Samacheer Kalvi 10th Maths Exercise 1.4 Solutions

10th Maths Exercise 1.4 solutions (Chapter 1 Relations and Functions) with step-by-step answers for all 32 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance29 answers · full steps below

Q1 (i)Not a function (fails the vertical line test — a vertical line meets the curve at two points).
Q1 (ii)It is a function (passes the vertical line test).
Q1 (iii)Not a function (fails the vertical line test).
Q1 (iv)It is a function (passes the vertical line test).
Q2 (i) $f=\{(2,0),(4,1),(6,2),(10,4),(12,5)\}$
Q2 (ii) $\begin{array}{|c|c|c|c|c|c|}\hline x & 2 & 4 & 6 & 10 & 12\\\hline f(x) & 0 & 1 & 2 & 4 & 5\\\hline\end{array}$
Q2 (iii)Arrow diagram: $2\to 0,\; 4\to 1,\; 6\to 2,\; 10\to 4,\; 12\to 5$ (the element $9\in B$ has no pre-image).
Q2 (iv)Plot the points $(2,0),(4,1),(6,2),(10,4),(12,5)$ ; they lie on a straight line.
Q3 (i)Arrow diagram: $1\to2,\;2\to2,\;3\to2,\;4\to3,\;5\to4$ .
Q3 (ii) $\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\\hline f(x) & 2 & 2 & 2 & 3 & 4\\\hline\end{array}$
Q3 (iii)Plot the points $(1,2),(2,2),(3,2),(4,3),(5,4)$ .
Q6 (i)Range of $f=\{1,8,27,64\}$
Q6 (ii) $f$ is a one-one and into function.
Q7 (i) $f$ is bijective (one-one and onto).
Q7 (ii) $f$ is not bijective (it is neither one-one nor onto).
Q8 $a=1$ or $a=-1$ , and $b=1$
Q9 (i) $f(3)=5$
Q9 (ii) $f(0)=2$
Q9 (iii) $f(-1\cdot 5)=-2\cdot 5$
Q9 (iv) $f(2)+f(-2)=1$
Q10 (i) $f(-3)+f(2)=2$
Q10 (ii) $f(7)-f(1)=10$
Q10 (iii) $2f(4)+f(8)=178$
Q10 (iv) $\dfrac{2f(-2)-f(6)}{f(4)+f(-2)}=\dfrac{-9}{17}$
Q12 (i) $t(0)=32$
Q12 (ii) $t(28)=82\cdot 4$
Q12 (iii) $t(-10)=14$
Q12 (iv) $C=100$
Q12 (v) $C=-40$ (i.e. $-40^\circ\text{C}=-40^\circ\text{F}$ )

Q1. Determine whether the graph given below represent functions. Give reason for your …

Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

(i)

Graph (i): a magenta sideways parabola opening to the left, symmetric about the X-axis with vertex just right of the origin O, on axes X'X and Y'Y.

Draw a vertical line on the curve.
The curve opens to the left, so a vertical line cuts it at two points.
A vertical line must meet the graph at only one point for a function.

Answer

Not a function (fails the vertical line test — a vertical line meets the curve at two points).

(ii)

Graph (ii): a magenta S-shaped increasing cubic-like curve through the origin O, descending into the third quadrant and rising into the first quadrant, with arrowheads at both ends.

Draw vertical lines anywhere on the curve.
Each vertical line meets the S‑shaped curve at exactly one point.
So every $x$ has only one image.

Answer

It is a function (passes the vertical line test).

(iii)

Graph (iii): a magenta S-shaped curve turned vertically about the Y-axis that doubles back on itself, so some vertical lines cut it more than once.

The curve bends back on itself.
A vertical line drawn in that region meets the curve at more than one point.
So one $x$ has more than one image.

Answer

Not a function (fails the vertical line test).

(iv)

Graph (iv): a straight magenta line with arrowheads at both ends passing through the origin with positive slope, from the third quadrant to the first quadrant.

The graph is a straight line through the origin.
Every vertical line meets it at exactly one point.

Answer

It is a function (passes the vertical line test).

Q2. Let $f : A \to B$ be a function defined by $f(x) = \dfrac{x}{2} - 1$ , where $A = \{2, 4, 6, 10, 12\}$ …

Let $f : A \to B$ be a function defined by $f(x) = \dfrac{x}{2} - 1$ , where $A = \{2, 4, 6, 10, 12\}$ , $B = \{0, 1, 2, 4, 5, 9\}$ . Represent $f$ by

(i)

set of ordered pairs

Show steps (6)

$f(x)=\dfrac{x}{2}-1$
$f(2)=\dfrac{2}{2}-1=0$
$f(4)=\dfrac{4}{2}-1=1$
$f(6)=\dfrac{6}{2}-1=2$
$f(10)=\dfrac{10}{2}-1=4$
$f(12)=\dfrac{12}{2}-1=5$

Answer

$f=\{(2,0),(4,1),(6,2),(10,4),(12,5)\}$

(ii)

a table

Use the images found from $f(x)=\dfrac{x}{2}-1$ .
Write $x$ in the first row and $f(x)$ in the second row.
$\begin{array}{|c|c|c|c|c|c|}\hline x & 2 & 4 & 6 & 10 & 12\\\hline f(x) & 0 & 1 & 2 & 4 & 5\\\hline\end{array}$

Answer

$\begin{array}{|c|c|c|c|c|c|}\hline x & 2 & 4 & 6 & 10 & 12\\\hline f(x) & 0 & 1 & 2 & 4 & 5\\\hline\end{array}$

(iii)

an arrow diagram

Draw two closed curves: $A=\{2,4,6,10,12\}$ on the left, $B=\{0,1,2,4,5,9\}$ on the right.
Join by arrows: $2\to 0$ , $4\to 1$ , $6\to 2$ , $10\to 4$ , $12\to 5$ .
The element $9$ in $B$ receives no arrow.

Answer

Arrow diagram: $2\to 0,\; 4\to 1,\; 6\to 2,\; 10\to 4,\; 12\to 5$ (the element $9\in B$ has no pre-image).

(iv)

a graph

Take the ordered pairs of $f$ as points.
Plot $(2,0),(4,1),(6,2),(10,4),(12,5)$ on the graph sheet.
All the points lie on a straight line (the line $y=\dfrac{x}{2}-1$ ).

Answer

Plot the points $(2,0),(4,1),(6,2),(10,4),(12,5)$ ; they lie on a straight line.

Q3. Represent the function $f = \{(1,2),(2,2),(3,2),(4,3),(5,4)\}$ through

(i)

an arrow diagram

Domain $=\{1,2,3,4,5\}$ , co-domain $=\{2,3,4\}$ .
Draw arrows: $1\to 2$ , $2\to 2$ , $3\to 2$ , $4\to 3$ , $5\to 4$ .

Answer

Arrow diagram: $1\to2,\;2\to2,\;3\to2,\;4\to3,\;5\to4$ .

(ii)

a table form

Write the first co-ordinates as $x$ and the second as $f(x)$ .
$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\\hline f(x) & 2 & 2 & 2 & 3 & 4\\\hline\end{array}$

Answer

$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\\hline f(x) & 2 & 2 & 2 & 3 & 4\\\hline\end{array}$

(iii)

a graph

Take each ordered pair as a point.
Plot $(1,2),(2,2),(3,2),(4,3),(5,4)$ on the graph sheet.

Answer

Plot the points $(1,2),(2,2),(3,2),(4,3),(5,4)$ .

Q4. Show that the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(x) = 2x - 1$ is one-one but not …

Show that the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(x) = 2x - 1$ is one-one but not onto.

Show steps (8)

One-one: Let $f(x_1)=f(x_2)$ for $x_1,x_2\in\mathbb{N}$ .
$2x_1-1=2x_2-1$
$2x_1=2x_2 \Rightarrow x_1=x_2$
So different elements have different images, hence $f$ is one-one.
Onto: $f(1)=1,\; f(2)=3,\; f(3)=5,\dots$
Range of $f=\{1,3,5,7,\dots\}$ , the set of odd natural numbers.
This is a proper subset of the co-domain $\mathbb{N}$ (e.g. $2$ has no pre-image, since $2x-1=2$ gives $x=\frac32\notin\mathbb{N}$ ).
Hence $f$ is not onto.

Answer

$f$ is one-one but not onto (it is an into function).

Q5. Show that the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(m) = m^2 + m + 3$ is one-one function.

Show steps (7)

Let $f(m_1)=f(m_2)$ for $m_1,m_2\in\mathbb{N}$ .
$m_1^2+m_1+3=m_2^2+m_2+3$
$m_1^2-m_2^2+m_1-m_2=0$
$(m_1-m_2)(m_1+m_2)+(m_1-m_2)=0$
$(m_1-m_2)(m_1+m_2+1)=0$
Since $m_1,m_2\in\mathbb{N}$ , $m_1+m_2+1\neq 0$ .
Therefore $m_1-m_2=0$ , that is $m_1=m_2$ .

Answer

$f(m_1)=f(m_2)\Rightarrow m_1=m_2$ , hence $f$ is a one-one function.

Q6. Let $A = \{1,2,3,4\}$ and $B = \mathbb{N}$ . Let $f : A \to B$ be defined by $f(x) = x^3$ then,

Let $A = \{1,2,3,4\}$ and $B = \mathbb{N}$ . Let $f : A \to B$ be defined by $f(x) = x^3$ then,

(i)

find the range of $f$

$f(x)=x^3$ , $A=\{1,2,3,4\}$
$f(1)=1^3=1$
$f(2)=2^3=8$
$f(3)=3^3=27$
$f(4)=4^3=64$

Answer

Range of $f=\{1,8,27,64\}$

(ii)

identify the type of function

Different elements of $A$ have different images, so $f$ is one-one.
Co-domain $=\mathbb{N}$ , but range $=\{1,8,27,64\}$ is a proper subset of $\mathbb{N}$ .
So $f$ is not onto; it is into.

Answer

$f$ is a one-one and into function.

Q7. In each of the following cases state whether the function is bijective …

In each of the following cases state whether the function is bijective or not. Justify your answer.

(i)

$f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 1$

One-one: $f(x_1)=f(x_2)\Rightarrow 2x_1+1=2x_2+1\Rightarrow x_1=x_2$ .
Onto: For any $y\in\mathbb{R}$ , take $x=\dfrac{y-1}{2}\in\mathbb{R}$ ; then $f(x)=y$ .
So every element of the co-domain has a pre-image; range $=\mathbb{R}$ .

Answer

$f$ is bijective (one-one and onto).

(ii)

$f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3 - 4x^2$

$f(1)=3-4(1)^2=-1$ and $f(-1)=3-4(-1)^2=-1$ .
Two different elements $1$ and $-1$ have the same image, so $f$ is not one-one.
Also $4x^2\ge 0$ , so $f(x)=3-4x^2\le 3$ .
Range $=(-\infty,3]$ , a proper subset of $\mathbb{R}$ ; so $f$ is not onto.

Answer

$f$ is not bijective (it is neither one-one nor onto).

Q8. Let $A = \{-1, 1\}$ and $B = \{0, 2\}$ . If the function $f : A \to B$ defined by $f(x) = ax + b$ …

Let $A = \{-1, 1\}$ and $B = \{0, 2\}$ . If the function $f : A \to B$ defined by $f(x) = ax + b$ is an onto function? Find $a$ and $b$ .

Show steps (9)

$f(x)=ax+b$ with $A=\{-1,1\}$ , $B=\{0,2\}$ .
Since $f$ is onto, the images $f(-1)$ and $f(1)$ must together cover all of $B=\{0,2\}$ , so two cases arise.
Case 1: $f(-1)=0$ and $f(1)=2$ .
$-a+b=0$ … (1) and $a+b=2$ … (2).
Adding (1) and (2): $2b=2\Rightarrow b=1$ ; then from (2), $a=1$ .
Case 2: $f(-1)=2$ and $f(1)=0$ .
$-a+b=2$ … (3) and $a+b=0$ … (4).
Adding (3) and (4): $2b=2\Rightarrow b=1$ ; then from (4), $a=-1$ .
Hence $a=1,\,b=1$ or $a=-1,\,b=1$ ; in both cases $f$ is onto.

Answer

$a=1$ or $a=-1$ , and $b=1$

Q9. If the function $f$ is defined by $f(x) = \begin{cases} x + 2; & x > 1 \\ 2; & -1 \le x \le 1 \\ x - 1; & -3 < x < -1 \end{cases}$ find the values of

(i)

$f(3)$

$x=3>1$ , so use $f(x)=x+2$ .
$f(3)=3+2=5$

Answer

$f(3)=5$

(ii)

$f(0)$

$x=0$ lies in $-1\le x\le 1$ , so $f(x)=2$ .
$f(0)=2$

Answer

$f(0)=2$

(iii)

$f(-1\cdot 5)$

$x=-1\cdot5$ lies in $-3<x<-1$ , so $f(x)=x-1$ .
$f(-1\cdot5)=-1\cdot5-1=-2\cdot5$

Answer

$f(-1\cdot 5)=-2\cdot 5$

(iv)

$f(2) + f(-2)$

$x=2>1$ , so $f(2)=2+2=4$ .
$x=-2$ lies in $-3<x<-1$ , so $f(-2)=-2-1=-3$ .
$f(2)+f(-2)=4+(-3)=1$

Answer

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

Q10. A function $f : [-5, 9] \to \mathbb{R}$ is defined as follows: $$f(x) = \begin{cases} 6x + …

A function $f : [-5, 9] \to \mathbb{R}$ is defined as follows: $f(x) = \begin{cases} 6x + 1; & -5 \le x < 2 \\ 5x^2 - 1; & 2 \le x < 6 \\ 3x - 4; & 6 \le x \le 9 \end{cases}$ Find

(i)

$f(-3) + f(2)$

$-3$ lies in $-5\le x<2$ , so $f(-3)=6(-3)+1=-17$ .
$2$ lies in $2\le x<6$ , so $f(2)=5(2)^2-1=19$ .
$f(-3)+f(2)=-17+19=2$

Answer

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

(ii)

$f(7) - f(1)$

$7$ lies in $6\le x\le 9$ , so $f(7)=3(7)-4=17$ .
$1$ lies in $-5\le x<2$ , so $f(1)=6(1)+1=7$ .
$f(7)-f(1)=17-7=10$

Answer

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

(iii)

$2f(4) + f(8)$

$4$ lies in $2\le x<6$ , so $f(4)=5(4)^2-1=79$ .
$8$ lies in $6\le x\le 9$ , so $f(8)=3(8)-4=20$ .
$2f(4)+f(8)=2(79)+20=158+20=178$

Answer

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

(iv)

$\dfrac{2f(-2) - f(6)}{f(4) + f(-2)}$

$-2$ lies in $-5\le x<2$ , so $f(-2)=6(-2)+1=-11$ .
$6$ lies in $6\le x\le 9$ , so $f(6)=3(6)-4=14$ .
$4$ lies in $2\le x<6$ , so $f(4)=5(4)^2-1=79$ .
$\dfrac{2f(-2)-f(6)}{f(4)+f(-2)}=\dfrac{2(-11)-14}{79+(-11)}$
$=\dfrac{-22-14}{68}=\dfrac{-36}{68}=\dfrac{-9}{17}$

Answer

$\dfrac{2f(-2)-f(6)}{f(4)+f(-2)}=\dfrac{-9}{17}$

Q11. The distance $S$ an object travels under the influence of gravity in …

The distance $S$ an object travels under the influence of gravity in time $t$ seconds is given by $S(t) = \dfrac{1}{2}gt^2 + at + b$ where, ( $g$ is the acceleration due to gravity), $a > 0$ , $b$ are constants. Verify wheather the function $S(t)$ is one-one or not.

Show steps (6)

Assume $S(t_1)=S(t_2)$ .
$\dfrac12 gt_1^2+at_1+b=\dfrac12 gt_2^2+at_2+b$
$\dfrac12 g\,(t_1^2-t_2^2)+a(t_1-t_2)=0$
$(t_1-t_2)\left[\dfrac{g}{2}(t_1+t_2)+a\right]=0$
Time $t_1,t_2\ge 0$ and $g>0,\;a>0$ , so $\dfrac{g}{2}(t_1+t_2)+a>0$ .
Hence $t_1-t_2=0$ , that is $t_1=t_2$ .

Answer

$S(t_1)=S(t_2)\Rightarrow t_1=t_2$ , so $S(t)$ is a one-one function.

Q12. The function ' $t$ ' which maps temperature in Celsius ( $C$ ) into temperature in …

The function ' $t$ ' which maps temperature in Celsius ( $C$ ) into temperature in Fahrenheit ( $F$ ) is defined by $t(C) = F$ where $F = \dfrac{9}{5}C + 32$ . Find,

(i)

$t(0)$

$F=\dfrac95 C+32$
$t(0)=\dfrac95(0)+32=32$

Answer

$t(0)=32$

(ii)

$t(28)$

$t(28)=\dfrac95(28)+32$
$=\dfrac{252}{5}+32=50\cdot4+32$
$=82\cdot4$

Answer

$t(28)=82\cdot 4$

(iii)

$t(-10)$

$t(-10)=\dfrac95(-10)+32$
$=-18+32=14$

Answer

$t(-10)=14$

(iv)

the value of $C$ when $t(C) = 212$

$t(C)=212\Rightarrow \dfrac95 C+32=212$
$\dfrac95 C=180$
$C=180\times\dfrac59=100$

Answer

$C=100$

(v)

the temperature when the Celsius value is equal to the Farenheit value.

Let $F=C$ .
$C=\dfrac95 C+32$
$5C=9C+160$
$-4C=160$
$C=-40$

Answer

$C=-40$ (i.e. $-40^\circ\text{C}=-40^\circ\text{F}$ )

Quick check

Let $f$ and $g$ be two functions given by $f = \{(0,1),(2,0),(3,-4),(4,2),(5,7)\}$ , $g = \{(0,2),(1,0),(2,4),(-4,2),(7,0)\}$ then the range of $f \circ g$ is
1. (1) $\{0,2,3,4,5\}$
2. (2) $\{-4,1,0,2,7\}$
3. (3) $\{1,2,3,4,5\}$
4. (4) $\{0,1,2\}$
Show answer
Answer: (4) $\{0,1,2\}$
1. $(f\circ g)(0)=f(2)=0$ , $(f\circ g)(1)=f(0)=1$ .
2. $(f\circ g)(2)=f(4)=2$ , $(f\circ g)(-4)=f(2)=0$ , $(f\circ g)(7)=f(0)=1$ .
3. Range $=\{0,1,2\}$ .
Question 1 of 5
If there are 1024 relations from a set $A = \{1, 2, 3, 4, 5\}$ to a set $B$ , then the number of elements in $B$ is
1. (1) 3
2. (2) 2
3. (3) 4
4. (4) 8
Show answer
Answer: (2) 2
1. Number of relations from $A$ to $B$ is $2^{n(A)\cdot n(B)}$ .
2. $2^{5\,n(B)}=1024=2^{10}$ .
3. $5\,n(B)=10\Rightarrow n(B)=2$ .
Question 2 of 5
Let $A = \{1,2,3,4\}$ and $B = \{4,8,9,10\}$ . A function $f : A \to B$ given by $f = \{(1,4),(2,8),(3,9),(4,10)\}$ is a
1. (1) Many-one function
2. (2) Identity function
3. (3) One-to-one function
4. (4) Into function
Show answer
Answer: (3) One-to-one function
1. Images are $4,8,9,10$ — all different.
2. Different elements of $A$ have different images.
3. So $f$ is one-to-one.
Question 3 of 5
If $f(x) = 2x^2$ and $g(x) = \dfrac{1}{3x}$ , then $f \circ g$ is
1. (1) $\dfrac{3}{2x^2}$
2. (2) $\dfrac{2}{3x^2}$
3. (3) $\dfrac{2}{9x^2}$
4. (4) $\dfrac{1}{6x^2}$
Show answer
Answer: (3) $\dfrac{2}{9x^2}$
1. $(f\circ g)(x)=f\big(g(x)\big)=f\left(\dfrac{1}{3x}\right)$ .
2. $=2\left(\dfrac{1}{3x}\right)^{2}$ .
3. $=\dfrac{2}{9x^{2}}$ .
Question 4 of 5
$A = \{a,b,p\}$ , $B = \{2,3\}$ , $C = \{p,q,r,s\}$ then $n[(A \cup C) \times B]$ is
1. (1) 8
2. (2) 20
3. (3) 12
4. (4) 16
Show answer
Answer: (3) 12
1. $A\cup C=\{a,b,p,q,r,s\}$ , so $n(A\cup C)=6$ .
2. $n(B)=2$ .
3. $n[(A\cup C)\times B]=6\times 2=12$ .
Question 5 of 5

Chapter 1 formulas

Cartesian product of A with B
$A \times B = \{(a,b) \mid \text{for all } a \in A,\ b \in B\}$
A relation R from A to B is a subset of A × B
$R \subseteq A \times B$
Condition for a relation R from X to Y to be a function
$\text{for every } x \in X \text{ there exists only one } y \in Y$
Identity function
$f(x) = x$
Reciprocal function
$f(x) = \dfrac{1}{x}$
Constant function
$f(x) = c$
Linear function
$f(x) = ax + b,\ a \neq 0$
Quadratic function
$f(x) = ax^{2} + bx + c,\ a \neq 0$
Cubic function
$f(x) = ax^{3} + bx^{2} + cx + d,\ a \neq 0$
Composition of two functions f : A → B and g : B → C
$g \circ f : A \to C,\quad (g \circ f)(x) = g(f(x)) \ \text{ for all } x \in A$
Composition is not commutative in general
$f \circ g \neq g \circ f$
Composition is associative
$f \circ (g \circ h) = (f \circ g) \circ h$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

Spotted a mistake? Email us at [email protected] and we will fix it.

Q1. Determine whether the graph given below represent functions. Give reason for your …

(i)

(ii)

(iii)

(iv)

Q2. Let f:A→Bf : A \to Bf:A→B be a function defined by f(x)=x2−1f(x) = \dfrac{x}{2} - 1f(x)=2x​−1 , where A={2,4,6,10,12}A = \{2, 4, 6, 10, 12\}A={2,4,6,10,12} …

(i)

(ii)

(iii)

(iv)

Q3. Represent the function f={(1,2),(2,2),(3,2),(4,3),(5,4)}f = \{(1,2),(2,2),(3,2),(4,3),(5,4)\}f={(1,2),(2,2),(3,2),(4,3),(5,4)} through

(i)

(ii)

(iii)

Q4. Show that the function f:N→Nf : \mathbb{N} \to \mathbb{N}f:N→N defined by f(x)=2x−1f(x) = 2x - 1f(x)=2x−1 is one-one but not …

Q5. Show that the function f:N→Nf : \mathbb{N} \to \mathbb{N}f:N→N defined by f(m)=m2+m+3f(m) = m^2 + m + 3f(m)=m2+m+3 is one-one function.

Q6. Let A={1,2,3,4}A = \{1,2,3,4\}A={1,2,3,4} and B=NB = \mathbb{N}B=N . Let f:A→Bf : A \to Bf:A→B be defined by f(x)=x3f(x) = x^3f(x)=x3 then,

(i)

(ii)

Q7. In each of the following cases state whether the function is bijective …

(i)

(ii)

Q8. Let A={−1,1}A = \{-1, 1\}A={−1,1} and B={0,2}B = \{0, 2\}B={0,2} . If the function f:A→Bf : A \to Bf:A→B defined by f(x)=ax+bf(x) = ax + bf(x)=ax+b …

Q9. If the function fff is defined by f(x)={x+2;x>12;−1≤x≤1x−1;−3<x<−1f(x) = \begin{cases} x + 2; & x > 1 \\ 2; & -1 \le x \le 1 \\ x - 1; & -3 < x < -1 \end{cases}f(x)=⎩⎨⎧​x+2;2;x−1;​x>1−1≤x≤1−3<x<−1​ find the values of

(i)

(ii)

(iii)

(iv)

Q10. A function f:[−5,9]→Rf : [-5, 9] \to \mathbb{R}f:[−5,9]→R is defined as follows: $$f(x) = \begin{cases} 6x + …

(i)

(ii)

(iii)

(iv)

Q11. The distance SSS an object travels under the influence of gravity in …

Q12. The function 'ttt' which maps temperature in Celsius (CCC) into temperature in …

(i)

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Quick check

Q2. Let $f : A \to B$ be a function defined by $f(x) = \dfrac{x}{2} - 1$ , where $A = \{2, 4, 6, 10, 12\}$ …

Q3. Represent the function $f = \{(1,2),(2,2),(3,2),(4,3),(5,4)\}$ through

Q4. Show that the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(x) = 2x - 1$ is one-one but not …

Q5. Show that the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(m) = m^2 + m + 3$ is one-one function.

Q6. Let $A = \{1,2,3,4\}$ and $B = \mathbb{N}$ . Let $f : A \to B$ be defined by $f(x) = x^3$ then,

Q8. Let $A = \{-1, 1\}$ and $B = \{0, 2\}$ . If the function $f : A \to B$ defined by $f(x) = ax + b$ …

Q9. If the function $f$ is defined by $f(x) = \begin{cases} x + 2; & x > 1 \\ 2; & -1 \le x \le 1 \\ x - 1; & -3 < x < -1 \end{cases}$ find the values of

Q10. A function $f : [-5, 9] \to \mathbb{R}$ is defined as follows: $$f(x) = \begin{cases} 6x + …

Q11. The distance $S$ an object travels under the influence of gravity in …

Q12. The function ' $t$ ' which maps temperature in Celsius ( $C$ ) into temperature in …