Samacheer Kalvi 10th Maths Unit Exercise 1 Solutions

10th Maths Unit Exercise 1 solutions (Chapter 1 Relations and Functions) with step-by-step answers for all 16 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance14 answers · full steps below

Q1 $x=1,\ 2$ and $y=1,\ -5$
Q2 $A=\{-1,0,1\}$ ; remaining elements are $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$
Q3 (i) $f(0)=4$
Q3 (ii) $f(3)=\sqrt{2}$
Q3 (iii) $f(a+1)=\sqrt{a}$
Q4 $f=\{(9,3),(10,5),(11,11),(12,3),(13,13),(14,7),(15,5),(16,2),(17,17)\}$ ; Range $=\{2,3,5,7,11,13,17\}$
Q5Domain $=[-1,1]$
Q7Yes, $A\times C\subseteq B\times D$
Q9 (i) $gg\left(\dfrac{1}{2}\right)=-\dfrac{5}{6}$
Q9 (ii) $gf(x)=2x+2$
Q10 (i)Domain $=\mathbb{R}-\{9\}$
Q10 (ii)Domain $=\mathbb{R}$
Q10 (iii)Domain $=[2,\infty)$
Q10 (iv)Domain $=\mathbb{R}$

Q1. If the ordered pairs $(x^2-3x,\ y^2+4y)$ and $(-2,5)$ are equal, then find $x$ …

If the ordered pairs $(x^2-3x,\ y^2+4y)$ and $(-2,5)$ are equal, then find $x$ and $y$ .

Show steps (7)

Equal ordered pairs have equal corresponding entries.
$x^2-3x=-2$
$x^2-3x+2=0$
$(x-1)(x-2)=0 \Rightarrow x=1$ or $x=2$
$y^2+4y=5$
$y^2+4y-5=0$
$(y+5)(y-1)=0 \Rightarrow y=-5$ or $y=1$

Answer

$x=1,\ 2$ and $y=1,\ -5$

Q2. The cartesian product $A \times A$ has 9 elements among which $(-1, 0)$ and $(0,1)$ …

The cartesian product $A \times A$ has 9 elements among which $(-1, 0)$ and $(0,1)$ are found. Find the set $A$ and the remaining elements of $A \times A$ .

Show steps (6)

$n(A\times A)=n(A)\times n(A)=9$ , so $n(A)=3$ .
$(-1,0)\in A\times A \Rightarrow -1\in A,\ 0\in A$
$(0,1)\in A\times A \Rightarrow 0\in A,\ 1\in A$
So $A=\{-1,0,1\}$ .
$A\times A=\{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1),(1,0),(1,1)\}$
Remove the two given pairs to get the rest.

Answer

$A=\{-1,0,1\}$ ; remaining elements are $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$

Q3. Given that $f(x) = \begin{cases} \sqrt{x-1} & x \geq 1 \\ 4 & x < 1 \end{cases}$ . Find

Given that $f(x) = \begin{cases} \sqrt{x-1} & x \geq 1 \\ 4 & x < 1 \end{cases}$ . Find

(i)

$f(0)$

Here $x=0<1$ , so use $f(x)=4$ .

Answer

$f(0)=4$

(ii)

$f(3)$

Here $x=3\ge 1$ , so use $f(x)=\sqrt{x-1}$ .
$f(3)=\sqrt{3-1}=\sqrt{2}$

Answer

$f(3)=\sqrt{2}$

(iii)

$f(a+1)$ in terms of $a$ . (Given that $a \geq 0$ )

Given $a\ge 0$ , so $a+1\ge 1$ .
Use $f(x)=\sqrt{x-1}$ .
$f(a+1)=\sqrt{(a+1)-1}=\sqrt{a}$

Answer

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

Q4. Let $A = \{9,10,11,12,13,14,15,16,17\}$ and let $f : A \to N$ be defined by $f(n) =$ the highest prime …

Let $A = \{9,10,11,12,13,14,15,16,17\}$ and let $f : A \to N$ be defined by $f(n) =$ the highest prime factor of $n \in A$ . Write $f$ as a set of ordered pairs and find the range of $f$ .

Find the highest prime factor of each element of $A$ .
$9=3\times3\Rightarrow 3$ ; $10=2\times5\Rightarrow 5$ ; $11\Rightarrow 11$
$12=2^2\times3\Rightarrow 3$ ; $13\Rightarrow 13$ ; $14=2\times7\Rightarrow 7$
$15=3\times5\Rightarrow 5$ ; $16=2^4\Rightarrow 2$ ; $17\Rightarrow 17$
$f=\{(9,3),(10,5),(11,11),(12,3),(13,13),(14,7),(15,5),(16,2),(17,17)\}$

Answer

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

Q5. Find the domain of the function $f(x) = \sqrt{1 + \sqrt{1 - \sqrt{1 - x^2}}}$

The innermost root needs $1-x^2\ge 0$ .
$x^2\le 1 \Rightarrow -1\le x\le 1$
For such $x$ , $0\le\sqrt{1-x^2}\le 1$ , so $1-\sqrt{1-x^2}\ge 0$ — second root is defined.
Then $1+\sqrt{1-\sqrt{1-x^2}}\ge 1>0$ — outer root is defined.
So the only restriction is $-1\le x\le 1$ .

Answer

Domain $=[-1,1]$

Q6. If $f(x) = x^2$ , $g(x) = 3x$ and $h(x) = x-2$ , Prove that $(f \circ g) \circ h = f \circ (g \circ h)$ .

$(f\circ g)(x)=f(g(x))=f(3x)=(3x)^2=9x^2$
$((f\circ g)\circ h)(x)=(f\circ g)(h(x))=9(x-2)^2$
$(g\circ h)(x)=g(h(x))=g(x-2)=3(x-2)=3x-6$
$(f\circ(g\circ h))(x)=f(3x-6)=(3x-6)^2=9(x-2)^2$
Both sides give $9(x-2)^2$ for every $x$ .

Answer

$(f\circ g)\circ h = f\circ(g\circ h) = 9(x-2)^2$ , hence composition is associative here.

Q7. Let $A = \{1,2\}$ and $B = \{1,2,3,4\}$ , $C = \{5,6\}$ and $D = \{5,6,7,8\}$ . Verify whether $A \times C$ …

Let $A = \{1,2\}$ and $B = \{1,2,3,4\}$ , $C = \{5,6\}$ and $D = \{5,6,7,8\}$ . Verify whether $A \times C$ is a subset of $B \times D$ ?

$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$
$B\times D$ contains all pairs with first entry from $\{1,2,3,4\}$ and second from $\{5,6,7,8\}$ .
Each of $(1,5),(1,6),(2,5),(2,6)$ lies in $B\times D$ .
Since $A\subseteq B$ and $C\subseteq D$ , every element of $A\times C$ is in $B\times D$ .

Answer

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

Q8. If $f(x) = \dfrac{x-1}{x+1},\ x \neq -1$ show that $f(f(x)) = -\dfrac{1}{x}$ , provided $x \neq 0$ .

$f(f(x))=f\left(\dfrac{x-1}{x+1}\right)=\dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x-1}{x+1}+1}$
Numerator: $\dfrac{(x-1)-(x+1)}{x+1}=\dfrac{-2}{x+1}$
Denominator: $\dfrac{(x-1)+(x+1)}{x+1}=\dfrac{2x}{x+1}$
$f(f(x))=\dfrac{-2}{x+1}\times\dfrac{x+1}{2x}=\dfrac{-2}{2x}=-\dfrac{1}{x}$

Answer

$f(f(x))=-\dfrac{1}{x}$ (for $x\neq 0,\ x\neq -1$ ) — hence proved.

Q9. The functions $f$ and $g$ are defined by $f(x) = 6x + 8$ ; $g(x) = \dfrac{x-2}{3}$ .

The functions $f$ and $g$ are defined by $f(x) = 6x + 8$ ; $g(x) = \dfrac{x-2}{3}$ .

(i)

Calculate the value of $gg\left(\dfrac{1}{2}\right)$

$g\left(\dfrac{1}{2}\right)=\dfrac{\frac{1}{2}-2}{3}=\dfrac{-\frac{3}{2}}{3}=-\dfrac{1}{2}$
$gg\left(\dfrac12\right)=g\left(-\dfrac12\right)=\dfrac{-\frac12-2}{3}=\dfrac{-\frac52}{3}=-\dfrac{5}{6}$

Answer

$gg\left(\dfrac{1}{2}\right)=-\dfrac{5}{6}$

(ii)

Write an expression for $gf(x)$ in its simplest form.

$gf(x)=g(f(x))=g(6x+8)$
$=\dfrac{(6x+8)-2}{3}=\dfrac{6x+6}{3}$
$=2x+2$

Answer

$gf(x)=2x+2$

Q10. Write the domain of the following real functions

(i)

$f(x) = \dfrac{2x+1}{x-9}$

$f(x)$ is not defined when the denominator is zero.
$x-9=0\Rightarrow x=9$

Answer

Domain $=\mathbb{R}-\{9\}$

(ii)

$p(x) = \dfrac{-5}{4x^2+1}$

$4x^2\ge 0$ for all real $x$ , so $4x^2+1\ge 1>0$ .
The denominator is never zero.

Answer

Domain $=\mathbb{R}$

(iii)

$g(x) = \sqrt{x-2}$

The square root needs $x-2\ge 0$ .
$x\ge 2$

Answer

Domain $=[2,\infty)$

(iv)

$h(x) = x + 6$

$h(x)=x+6$ is a polynomial (linear) function.
It is defined for every real value of $x$ .

Answer

Domain $=\mathbb{R}$

Quick check

The range of the relation $R = \{(x, x^2) \mid x$ is a prime number less than 13 $\}$ is
1. (1) $\{2,3,5,7\}$
2. (2) $\{2,3,5,7,11\}$
3. (3) $\{4,9,25,49,121\}$
4. (4) $\{1,4,9,25,49,121\}$
Show answer
Answer: (3) $\{4,9,25,49,121\}$
1. Primes less than 13 are $2,3,5,7,11$ .
2. Range consists of the squares $x^{2}$ .
3. Range $=\{4,9,25,49,121\}$ .
Question 1 of 5
Let $f(x) = \sqrt{1 + x^2}$ then
1. (1) $f(xy) = f(x).f(y)$
2. (2) $f(xy) \geq f(x).f(y)$
3. (3) $f(xy) \leq f(x).f(y)$
4. (4) None of these
Show answer
Answer: (3) $f(xy) \leq f(x).f(y)$
1. $f(xy)=\sqrt{1+x^{2}y^{2}}$ .
2. $f(x)\cdot f(y)=\sqrt{(1+x^{2})(1+y^{2})}=\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}$ .
3. Since $x^{2}+y^{2}\ge 0$ , $\;1+x^{2}y^{2}\le 1+x^{2}+y^{2}+x^{2}y^{2}$ .
4. Hence $f(xy)\le f(x)\cdot f(y)$ .
Question 2 of 5
If $\{(a,8),(6,b)\}$ represents an identity function, then the value of $a$ and $b$ are respectively
1. (1) $(8,6)$
2. (2) $(8,8)$
3. (3) $(6,8)$
4. (4) $(6,6)$
Show answer
Answer: (1) $(8,6)$
1. For an identity function $f(x)=x$ , each pair is of the form $(x,x)$ .
2. $(a,8)\Rightarrow a=8$ .
3. $(6,b)\Rightarrow b=6$ .
Question 3 of 5
If $A = \{1,2\}$ , $B = \{1,2,3,4\}$ , $C = \{5,6\}$ and $D = \{5, 6, 7, 8\}$ then state which of the following statement is true.
1. (1) $(A \times C) \subset (B \times D)$
2. (2) $(B \times D) \subset (A \times C)$
3. (3) $(A \times B) \subset (A \times D)$
4. (4) $(D \times A) \subset (B \times A)$
Show answer
Answer: (1) $(A \times C) \subset (B \times D)$
1. $A=\{1,2\}\subset B=\{1,2,3,4\}$ and $C=\{5,6\}\subset D=\{5,6,7,8\}$ .
2. So every pair of $A\times C$ is also a pair of $B\times D$ .
3. Hence $(A\times C)\subset (B\times D)$ .
Question 4 of 5
If $f : A \to B$ is a bijective function and if $n(B) = 7$ , then $n(A)$ is equal to
1. (1) 7
2. (2) 49
3. (3) 1
4. (4) 14
Show answer
Answer: (1) 7
1. A bijection pairs each element of $A$ with exactly one element of $B$ .
2. So $n(A)=n(B)$ .
3. $n(A)=7$ .
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

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Q1. If the ordered pairs (x2−3x, y2+4y)(x^2-3x,\ y^2+4y)(x2−3x, y2+4y) and (−2,5)(-2,5)(−2,5) are equal, then find xxx …

Q2. The cartesian product A×AA \times AA×A has 9 elements among which (−1,0)(-1, 0)(−1,0) and (0,1)(0,1)(0,1) …

Q3. Given that f(x)={x−1x≥14x<1f(x) = \begin{cases} \sqrt{x-1} & x \geq 1 \\ 4 & x < 1 \end{cases}f(x)={x−1​4​x≥1x<1​ . Find

(i)

(ii)

(iii)

Q4. Let A={9,10,11,12,13,14,15,16,17}A = \{9,10,11,12,13,14,15,16,17\}A={9,10,11,12,13,14,15,16,17} and let f:A→Nf : A \to Nf:A→N be defined by f(n)=f(n) =f(n)= the highest prime …

Q5. Find the domain of the function f(x)=1+1−1−x2f(x) = \sqrt{1 + \sqrt{1 - \sqrt{1 - x^2}}}f(x)=1+1−1−x2​​​

Q6. If f(x)=x2f(x) = x^2f(x)=x2 , g(x)=3xg(x) = 3xg(x)=3x and h(x)=x−2h(x) = x-2h(x)=x−2 , Prove that (f∘g)∘h=f∘(g∘h)(f \circ g) \circ h = f \circ (g \circ h)(f∘g)∘h=f∘(g∘h) .

Q7. Let A={1,2}A = \{1,2\}A={1,2} and B={1,2,3,4}B = \{1,2,3,4\}B={1,2,3,4} , C={5,6}C = \{5,6\}C={5,6} and D={5,6,7,8}D = \{5,6,7,8\}D={5,6,7,8} . Verify whether A×CA \times CA×C …

Q8. If f(x)=x−1x+1, x≠−1f(x) = \dfrac{x-1}{x+1},\ x \neq -1f(x)=x+1x−1​, x=−1 show that f(f(x))=−1xf(f(x)) = -\dfrac{1}{x}f(f(x))=−x1​ , provided x≠0x \neq 0x=0 .

Q9. The functions fff and ggg are defined by f(x)=6x+8f(x) = 6x + 8f(x)=6x+8 ; g(x)=x−23g(x) = \dfrac{x-2}{3}g(x)=3x−2​ .

(i)

(ii)

Q10. Write the domain of the following real functions

(i)

(ii)

(iii)

(iv)

Quick check

Q1. If the ordered pairs $(x^2-3x,\ y^2+4y)$ and $(-2,5)$ are equal, then find $x$ …

Q2. The cartesian product $A \times A$ has 9 elements among which $(-1, 0)$ and $(0,1)$ …

Q3. Given that $f(x) = \begin{cases} \sqrt{x-1} & x \geq 1 \\ 4 & x < 1 \end{cases}$ . Find

Q4. Let $A = \{9,10,11,12,13,14,15,16,17\}$ and let $f : A \to N$ be defined by $f(n) =$ the highest prime …

Q5. Find the domain of the function $f(x) = \sqrt{1 + \sqrt{1 - \sqrt{1 - x^2}}}$

Q6. If $f(x) = x^2$ , $g(x) = 3x$ and $h(x) = x-2$ , Prove that $(f \circ g) \circ h = f \circ (g \circ h)$ .

Q7. Let $A = \{1,2\}$ and $B = \{1,2,3,4\}$ , $C = \{5,6\}$ and $D = \{5,6,7,8\}$ . Verify whether $A \times C$ …

Q8. If $f(x) = \dfrac{x-1}{x+1},\ x \neq -1$ show that $f(f(x)) = -\dfrac{1}{x}$ , provided $x \neq 0$ .

Q9. The functions $f$ and $g$ are defined by $f(x) = 6x + 8$ ; $g(x) = \dfrac{x-2}{3}$ .