \mathrm{f}\left(1\right)=1^2=1 \\ \text{On putting these values, we get} \\ \frac{\mathrm{f}\left(2.1\right)–\mathrm{f}\left(1\right)}{2.1–1}=\frac{4.41–1}{1.1}=\frac{341}{110} \end{gathered}$$

Q.13 Determine the domain and range of the relation R defined by R = {(x, x²): x ∈ (1, 2, 3, 4)}.

Ans

R = {(1, 1), (2, 4), (3, 9), (4, 16)}
Since domain is the set of first components in R and range is the set of second components in R .
Therefore,
Domain = {1, 2, 3, 4}
Range = {1, 4, 9, 16}

Q.14 If A = {–1, 0}, find A × A × A.

Ans

A × A × A = {(–1, –1, –1), (–1, –1, 0), (–1, 0, –1), (–1, 0, 0), (0, –1, –1), (0, –1, 0), (0, 0, –1), (0, 0, 0)}

Q.15 Let A = {1, 2, 3} and B = {1, 4}. Find A × B and B × A.

Ans

Q.16 Under which condition a relation from A to B is said to be a function?

Ans

If every element of A has one and only one Image in B, then f is a function from A to B.

Q.17

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}function\hspace{0.33em}f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\text{x}}^{\text{3}}-{\text{6x}}^{\text{2}}\text{+11x}-\text{6}}$

Ans

$$\begin{gathered} \text{We know that all the values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \therefore \text{For function to be defined,} \\ {\mathrm{x}}^3–6{\mathrm{x}}^2+11\mathrm{x}–6\ge 0\dots \left(1\right) \\ \text{Now, for getting its factor} \\ \text{put}\mathrm{x}=–1,1,0\dots \\ \text{we find that}\mathrm{x}=1,\text{satisfies}\left(1\right) \\ {\text{Hence}}_{\mathrm{r}}\left(\mathrm{x}–1\right)\text{is one of its factors} \\ \text{Now, on dividing}\left(1\right)\text{by}\left(\mathrm{x}–1\right),\text{we get} \\ {\mathrm{x}}^2–5\mathrm{x}+6,\text{which is a quadratic poynomial} \\ \text{On splitting the middle term, we get} \\ \left(\mathrm{x}–2\right)\left(\mathrm{x}–3\right) \\ \text{Finally, we get} \\ {\mathrm{x}}^3–6{\mathrm{x}}^2+11\mathrm{x}–6=\left(\mathrm{x}–1\right)\left(\mathrm{x}–2\right)\left(\mathrm{x}–3\right) \\ \text{eqn (1) becomes} \\ \left(\mathrm{x}–1\right)\left(\mathrm{x}–2\right)\left(\mathrm{x}–3\right)\ge 0 \\ \text{Using number line, we get} \end{gathered}$$

$$\begin{gathered} \text{Expression is +ve in the interval}[1,2]\text{and}[3,+\infty ), \\ \text{Hence, the domain of the given function is} \\ [1,2]\cup [3,+\infty ). \end{gathered}$$

Q.18 Define a relation.

Ans

Let A and B be two sets. Then a relation from A to B is a subset of A × B, i.e., R ⊆ A × B.

Q.19 If R is a relation defined by R = {(x, 2x): x ∈ (1, 2, 3, 4)}, find the range and domain of R.

Ans

Range of R = (2, 4, 6, 8)
and Domain of R = (1, 2, 3, 4)

Q.20 If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Define a relation on a set A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Show this relationship using arrow diagram.

Ans

R = {(1, 3), (2, 6), (3, 9)}

Q.21 If n(A) = 3 and n(B) = 3, then find n(A × B).

Ans

n(A × B) = n(A) × n(B) = 9

Q.22

$\text{If a relation\hspace{0.33em}}\mathrm{f}\text{\hspace{0.33em}is defined as}\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3,\text{then find the value of}\mathrm{f}\left(\mathrm{f}\left(\frac{3}{2}\right)\right)$

Ans

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3 \\ \mathrm{f}\left(\frac{3}{2}\right)=2\times \frac{3}{2}+3=6 \\ \mathrm{f}\left(\mathrm{f}\left(\frac{3}{2}\right)\right)=\mathrm{f}\left(6\right)=2\times 6+3=15 \end{gathered}$$

Q.23 Let f, g: R → R defined by f(x) = 6x and g(x) = 3x. Find f – g.

Ans

(f – g)(x) = f(x) – g(x) = 6x – 3x = 3x

Q.24 Let A = {a, b, c, d}, B = {e, f, g, h, i, j} and f = {(a, e), (b, f), (c, g), (d, h), (a, i)}. Check whether f is a function from A to B or not.

Ans

No. f is not a function from A to B since element ‘a’ of set A has more than one image in set B.

Q.25

$$\begin{gathered} \text{Let\hspace{0.33em}}\mathrm{f}\text{\hspace{0.33em}and\hspace{0.33em}}\mathrm{g}\text{\hspace{0.33em}be real function defined by\hspace{0.33em}}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+2} \\ \text{and\hspace{0.33em}}\mathrm{g}\left(\mathrm{x}\right)=\sqrt{4–{\mathrm{x}}^2}.\text{Then find}\frac{\mathrm{f}}{\mathrm{g}}\text{and}\mathrm{ff}\text{.} \end{gathered}$$

Ans

$$\begin{gathered} \text{a)}\frac{\mathrm{f}}{\mathrm{g}}=\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\sqrt{\mathrm{x}+2}}{\sqrt{4–{\mathrm{x}}^2}}=\frac{1}{\sqrt{2–\mathrm{x}}} \\ \text{b)}\mathrm{ff}=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{x}\right)=[\mathrm{f}\left(\mathrm{x}\right){]}^2={\left(\sqrt{\mathrm{x}+2}\right)}^2=\mathrm{x}+2 \end{gathered}$$

Q.26 If P = {a, b} and Q = {2, 3}, find P × Q.

Ans

P × Q = {(a, 2), (a, 3), (b, 2), (b, 3)}

Q.27 Given that f: X → Y, the image of x is f(x). Write the preimage of y.

Ans

Preimage of y is f^–1(y) = {x: f(x) = y}, or all x whose image is y.

Q.28 Define the signum function.

Ans

$$\begin{gathered} \text{The function}\mathrm{f}:\mathrm{R}\to \mathrm{R}\text{defined by}\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}1& \mathrm{if}\mathrm{x}>0\\ 0& \mathrm{if}\mathrm{x}=0\\ –1& \mathrm{if}\mathrm{x}<0\end{array}\right. \\ \mathrm{is}\mathrm{called}\mathrm{signum}\mathrm{function}. \end{gathered}$$

Q.29

$\text{Find the domain of}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{4–{\mathrm{x}}^2}.$

Ans

f(x) is defined for 4 – x² ≥ 0
or x² – 4 ≤ 0
or (x – 2)(x + 2) ≤ 0
Therefore, Domain of function = {x: – 2 ≤ x ≤ 2}

Q.30 The cartesian product of A × A has 9 elements among which (–1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.

Ans

A = {–1, 0, 1}
The remaining elements of A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1).

Q.31 Find the range of function, f(x) = |x|.

Ans

Since function, f(x) = |x| is defined for all the real values of x ∈ R and value of y = f(x) = |x| is always a positive real number ∀ x ∈ R.
Hence, its range is the set of all positive real values, i.e.,
[0, ∞).

Q.32

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}function\hspace{0.33em}f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{x}}{\text{x}-\text{1}}$

Ans

$$\begin{gathered} \text{We know that all values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \because \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}–1} \\ \text{For function to be defined, denominator should be non-zero.} \\ \text{Here, function is not defined for}\mathrm{x}=1\text{.} \\ \text{Hence, its domain is all the real values of}\mathrm{x}\text{except}1,\text{i.e.,} \\ \mathrm{R}–\left\{1\right\}. \end{gathered}$$

Q.33

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}function\hspace{0.33em}f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{36}-{\text{x}}^{\text{2}}}\text{.}$

Ans

$$\begin{gathered} \text{We know that all the values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \because \mathrm{f}\left(\mathrm{x}\right)=\sqrt{36–{\mathrm{x}}^2} \\ \text{For function to be defined,} \\ 36–{\mathrm{x}}^2\ge 0 \\ \Rightarrow {\mathrm{x}}^2–36\le 0 \\ \Rightarrow {\mathrm{x}}^2–6^2\le 0 \\ \Rightarrow \left(\mathrm{x}–6\right)\left(\mathrm{x}+6\right)\le 0 \\ \Rightarrow \mathrm{x}\ge –6\text{and}\mathrm{x}\le 6 \\ \Rightarrow –6\le \mathrm{x}\le 6 \\ \Rightarrow \mathrm{X}\in [–6,6] \\ \text{Hence, the domain of the given function is}[–6,6]\text{.} \end{gathered}$$

Q.34 Find the domain of function, f(x) = |X|.

Ans

Since function, f(x) = |X| is defined for all values of x ∈ R.
Hence, its domain is the set of all real values of x, i.e., R or (– ∞, ∞).

Q.35 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).

Ans

Since f is a linear function,
Therefore, let us suppose f(x) = mx + c… (1)
Also, since (1, 1), (0, – 1) ∈ R,
⇒ f (1) = m + c = 1 …(2)
and f (0) = c = –1.
On putting the value of c in (2), we get
m – 1 = 1
⇒ m = 2
On putting the value of c and m in (1), we get
f(x) = 2x – 1.

Q.36 If (x + y, x – y) = (6, 2), find the values of x and y.

Ans

$\begin{array}{l}\text{Since the ordered pairs are equal, therefore,}\\ \text{the corresponding elements are also equal.}\\ \text{x}+\text{y}=6\dots \dots .\left(1\right)\text{}\\ \text{x}-\text{y}=2\dots \dots \dots \left(2\right)\text{}\\ \text{On adding equations (1) and (2), we get}\\ \text{x}+\cancel{\text{y}}=6\\ \underset{¯}{\text{x}-\cancel{\text{y}}=2}\\ 2\text{x}=8\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}X}=4\\ \text{On putting the value of x in}\left(1\right),\text{we get}\\ 4+\text{y}=6\text{}\\ \text{or y}=2\text{}\\ \text{Hence, x}=4,\text{y}=2\end{array}$

Q.37

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}function f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{x}}{\sqrt{\text{49}-{\text{x}}^{\text{2}}}}\text{.}$

Ans

$$\begin{gathered} \text{We know that all the values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\sqrt{49–{\mathrm{x}}^2}} \\ \text{For function to be defined,} \\ 49–{\mathrm{x}}^2>0 \\ \Rightarrow {\mathrm{x}}^2–49<0 \\ \Rightarrow {\mathrm{x}}^2–7^2<0 \\ \Rightarrow \left(\mathrm{x}–7\right)\left(\mathrm{x}+7\right)<0 \\ \Rightarrow \mathrm{x}>–7\text{and}\mathrm{x}<7 \\ \Rightarrow –7<\mathrm{x}<7 \\ \Rightarrow \mathrm{x}\in \left(–7,7\right) \\ \text{Hence, the domain of the given function is}\left(–7,7\right)\text{.} \end{gathered}$$

Q.38

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}function\hspace{0.33em}f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{{\text{3x}}^{\text{2}}\text{+7}}{\sqrt{{\text{x}}^{\text{2}}-\text{5x+6}}}\text{.}$

Ans

$$\begin{gathered} \text{We know that all the values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \because \mathrm{f}\left(\mathrm{x}\right)=\frac{3{\mathrm{x}}^2+7}{\sqrt{{\mathrm{x}}^2–5\mathrm{x}+6}} \\ \text{For function to be defined,} \\ {\mathrm{x}}^2–5\mathrm{x}+6>0 \\ \Rightarrow \left(\mathrm{x}–2\right)\left(\mathrm{x}–3\right)>0 \\ \Rightarrow \left(\mathrm{x}–2\right)>0\text{and}\left(\mathrm{x}–3\right)>0 \\ \Rightarrow \mathrm{x}>2\text{and}\mathrm{x}>3\Rightarrow \mathrm{x}>3\dots \left(1\right) \\ \Rightarrow \text{Also,}\left(\mathrm{x}–2\right)<0\text{and}\left(\mathrm{x}–3\right)<0 \\ \Rightarrow \mathrm{x}<2\text{and}\mathrm{x}<3\Rightarrow \mathrm{x}<2\text{…}\left(2\right) \\ \text{From}\left(1\right)\text{and,}\left(2\right)\text{get} \\ –\infty <\mathrm{x}<2\text{and}3<\mathrm{x}<\infty \\ \text{Hence, the domain of the given function is} \\ \left(–\infty ,2\right)\cup \left(3,\infty \right) \end{gathered}$$

Q.39

$$\begin{gathered} \text{Find the range of the function} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1–{\mathrm{x}}^2}{1+{\mathrm{x}}^2} \end{gathered}$$

Ans

$$\begin{gathered} \text{Since, the value of}\mathrm{y}\text{for which domain}\left(\mathrm{x}\right)\text{is defined is called range,} \\ \text{Let}\mathrm{y}=\frac{1–{\mathrm{x}}^2}{1+{\mathrm{x}}^2} \\ \Rightarrow \mathrm{y}+{\mathrm{yx}}^2=1–{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{x}}^2\mathrm{y}=1–\mathrm{y} \\ \Rightarrow \left(1+\mathrm{y}\right){\mathrm{x}}^2=1–\mathrm{y} \\ \Rightarrow {\mathrm{x}}^2=\frac{1–\mathrm{y}}{1+\mathrm{y}} \\ \Rightarrow \mathrm{x}=\sqrt{\frac{1–\mathrm{y}}{1+\mathrm{y}}} \\ \text{For domain}\left(\mathrm{x}\right)\text{to be defined,} \\ \frac{1–\mathrm{y}}{1+\mathrm{y}}\ge 0 \\ \text{On using number line, we get} \end{gathered}$$

$$\begin{gathered} –1<\mathrm{y}\le 1\left[\begin{array}{c}\because \text{denominator}\ne 0\\ \Rightarrow \left(1+\mathrm{y}\right)\ne 0\\ \Rightarrow \mathrm{y}\ne –1\end{array}\right] \\ \Rightarrow \mathrm{y}\in (–1,1] \\ \text{Hence, range is}(–1,1]. \end{gathered}$$

Q.40

$$\begin{gathered} \text{Find the range of the function} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1–\mathrm{x}}. \end{gathered}$$

Ans

$$\begin{gathered} \text{Since, the value of}\mathrm{y}\text{for which domain}\left(\mathrm{x}\right)\text{is defined is called range,} \\ \text{Let}\mathrm{y}=\frac{\mathrm{X}}{1–\mathrm{x}} \\ \Rightarrow \mathrm{x}=\mathrm{y}–\mathrm{yx} \\ \Rightarrow \mathrm{x}+\mathrm{xy}=\mathrm{y} \\ \Rightarrow \left(1+\mathrm{y}\right)\mathrm{x}=\mathrm{y} \\ \Rightarrow \mathrm{X}=\frac{\mathrm{y}}{1+\mathrm{y}} \\ \text{For domain}\left(\mathrm{x}\right)\text{to be defined, denominator}\ne 0 \\ \Rightarrow \left(1+\mathrm{y}\right)\ne 0 \\ \Rightarrow \mathrm{y}\ne –1 \\ \text{Hence, the range is set of all real numbers}\mathrm{R}\text{except}–1_1 \\ \text{i.e.,}\mathrm{R}–\{–1\}\text{.} \end{gathered}$$

Q.41

$\text{Find\hspace{0.33em}the\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}function\hspace{0.33em}f}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\text{x}}^{\text{2}}-\text{7x+10}}\text{.}$

Ans

$$\begin{gathered} \text{We know that all the values of}\mathrm{x}\text{for which function is defined is called domain.} \\ \because \mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2–7\mathrm{x}+10} \\ \text{For function to be defined,} \\ {\mathrm{x}}^2–7\mathrm{x}+10\ge 0 \\ \Rightarrow \left(\mathrm{x}–2\right)\left(\mathrm{x}–5\right)\ge 0 \\ \Rightarrow \left(\mathrm{x}–2\right)\ge 0\text{and}\left(\mathrm{x}–5\right)\ge 0 \\ \Rightarrow \mathrm{x}\ge 2\text{and}\mathrm{x}\ge 5\Rightarrow \mathrm{x}\ge 5\dots \left(1\right) \\ \Rightarrow \text{Also,}\left(\mathrm{x}–2\right)\le 0\text{and}\left(\mathrm{x}–5\right)\le 0 \\ \Rightarrow \mathrm{x}\le 2\text{and}\mathrm{x}\le 5\Rightarrow \mathrm{x}\le 2\dots \left(2\right) \\ \text{Froma}\left(1\right)\text{nd,}\left(2\right)\text{we get} \\ –\infty <\mathrm{x}\le 2\text{and}5\le \mathrm{x}<\infty \\ \text{Hence, the domain of the given function is} \\ \left(–\infty ,2]\cup [5,\infty \right). \end{gathered}$$

Q.42 If Set A has m elements and Set B has n elements, find the number of relations defined from Set A to Set B.

Ans

Since the number of elements in set A = m
Number of elements in set B = n
∴ Number of elements in set (A × B) = mn
The number of all possible subsets of Set (A × B) = 2^mn
⇒ The number of relations defined from Set A to B = 2^mn

Q.43 If Set A has m elements, find the number of all possible subsets of Set (A × A).

Ans

$$\begin{gathered} \text{Since, the number of elements in set}\mathrm{A}=\mathrm{m} \\ \Rightarrow \text{Number of elements in}set\left(\mathrm{A}\times \mathrm{A}\right)=\mathrm{m}·\mathrm{m}={\mathrm{m}}^2 \\ \text{The number of all possible subsets of}set\left(\mathrm{A}\times \mathrm{A}\right)=2^{{\mathrm{m}}^2} \end{gathered}$$

Q.44 If Set A has m elements and Set B has n elements, find the elements in Set (AXB).

Ans

Since the number of elements in set A = m
Number of elements in set B = n
∴ Number of elements in set (AXB) = mn

Q.45

$\begin{array}{l}\text{Express the following functions as a sets of ordered}\\ \text{pairs and determine their ranges}\\ \text{(a)\hspace{0.33em}f\hspace{0.33em}:\hspace{0.33em}A}\to \text{R, f}\left(\text{x}\right){\text{\hspace{0.33em}=\hspace{0.33em}x}}^{\text{2}}\text{+1, where\hspace{0.33em}A\hspace{0.33em}=\hspace{0.33em}}\left\{\text{-1,0,2,4}\right\}\text{}\\ \text{(b)\hspace{0.33em}g\hspace{0.33em}:\hspace{0.33em}A}\to \text{N, g}\left(\text{x}\right)\text{\hspace{0.33em}=\hspace{0.33em}2x, where\hspace{0.33em}A}=\left\{\text{x:x}\in \text{N,x}\le \text{10}\right\}\end{array}$

Ans

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{We}\mathrm{have}, \\ \mathrm{f}(–1)=(–1{)}^2+1=2, \\ \mathrm{f}\left(0\right)=0^2+1,\mathrm{f}(2)=2^2+1=\text{and}\mathrm{f}\left(4\right)=4^2+1=17 \\ \therefore \mathrm{f}=\{\mathrm{x},\mathrm{f}\left(\mathrm{x}\right)):\mathrm{x}\in \mathrm{A}\}=\left\{\right(–1,2),(0,1),(2,5),(4,17\left)\right\} \\ \mathrm{Hence},\text{Range of}\left(\mathrm{f}\right)=\{2,1,5,17\} \\ \left(\mathrm{b}\right)\text{We have,} \\ \mathrm{A}=\{1,2,3,\dots ,10\}\text{. Therefore,} \\ \mathrm{g}\left(1\right)=2\times 1=2,\mathrm{g}\left(2\right)=2\times 2=4,\mathrm{g}\left(3\right)=2\times 3=6, \\ \mathrm{g}\left(4\right)=2\times 4=8,\mathrm{g}\left(5\right)=2\times 5=10,\mathrm{g}\left(6\right)=2\times 6=12, \\ \mathrm{g}\left(7\right)=2\times 7=14,\mathrm{g}\left(8\right)=2\times 8=16,\mathrm{g}\left(9\right)=2\times 9=18, \\ \mathrm{g}\left(10\right)=2\times 10=20 \\ \therefore \mathrm{g}=\{\mathrm{x},\mathrm{g}\left(\mathrm{x}\right)):\mathrm{x}\in \mathrm{A}\}=\{(1,2),(2,4),(3,6),\dots ,(10,20)\} \\ \text{Range of}\mathrm{g}=\mathrm{g}(\mathrm{A})=\{\mathrm{g}\left(\mathrm{x}\right):\mathrm{x}\in \mathrm{A}\} \\ =\{2,4,6,8,10,12,14,16,18,20\} \end{gathered}$$

Q.46 Let f and g be real function defined by f(x) = 4x + 1 and g(x) = 6x – 7.

(a) For what real numbers x, f(x) = g(x)?
(b) For what real numbers x, f(x) < g(x)

Ans

$$\begin{gathered} \text{We have}\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+1\text{and}\mathrm{g}\left(\mathrm{x}\right)=6\mathrm{x}–7\sqrt{{\mathrm{b}}^2–4\mathrm{ac}} \\ \mathrm{Now}, \\ \left(\mathrm{a}\right)\text{For}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right),\text{we must have,} \\ 4\mathrm{x}+1=6\mathrm{x}–7 \\ \text{Solving above equation for}\mathrm{x},\text{we get,} \\ 4\mathrm{x}+1=6\mathrm{x}–7 \\ \Rightarrow 6\mathrm{x}–4\mathrm{x}=7+1 \\ \Rightarrow 2\mathrm{x}=8 \\ \Rightarrow \mathrm{x}=\frac{8}{2}=4 \\ \therefore \mathrm{x}=4 \\ \left(\mathrm{b}\right)\text{Now,} \\ \text{For}\mathrm{f}\left(\mathrm{x}\right)<\mathrm{g}\left(\mathrm{x}\right),\text{we must have,} \\ 4\mathrm{x}+1<6\mathrm{x}–7 \\ \text{Solving above inequality for}\mathrm{x},\text{we get} \\ 4\mathrm{x}+1<6\mathrm{x}–7 \\ \Rightarrow 2\mathrm{x}>8 \\ \Rightarrow \mathrm{x}>4 \\ \text{Hence, for}\mathrm{x}=4,\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\text{and,} \\ \text{for}\mathrm{x}>4,\mathrm{f}\left(\mathrm{x}\right)<\mathrm{g}\left(\mathrm{x}\right) \end{gathered}$$

Q.47

$\begin{array}{l}\text{If X\hspace{0.33em}=\hspace{0.33em}}\left\{\text{1,3,5}\right\}\text{,\hspace{0.33em}Y\hspace{0.33em}=\hspace{0.33em}}\left\{\text{a,b}\right\}\text{,\hspace{0.33em}then\hspace{0.33em}represent\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}products\hspace{0.33em}by\hspace{0.33em}arrow\hspace{0.33em}diagrams:}\\ \left(\text{i}\right)\text{X×Y}\\ \left(\text{ii}\right)\text{Y×X}\\ \left(\text{iii}\right)\text{X×X}\\ \left(\text{iv}\right)\text{Y×Y}\end{array}$

Ans

$$\begin{gathered} \left(\text{i}\right)\text{We have,} \\ \mathrm{X}\times \mathrm{Y}=\{1,3,5\}\times \{\mathrm{a},\mathrm{b}\} \\ =\left\{\right(1,\mathrm{a}),(1,\mathrm{b}),(3,\mathrm{a}),(3,\mathrm{b}),(5,\mathrm{a}),(5,\mathrm{b}\left)\right\} \\ \text{Following arrow diagram represent}\mathrm{X}\times \mathrm{Y}\text{.} \\ \left(\text{ii}\right)\text{We have,} \\ \mathrm{Y}\times \mathrm{X}=\{\mathrm{a},\mathrm{b}\}\times \{1,3,5\} \\ =\left\{\right(\mathrm{a},1),(\mathrm{a},3),(\mathrm{a},5),(\mathrm{b},1),(\mathrm{b},3),(\mathrm{b},5\left)\right\} \\ \text{Following arrow diagram represent}\mathrm{Y}\times \mathrm{X}\text{.} \\ \left(\text{iii}\right)\text{We have,} \\ \mathrm{X}\times \mathrm{X}=\{1,3,5\}\times \{1,3,5\} \\ =\left\{\right(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5\left)\right\} \\ \text{Following is the arrow diagram represent}\mathrm{X}\times \mathrm{X}\text{.} \\ \left(\text{iii}\right)\text{We have,} \\ \mathrm{Y}\times \mathrm{Y}=\{\mathrm{a},\mathrm{b}\}\times \{\mathrm{a},\mathrm{b}\} \\ =\left\{\right(\mathrm{a},\mathrm{a}),(\mathrm{a},\mathrm{b}),(\mathrm{b},\mathrm{a}),(\mathrm{b},\mathrm{b}\left)\right\} \\ \text{Following is the arrow diagram represent}\mathrm{Y}\times \mathrm{Y}\text{.} \end{gathered}$$

Q.48

$$\begin{gathered} \text{A relation R is\hspace{0.33em}defined\hspace{0.33em}on\hspace{0.33em}the\hspace{0.33em}set\hspace{0.33em}Z\hspace{0.33em}of\hspace{0.33em}integers\hspace{0.33em}as\hspace{0.33em}follows:}\underset{\mathrm{x}\to \infty }{\text{lim}} \\ \left(\mathrm{x},\mathrm{y}\right)\in \mathrm{R}\iff {\mathrm{x}}^2+{\mathrm{y}}^2=100 \\ {\text{Express\hspace{0.33em}R\hspace{0.33em}and\hspace{0.33em}R}}^{-\text{1}}\text{\hspace{0.33em}as\hspace{0.33em}the\hspace{0.33em}sets\hspace{0.33em}of\hspace{0.33em}ordered\hspace{0.33em}pairs\hspace{0.33em}and\hspace{0.33em}hence} \\ \text{find their respective domains.} \end{gathered}$$

Ans

$$\begin{gathered} \text{We have,} \\ \left(\mathrm{x},\mathrm{y}\right)\in \mathrm{R}\iff {\mathrm{x}}^2+{\mathrm{y}}^2=100 \\ \Rightarrow \mathrm{y}=\pm \sqrt{100–{\mathrm{x}}^2} \\ \text{We observe that}\mathrm{x}=0\Rightarrow \mathrm{y}=\pm 10 \\ \therefore \left(0,10\right)\in \mathrm{R}\text{and}\left(0,–10\right)\in \mathrm{R} \\ \mathrm{x}=\pm 6\Rightarrow \mathrm{y}=\sqrt{100–36}=\pm 8 \\ \therefore \left(6,8\right)\in \mathrm{R},\left(–6,8\right)\in \mathrm{R},\left(6,–8\right)\in \mathrm{R}\text{and}\left(–6,–8\right)\in \mathrm{R} \\ \mathrm{x}=\pm 8\Rightarrow \mathrm{y}=\sqrt{100–64}=\pm 6 \\ \therefore \left(8,6\right)\in \mathrm{R},\left(–8,6\right)\in \mathrm{R},\left(8,–6\right)\in \mathrm{R}\text{and}\left(–8,–6\right)\in \mathrm{R} \\ \mathrm{x}=\pm 10\Rightarrow \mathrm{y}=\sqrt{100–100}=0 \\ \left(10,0\right)\in \mathrm{R}\text{and}\left(–10,0\right)\in \mathrm{R} \\ \mathrm{We}\mathrm{also}\mathrm{notice}\mathrm{that}\mathrm{for}\mathrm{any}\mathrm{other}\mathrm{integral}\mathrm{value}\mathrm{of}\mathrm{x},\mathrm{y}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer} \\ \therefore \mathrm{R}=\{\left(0,10\right),\left(0,–10\right),\left(6,8\right),\left(–6,8\right),\left(6,–8\right),\left(–6,–8\right),\left(8,6\right) \\ \left(–8,6\right),\left(8,–6\right),\left(–8,–6\right),\left(10,0\right),\left(–10,0\right)\} \\ {\mathrm{R}}^{–1}=\{\left(10,0\right),\left(–10,0\right),\left(8,6\right),\left(8,–6\right),\left(–8,6\right),\left(–8,–6\right),\left(6,8\right) \\ \left(6,–8\right),\left(–6,8\right),\left(–6,–8\right),\left(0,10\right),\left(0,–10\right)\} \\ \text{Clearly domain}\left(\mathrm{R}\right)=\{0,6,–6,8,–8,10,–10\}=\text{domain}\left({\mathrm{R}}^{–1}\right) \end{gathered}$$

Q.49

$\begin{array}{l}\text{If A\hspace{0.33em}}=\{\text{x}:\text{x}<6,\text{x}\in \text{N}\},\text{B}=\left\{\text{x}:\text{x}\le 4,\text{x}\in \text{W}\right\}.\mathrm{Find}\left(\text{A}\cup \text{B}\right)\times \left(\text{A}\cap \text{B}\right)\text{}\\ \text{where W is the set of whole numbers, and N is the set of natural numbers.}\end{array}$

Ans

$$\begin{gathered} \text{It is given that}\mathrm{A}=\{\mathrm{x}:\mathrm{x}<4,\mathrm{x}\in \mathrm{N}\}\text{where}\mathrm{N}\text{is the set of natural numbers.} \\ \text{Thus, we have}\mathrm{A}=\{1,2,3\} \\ \text{Similarly, we are given that}\mathrm{B}=\{\mathrm{x}:\mathrm{x}\le 3,\mathrm{x}\in \mathrm{W}\}\text{where}\mathrm{W}\text{is the} \\ \text{set of whole numbers.} \\ \text{Thus, we have,}\mathrm{B}=\{0,1,2,3\} \\ \text{Now,} \\ \mathrm{A}\cup \mathrm{B}=\mathrm{B}=\{0,1,2,3\},\text{and} \\ \mathrm{A}\cap \mathrm{B}=\mathrm{A}=\{1,2,3\} \\ \text{For finding}\left(\mathrm{A}\cup \mathrm{B}\right)\times \left(\mathrm{A}\cap \mathrm{B}\right)\text{we will make following table :} \end{gathered}$$

$$\begin{gathered} \mathrm{Thus}, \\ \left(\mathrm{A}\cup \mathrm{B}\right)\times \left(\mathrm{A}\cap \mathrm{B}\right)=\{\left(0,1\right),\left(0,2\right),\left(0,3\right),\left(1,1\right),\left(1,2\right),\left(1,3\right) \\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(3,1\right),\left(3,2\right),\left(3,3\right)\}, \end{gathered}$$

Q.50

$\text{If\hspace{0.33em}4f}\left(\text{x}\right)\text{\hspace{0.33em}+\hspace{0.33em}6f}\left(\frac{\text{1}}{\text{x}}\right)\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{x}}-\text{5,\hspace{0.33em}then\hspace{0.33em}find\hspace{0.33em}f}\left(\text{x}\right)\text{.}$

Ans

$$\begin{gathered} \text{We have been given,} \\ 4\mathrm{f}\left(\mathrm{x}\right)+6\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{1}{\mathrm{x}}–5...\left(1\right) \\ \text{Replace x by}\frac{1}{\mathrm{x}}\text{in the above expression} \\ 4\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)+6\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}–5...\left(2\right) \\ \text{On adding equationsa}\left(1\right)\text{and}\left(2\right) \\ 10\mathrm{f}\left(\mathrm{x}\right)+10\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}–10 \\ \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\left(\frac{1}{10}\right)\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)–1...\left(3\right) \\ \mathrm{Now},\mathrm{subtracting}\left(1\right)\mathrm{from}\left(2\right) \\ 2\mathrm{f}\left(\mathrm{x}\right)–2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}–\frac{1}{\mathrm{x}}...\left(4\right) \\ \text{Now, multiply equation}\left(3\right)\text{by}\left(2\right) \\ 2\mathrm{f}\left(\mathrm{x}\right)+2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\left(\frac{1}{5}\right)\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)–2...\left(5\right) \\ \text{Adding equationa}\left(4\right)\text{nd equation}\left(5\right) \\ 4\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{5}+\frac{1}{5\mathrm{x}}+\mathrm{x}–\frac{1}{\mathrm{x}}–2 \\ =\frac{\mathrm{x}}{5}+\mathrm{x}+\frac{1}{5\mathrm{x}}–\frac{1}{\mathrm{x}}–2 \\ =\frac{6\mathrm{x}}{5}–\frac{4}{5\mathrm{x}}–2 \\ 4\mathrm{f}\left(\mathrm{x}\right)=\frac{6\mathrm{x}}{5}–\frac{4}{5\mathrm{x}}–2 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{3\mathrm{x}}{10}–\frac{1}{5\mathrm{x}}–\frac{1}{2} \end{gathered}$$

Q.51 There are 4 elements in a set A and 5 elements in the set B, then there will be ___ elements in A × B.

Ans

It is given that n(A) = 4
and n(B) = 5
Therefore, n(A × B) = 4 × 5
= 20
There are 4 elements in a set A and 5 elements in the set B, then there will be 20 elements in A × B.