10th Maths - Book Back Answers - Chapter 1 Exercise 1.5 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 1 Exercise 1.5 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 1 Exercise 1.5 : Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 1 Exercise 1.5 – 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 Exercise 1.5 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

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

Chapter 1 Relations and functions Ex 1.5

1. Using the functions f and g given below, find fog and gof Check whether fog = gof.

(i) f(x) = x – 6, g(x) = x²

(ii) f(x) = $\frac{\mathbf{2}}{\mathbf{x}}$, g(x) = 2x² – 1

(iii) f(x) =$\frac{\mathbf{x}\mathbf{+}\mathbf{6}}{\mathbf{3}}$3, g(x) = 3 – x

(iv) f(x) = 3 + x, g(x) = x – 4

(v) f(x) = 4x² – 1,g(x) = 1 + x

Answer Key:

(i) f(x) = x – 6, g(x) = x²
f(x) = x – 6, g(x) = x²
fog = fog (x)
= f(g(x))

fog = f(x)²
= x² – 6
gof = go f(x)
= g(x – 6)
= (x – 6)²
= x² – 12x + 36
fog ≠ gof

 

(ii) f(x) = $\frac{\mathbf{2}}{\mathbf{x}}$, g(x) = 2x² – 1
f(x) – $\frac{\mathbf{2}}{\mathbf{x}}$; g(x) = 2x² – 1
fag = f[g (x)]
= f(2x² – 1)
$\mathbf{=}\frac{\mathbf{2}}{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$
gof = g [f(x)]

= g ($\frac{\mathbf{2}}{\mathbf{x}}$)
= 2 ($\frac{\mathbf{2}}{\mathbf{x}}$)² – 1
=2×$\frac{\mathbf{4}}{{\mathbf{x}}^{\mathbf{2}}}$−1
=$\frac{\mathbf{8}}{{\mathbf{x}}^{\mathbf{2}}}$−1
fog ≠ gof

(iii) f(x) =$\frac{\mathbf{x}\mathbf{+}\mathbf{6}}{\mathbf{3}}$, g(x) = 3 – x
f(x) =$\frac{\mathbf{x}\mathbf{+}\mathbf{6}}{\mathbf{x}}$, g(x) = 3 – x
fog = f[g(x)]
= f(3 – x) 

$\mathbf{=}\frac{\mathbf{3}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{6}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{9}\mathbf{-}\mathbf{x}}{\mathbf{3}}$

$\frac{\mathbf{gof=g[f(x)]}}{}$

$\frac{\mathbf{<span>fog</span>}}{}$≠gof

(iv) f(x) = 3 + x, g(x) = x – 4
f(x) = 3 + x; g(x) = x – 4
fog = f[g(x)]
= f(x – 4)
= 3 + x – 4
= x – 1
gof = g[f(x)]
= g(3 + x)
= 3 + x – 4
= x – 1
fog = gof

 

(v) f(x) = 4x² – 1, g(x) = 1 + x
f(x) = 4x² – 1; g(x) = 1 + x
fog = f[g(x)]
= 4(1 + x)
= 4(1 + x)² – 1
= 4[1 + x² + 2x] – 1
= 4 + 4x² + 8x – 1
= 4x² + 8x + 3
gof = g [f(x)]
= g (4x² – 1)
= 1 + 4x² – 1
= 4x²
fog ≠ gof

 

2. Find the value of k, such that fog = gof
(i) f(x) = 3x + 2, g(x) = 6x – k
(ii) f(x) = 2x – k, g(x) = 4x + 5

Answer Key:

(i) f(x) = 3x + 2, g(x) = 6x – k
fog(x) = f(g(x)) = f(6x – k) = 3(6x – k) + 2
= 18x – 3k + 2 …………… (1)
gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) – k
= 18x + 12 – k ……………. (2)
(1) = (2)
⇒ 18x – 3k + 2 = 18x + 12 – k
2k = -10
k = -5

 

(ii) f(x) = 2x – k, g(x) = 4x + 5
fog(x) = f(g(x)) = f(4x + 5) = 2(4x + 5) – k
= 8x + 10 – k ……………… (1)
gof(x) = g(f(x)) = g(2x – k) = 4(2x – k) + 5
= 8x – 4k + 5 ……………. (2)
(1) = (2)
⇒ 8x + 10 – k = 8x – 4k + 5
3k = -5
k =$\frac{\mathbf{-}\mathbf{5}}{\mathbf{3}}$

 

3. If f(x) = 2x – 1, g(x) =$\frac{\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{2}}$, show that fog = gof = x
f(x) = 2x – 1 ; g(x) =$\frac{\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{2}}$

fog = f[g(x)]

=f[$\frac{\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{2}}$]

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

=x+1-1

=x

gof=g[f(x)]

=g(2x-1)

$\frac{\mathbf{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mi\; mathvariant="bold"><br></mi></math>}}{}$

∴ fog=gof=x

Hence it is proved.

 

4. If f (x) = x² – 1, g(x) = x – 2 find a, if gof(a) = 1.
Answer Key:
f(x) = x² – 1, g(x) = x – 2
Given gof(a) = 1
gof(x) = g(f(x)
= g(x² – 1) = x² – 1 – 2
= x² – 3
gof(a) ⇒ a² – 3 = 1 =+ a² = 4
a = ± 2

5.  Let A, B, C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g: B → C be defined by g(x) = x². Find the range of fog and gof.
Answer Key:
f(x) = 2x + 1; g(x) = x²
fog = f[g(x)]
= f(x²)
= 2x² + 1
2x² + 1 ∈ N
g o f = g [f(x)]
= g (2x + 1)
g o f = (2x + 1)²
(2x + 1)² ∈ N
Range = {y/y = 2x² + 1, x ∈ N};
{y/y = (2x + 1)², x ∈ N)

 

6. If f(x) = x² – 1. Find (i)fof, (ii)fofof
Answer Key:
(i) fof
fof(x) = f(fx)) = f(x² – 1)
= (x² – 1 )² – 1;
= x⁴ – 2x² + 1 – 1
= x⁴ – 2x²
 

(ii) fofof = f o f(f(x))
= f o f (x⁴ – 2x²)
= f(f(x⁴ – 2x²))
= (x⁴ – 2x²)² – 1
= x⁸ – 4x⁶ + 4x⁴ – 1

 

7. If f:R → R and g:R → R are defined by f(x) = x⁵ and g(x) = x⁴ then check if f, g are one – one and fog is one – one?
Answer key:
f(x) = x⁵ – It is one – one function
g(x) = x⁴ – It is one – one function
fog = f[g(x)]
= f(x⁴)
= (x⁴)⁵
fag = x²⁰
It is also one-one function. 

 

8. Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case.
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x²
(ii) f(x) = x², g(x) = 2x and h(x) = x + 4
(iii) f(x) = x – 4, g(x) = x² and h(x) = 3x – 5
Answer key:
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x²
f(x) = x – 1
g(x) = 3x + 1
f(x) = x²
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x
(fog)oh = (fog)(h(x)) = (fog)(x² ) = 3²  ……………. (1)
RHS = fo(goh)
goh = g(h(x)) = g(x²) = 3x² + 1
fo(goh) = f(3x² + 1) = 3x² + 1 – 1= 3x² ………… (2)
LHS = RHS Hence it is verified.

 

(ii) f(x) = x², g(x) = 2x, h(x) = x + 4
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(2x) = (2x)² = 4x²
(fog)oh = (fog) h(x) = (fog) (x + 4)
= 4(x + 4)² = 4(x² + 8x+16)
= 4x² + 32x + 64 ………….. (1)
RHS = fo(goh) goh = g(h(x)) = g(x + 4)
= 2(x + 4) = (2x + 8)
fo(goh) = f(goh) = f(2x + 8) = (2x + 8)²
= 4x² + 32x + 64 ……………… (2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh) It is proved.

 

(iii) f(x) = x – 4, g(x) = x², h(x) = 3x – 5
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(x²) = x² – 4
(fog)oh = (fog)(3x – 5) = (3x – 5)² – 4
= 9x² – 30x + 25 -4
= 9x² – 30x + 21 ………….. (1)
∴ RHS = fo(goh)
(goh) = g(h(x)) = g(3x – 5) = (3x – 5)²
= 9x² – 30x + 25
fo(goh) = f(9x² – 30 x + 25)
= 9x² – 30x + 25 – 4
= 9x² – 30x + 21 …………… (2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh)
It is proved.

 

9. Let f = {(-1, 3), (0, -1), (2, -9)} be a linear function from Z into Z. Find f(x).
Answer Key:
The linear equation is f(x) = ax + b
f(-1) = 3
a(-1) + b = 3
-a + b = 3 ….(1)
f(0) = -1
a(0) + b = -1
0 + b = -1
b = -1
Substitute the value of b = -1 in (1)
-a – 1 = 3
-a = 3 + 1
-a = 4
a = -4
∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)

 

10. In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at₁ + bt₂) = aC(t₁) + bC(t₂), where a,b are constants. Show that the circuit C(t) = 3t is linear.
Answer Key:

Given C(t) = 3t. To prove that the function is linear
C(at₁) = 3a(t₁)
C(bt₂) = 3 b(t₂)
C(at₁ + bt₂) = 3 [at₁ + bt₂] = 3at₁ + 3bt₂
= a(3t₁) + b(3t₂) = a[C(t₁) + b(Ct₂)]
∴ Superposition principle is satisfied.
Hence C(t) = 3t is linear function.

 

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