Samacheer Class 10 Maths - Relations and Functions

$$ B=\{2,3,5,7\} $$

$$ A \times B = \{ (1,2),(1,3),(1,5),(1,7), (2,2),(2,3),(2,5),(2,7), (3,2),(3,3),(3,5),(3,7) \} $$

$$ B \times A = \{ (2,1),(2,2),(2,3), (3,1),(3,2),(3,3), (5,1),(5,2),(5,3), (7,1),(7,2),(7,3) \} $$

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$$ A=\{3,4\} $$

$$ B=\{-2,0,3\} $$

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$$ A \times A = \{(5,5),(5,6),(6,5),(6,6)\} $$

$$ B \times B = \{ (4,4),(4,5),(4,6), (5,4),(5,5),(5,6), (6,4),(6,5),(6,6) \} $$

$$ C \times C = \{ (5,5),(5,6),(5,7), (6,5),(6,6),(6,7), (7,5),(7,6),(7,7) \} $$

$$ (B \times B)\cap(C \times C) = \{(5,5),(5,6),(6,5),(6,6)\} $$

Hence,

$$ A \times A = (B \times B)\cap(C \times C) $$

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$$ A \cap C = \{3\} $$

$$ B \cap D = \{3,5\} $$

$$ (A \cap C)\times(B \cap D) = \{(3,3),(3,5)\} $$

Also,

$$ A \times B = \{ (1,2),(1,3),(1,5), (2,2),(2,3),(2,5), (3,2),(3,3),(3,5) \} $$

$$ C \times D = \{ (3,1),(3,3),(3,5), (4,1),(4,3),(4,5) \} $$

$$ (A \times B)\cap(C \times D) = \{(3,3),(3,5)\} $$

Hence the statement is true.

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$$ A=\{x \in W \mid x<2\}, B=\{x \in N \mid 1

Verify:

- $A$ = set of all natural numbers less than 8 - $B$ = set of all prime numbers less than 8 - $C$ = set of even prime number

Verify:

$$ A=\{1,2,3,7\}, \quad B=\{3,0,-1,7\} $$

Which of the following are relations from $A$ to $B$?

Squares within 45 are:

$$ 1,4,9,16,25,36 $$

Hence,

$$ R= \{ (1,1), (2,4), (3,9), (4,16), (5,25), (6,36) \} $$

### Domain

$$ \{1,2,3,4,5,6\} $$

### Range

$$ \{1,4,9,16,25,36\} $$

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For each value of $x$:

$$ \begin{aligned} x=0 &\Rightarrow y=3\\ x=1 &\Rightarrow y=4\\ x=2 &\Rightarrow y=5\\ x=3 &\Rightarrow y=6\\ x=4 &\Rightarrow y=7\\ x=5 &\Rightarrow y=8 \end{aligned} $$

Thus,

$$ R= \{ (0,3), (1,4), (2,5), (3,6), (4,7), (5,8) \} $$

### Domain

$$ \{0,1,2,3,4,5\} $$

### Range

$$ \{3,4,5,6,7,8\} $$

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- (a) Arrow diagram - (b) Graph - (c) Set in roster form

wherever possible.

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$$ \begin{aligned} R= \{ &(10000,A_1), (10000,A_2), (10000,A_3), (10000,A_4), (10000,A_5),\\ &(25000,C_1), (25000,C_2), (25000,C_3), (25000,C_4),\\ &(50000,M_1), (50000,M_2), (50000,M_3),\\ &(100000,E_1), (100000,E_2) \} \end{aligned} $$

### Arrow Representation

$$ 10000 \to A_1,A_2,A_3,A_4,A_5 $$

$$ 25000 \to C_1,C_2,C_3,C_4 $$

$$ 50000 \to M_1,M_2,M_3 $$

$$ 100000 \to E_1,E_2 $$

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# UNIT 1 : Relations and Functions

Since $y=2x$,

$$ f=\{(1,2),(2,4),(3,6),(4,8),\dots\} $$

### Domain

$$ \{1,2,3,4,\dots\} $$

### Co-domain

$$ N=\{1,2,3,4,\dots\} $$

### Range

$$ \{2,4,6,8,\dots\} $$

Each input has exactly one output.

Hence, it is a function.

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For each element of $X$:

$$ \begin{aligned} f(3)&=10\\ f(4)&=17\\ f(6)&=37\\ f(8)&=65 \end{aligned} $$

Thus,

$$ R=\{(3,10),(4,17),(6,37),(8,65)\} $$

Every element in $X$ has exactly one image in $N$.

Hence, it is a function.

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$$ f(x)=x^2-5x+6 $$

evaluate:

It is clear that:

$$ f(9)=2 $$

$$ \frac{(2x+5)-5}{x} = \frac{2x}{x} =2 $$

### Answer

$$ 2 $$

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$$ f(x)=2x-3 $$

Length:

$$ 24-2x $$

Breadth:

$$ 24-2x $$

Height:

$$ x $$

Volume:

$$ V=x(24-2x)^2 $$

$$ =x(576-96x+4x^2) $$

$$ =576x-96x^2+4x^3 $$

### Answer

$$ V(x)=4x^3-96x^2+576x $$

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$$ f(x^2)=3-2x^2 $$

$$ (f(x))^2=(3-2x)^2 $$

Equating:

$$ 3-2x^2=9-12x+4x^2 $$

$$ 6x^2-12x+6=0 $$

$$ x^2-2x+1=0 $$

$$ (x-1)^2=0 $$

$$ x=1 $$

### Answer

$$ 1 $$

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$$ \text{Distance}=\text{Speed}\times\text{Time} $$

$$ d=500t $$

### Answer

$$ d(t)=500t $$

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$$ y=ax+b $$

where $a,b$ are constants.

1. Not a function 2. Function 3. Not a function 4. Function

### Reason

A graph represents a function only if every vertical line intersects the graph at most once (Vertical Line Test).

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$$ f:A\to B $$

be defined by

$$ f(x)=\frac{x}{2}-1 $$

where

$$ A=\{2,4,6,10,12\} $$

$$ B=\{0,1,2,4,5,9\} $$

Represent $f$ by:

$$ f=\{(1,2),(2,2),(3,2),(4,3),(5,4)\} $$

through:

Suppose

$$ f(a)=f(b) $$

Then,

$$ 2a-1=2b-1 $$

$$ 2a=2b $$

$$ a=b $$

Hence $f$ is one-one.

The range is

$$ \{1,3,5,7,\dots\} $$

Even natural numbers are not images of any element.

Hence the function is not onto.

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Suppose

$$ f(a)=f(b) $$

Then,

$$ a^2+a+3=b^2+b+3 $$

$$ a^2-b^2+a-b=0 $$

$$ (a-b)(a+b+1)=0 $$

Since $a+b+1\neq0$,

$$ a-b=0 $$

$$ a=b $$

Hence the function is one-one.

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$$ A=\{1,2,3,4\}, \quad B=N $$

and

$$ f(x)=x^3 $$

Since function is onto,

$$ f(-1)=0,\quad f(1)=2 $$

Thus,

$$ -a+b=0 $$

$$ a+b=2 $$

Adding:

$$ 2b=2 $$

$$ b=1 $$

Substitute:

$$ a=1 $$

### Answer

$$ a=1,\quad b=1 $$

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$$ f(x)= \begin{cases} x+2, & x\geq0\\ 2x+1, & x<0 \end{cases} $$

find:

1. $f(-3)+f(2)=2$

2. $f(7)-f(1)=10$

3. $2f(4)+f(8)=178$

4.

$$ -\frac{9}{17} $$

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Yes.

For practical positive values of time, distance increases continuously with time.

Hence it is one-one.

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$$ F=\frac95C+32 $$

Find:

$$ f\circ g=g\circ f $$

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$$ f\circ g=g\circ f $$

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$$ (f\circ g)(x)=f\left(\frac{x+1}{2}\right) $$

$$ =2\left(\frac{x+1}{2}\right)-1 $$

$$ =x+1-1=x $$

Now,

$$ (g\circ f)(x)=g(2x-1) $$

$$ =\frac{(2x-1)+1}{2} $$

$$ =\frac{2x}{2}=x $$

### Answer

$$ f\circ g=g\circ f=x $$

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If

$$ f(x)=x^2-1,\quad g(x)=x-2 $$

find $a$, if

$$ (g\circ f)(a)=1 $$

### Solution

$$ (g\circ f)(a)=g(a^2-1) $$

$$ =(a^2-1)-2 $$

$$ =a^2-3 $$

Given:

$$ a^2-3=1 $$

$$ a^2=4 $$

$$ a=\pm2 $$

### Answer

$$ a=2,-2 $$

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$$ (f\circ g)(x)=2x^2+1 $$

Range:

$$ \{2x^2+1\mid x\in A\} $$

Now,

$$ (g\circ f)(x)=(2x+1)^2 $$

$$ =4x^2+4x+1 $$

Range:

$$ \{4x^2+4x+1\mid x\in A\} $$

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$$ f(x)=x^2-1 $$

Find:

$f(x)=x^5$ is strictly increasing.

Hence $f$ is one-one.

But,

$$ g(2)=16 $$

and

$$ g(-2)=16 $$

Hence $g$ is not one-one.

Now,

$$ (f\circ g)(x)=f(x^4)=x^{20} $$

$$ (f\circ g)(2)=(f\circ g)(-2) $$

Hence $f\circ g$ is not one-one.

### Answer

- $f$ is one-one - $g$ is not one-one - $f\circ g$ is not one-one

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$$ (f\circ g)\circ h=f\circ(g\circ h) $$

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Assume

$$ f(x)=ax+b $$

Using $(0,-1)$:

$$ b=-1 $$

Using $(-1,3)$:

$$ -a-1=3 $$

$$ a=-4 $$

Thus,

$$ f(x)=-4x-1 $$

Check:

$$ f(2)=-8-1=-9 $$

Correct.

### Answer

$$ f(x)=-4x-1 $$

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$$ C(at_1+bt_2)=3(at_1+bt_2) $$

$$ =3at_1+3bt_2 $$

$$ =a(3t_1)+b(3t_2) $$

$$ =aC(t_1)+bC(t_2) $$

Hence the circuit is linear.

### Answer

$$ C(t)=3t $$

satisfies the superposition principle and is linear.

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# UNIT 1 : Relations and Functions

$$ (x,y)=(1,-5),(1,1),(2,-5),(2,1) $$

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$$ A=\{-1,0,1\} $$

Remaining elements:

$$ \{ (-1,-1),(-1,1), (0,-1),(0,0), (1,-1),(1,0),(1,1) \} $$

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$$ f(0)=2 $$

$$ f(3)=\sqrt7 $$

$$ f(a+1)=\sqrt{a+5} $$

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$$ \{ (9,3), (10,5), (11,11), (12,3), (13,13), (14,7), (15,5), (16,2), (17,17) \} $$

Range:

$$ \{2,3,5,7,11,13,17\} $$

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$$ [-1,1] $$

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$$ (f\circ g)\circ h=f\circ(g\circ h)=9(x-2)^2 $$

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Verified.

$$ A\times C\subseteq B\times D $$

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$$ f(f(x))=-\frac1x $$

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$$ g(g(1/2))=-\frac56 $$

$$ g(f(x))=2(x+1) $$

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(i)

$$ R-\{9\} $$

(ii)

$$ R $$

(iii)

$$ [2,\infty) $$

(iv)

$$ R $$

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# UNIT 1 : Relations and Functions

# Multiple Choice Questions

$$ n(A)\times n(B)=6 $$

$$ 2\times n(B)=6 $$

$$ n(B)=3 $$

Correct option: (3)

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$$ A\cup C=\{a,b,p,q,r,s\} $$

$$ n(A\cup C)=6 $$

$$ n(B)=2 $$

$$ 6\times2=12 $$

Correct option: (3)

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Every element of $A\times C$ belongs to $B\times D$.

Correct option: (1)

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Number of relations:

$$ 2^{mn}=1024 $$

$$ 2^{5n}=2^{10} $$

$$ 5n=10 $$

$$ n=2 $$

Correct option: (2)

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Primes less than 13:

$$ 2,3,5,7,11 $$

Squares:

$$ 4,9,25,49,121 $$

Correct option: (3)

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$$ a+2=5 $$

$$ a=3 $$

$$ 2a+b=4 $$

$$ 6+b=4 $$

$$ b=-2 $$

Correct option: (4)

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Total relations:

$$ 2^{mn} $$

Non-empty relations:

$$ 2^{mn}-1 $$

Correct option: (3)

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Identity function means:

$$ (x,x) $$

Thus,

$$ a=8,\quad b=6 $$

Correct option: (1)

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Distinct inputs give distinct outputs.

Correct option: (3)

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$$ (f\circ g)(x)=2\left(\frac{x}{3}\right)^2 $$

$$ =\frac{2x^2}{9} $$

Correct option: (3)

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Bijective functions have equal cardinalities.

$$ n(A)=n(B)=7 $$

Correct option: (1)

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Compute:

$$ g(0)=2 \Rightarrow f(2)=0 $$

$$ g(1)=0 \Rightarrow f(0)=1 $$

$$ g(2)=4 \Rightarrow f(4)=2 $$

$$ g(-4)=2 \Rightarrow f(2)=0 $$

$$ g(7)=0 \Rightarrow f(0)=1 $$

Range:

$$ \{0,1,2\} $$

Correct option: (4)

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$$ |xy|=|x||y| $$

Correct option: (1)

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Using points:

$$ g(1)=1 $$

$$ \alpha+\beta=1 $$

$$ g(2)=3 $$

$$ 2\alpha+\beta=3 $$

Subtract:

$$ \alpha=2 $$

$$ \beta=-1 $$

Correct option: (2)

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Expand:

$$ (x+1)^3=x^3+3x^2+3x+1 $$

$$ (x-1)^3=x^3-3x^2+3x-1 $$

Subtracting:

$$ f(x)=6x^2+2 $$

This is quadratic.

Correct option: (4)