10th Maths - Book Back Answers - Chapter 1 Unit Exercise 1 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 1 unit Exercise 1 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 1 Unit Exercise 1: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 1 Unit Exercise 1 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 1 unit Exercise 1 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 1 Relations and functions Unit Ex 1

1. If the ordered pairs (x² – 3x, y² + 4y) and (-2, 5) are equal, then find x and y.
Answer Key:
(x² – 3x, y² + 4y) = (-2, 5)
x² – 3x = -2
x² – 3x + 2 = 0
(x – 2) (x – 1) = 0
x – 2 = 0 or x – 1 = 0
x = 2 or 1

y² + 4y = 5
y² + 4y – 5 = 0
(y + 5) (y – 1) = 0
y + 5 = 0 or y – 1 = 0
y = -5 or y = 1
The value of x = 2, 1
and 7 = -5, 1

2. The Cartesian product 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.

Answer Key:

A = {-1, 0, 1}, B = {1, 0, -1}
A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}

 

3. 

Find

(i) f(0)      (ii) f(3)    (iii) f(a+1) in terms of a (Given that a >0)

Answer Key:

f(x)=$\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{;}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{4}$

(i) f(0)=4

(ii) f(3)=$\sqrt{\mathbf{3}\mathbf{-}\mathbf{1}}\mathbf{=}\sqrt{\mathbf{2}}$

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

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

Answer Key:

A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
Range = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}

 

5. Find the domain of the function

$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{1}\mathbf{+}\sqrt{\mathbf{1}\mathbf{-}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}}$ 

Answer Key:

$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{1}\mathbf{+}\sqrt{\mathbf{1}\mathbf{-}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}}$  

$\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\sqrt{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}$

⇒x=-1(or)x=1

=-1≤x≤ 1

Domain of f(x) = {-1, 0, 1}

6. If f(x)= x², g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).

Answer Key:

f(x) = x²
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)² = 9x²
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)² = 9(x² – 4x + 4)
= 9x² – 36x + 36 ……………. (1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x – 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)²
= 9x² – 36x + 36 ………….. (2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.

 

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

Answer Key:

Given A = {1, 2}
B = {1, 2, 3, 4}
C = {5,6}
D = {5,6, 7,8}
A × C = {1,2} × {5,6}
= {(1,5) (1,6) (2, 5) (2, 6)}
B × D = {1,2, 3, 4} × {5, 6, 7, 8}
= {(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)}
∴ A × C ⊂ B × D
Hence it is verified

8. If f(x) = $\frac{\mathbf{x}\mathbf{-}\mathbf{1}}{\mathbf{x}\mathbf{+}\mathbf{1}}$,x≠1 Show that f(f(x)) = -$\frac{\mathbf{1}}{\mathbf{x}}$, Provided x ≠ 0.
Answer Key:

9. The functions f and g are defined by f{x) = 6x + 8; g(x) = $\frac{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{3}}$
(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]
(a) Write an expression for gf (x) in its simplest form.
Answer Key:
f(x) = 6x + 8 ; g(x) =$\frac{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{3}}$

10. Write the domain of the following real functions

(i)$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}}{\mathbf{x}\mathbf{-}\mathbf{9}}$

(ii)$\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{5}}{\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}$

(iii) $\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}$

(iv) h(x)=x+6

Answer Key:

(i)$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}}{\mathbf{x}\mathbf{-}\mathbf{9}}$

If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]

(ii)$\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{5}}{\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}$

 p(x) is defined for all values of x. So domain is x ∈ R.

 

(iii) $\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}$ 

When x < 2 g(x) becomes complex. But given "g" is real valued function .

So x>2

Domain x ∈ (2, ∝)

 

(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.