# 10th Maths Chapter 1 Exercise 1.5 Solutions

10th Standard Maths Chapter 1 Exercise 1.5 Relations and Functions Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

## 10th Maths Chapter 1 Exercise 1.5 Solutions

### 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) = 2x, g(x) = 2×2 – 1
- (iii) f(x) = x+63g(x) = 3 – x
- (iv) f(x) = 3 + x, g(x) = x – 4
- (v) f(x) = 4×2– 1,g(x) = 1 + x

**Solu.:**

(i) f(x) = x – 6, g(x) = x2

fog(x) = f(g(x)) = f(x2) = x2– 6 …………….. (1)

gof(x) = g(f(x)) = g(x – 6) = (x – 6)2

= x2 + 36 – 12x = x2 – 12x + 36 ……………… (2)

(1) ≠ (2)

∴ fog(x) ≠ gof(x)

(iii) f(x) = x+63 g(x) = 3 – x

(iv) f(x) = 3 + x, g(x) = x – 4

fog(x) = f(g(x)) = f(x – 4) = 3 + x – 4

= x – 1 ………… (1)

gof(x) = g(f(x)) = g(3 + x) = 3 + x – 4

= x – 1 ……………… (2)

Here fog(x) = gof(x)

(v) f(x) = 4×2 – 1, g(x) = 1 + x

fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1

= 4(1 + x2 + 2x) – 1 = 4 + 4×2 + 8x – 1

= 4×2 + 8x + 3 ……………. (1)

gof(x) = g(f(x)) = g(4×2 – 1)

= 1 + 4×2 – 1 = 4×2 …………….. (2)

(1) ≠ (2)

∴ fog(x) ≠ gof(x)

### 2. Find the value of k, such that f o g = g o f

**(i) f(x) = 3x + 2, g(x) = 6x – k
**

**Answer:**

f(x) = 3x + 2 ;g(x) = 6x – k

fog = f[g(x)]

= f (6x – k)

= 3(6x – k) + 2

= 18x – 3K + 2

g0f= g [f(x)]

= g (3x + 2)

= 6(3x + 2) – k

= 18x + 12 – k

But given fog = gof.

18x – 3x + 2 = 18x + 12 – k

-3k + 2 = 12 – k

-3 k + k = 12-2

-2k = 10

k = −102 = -5

The value of k = -5

**(ii) f(x) = 2x – k, g(x) = 4x + 5
**

**Answer:**

f(x) = 2x – k ; g(x) = 4x + 5

fog = f[g(x)]

= f(4x + 5)

= 2(4x + 5) – k

= 8x + 10 – k

gof = g [f(x)]

= g(2x – k)

= 4(2x – k) + 5

= 8x – 4k + 5

But fog = gof

8x + 10 – k = 8x – 4k + 5

-k + 4k = 5 – 10

3k = -5

k = −53

The value of k = −53

### 3. if f(x) = 2x – 1, g(x) = x+12, show that fog = gof = x

**Solu.:**

### 4. (i) If f (x) = x2 – 1, g(x) = x – 2 find a, if g o f(a) = 1.

(a) Find k, if f(k) = 2k -1 and

fof (k) = 5.

**Solu:
**(i) f(x) = x2 – 1 ; g(x) = x – 2 .

gof = g [f(x)]

= g(x2 – 1)

= x2 – 1 – 2

= x2 – 3

given gof (a) = 1

a2 – 3 = 1 [But go f(x) = x2 – 3]

a2 = 4

a = 4–√ = ± 2

The value of a = ± 2

**(ii) f(k) = 2k – 1 ; fof(k) = 5
**fof = f[f(k)]

= f(2k – 1)

= 2(2k – 1) – 1

= 4k – 2 – 1

= 4k – 3

fof (k) = 5

4k – 3 = 5

4k = 5 + 3

4k = 8

k = 84 = 2

The value of k = 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

**Solu.:
**f(x) = 2x + 1

g(x) = x2

fog(x) = fg(x)) = f(x2) = 2×2 + 1

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

= 4×2 + 4x + 1

Range of fog is

{y/y = 2×2 + 1, x ∈ N}

Range of gof is

{y/y = (2x + 1)2, x ∈ N}.

### 6. Let f(x) = x2 – 1. Find (i) fof (ii) fofof

**Solu.:
**f(x) = x2 – 1

(i) fof = f[f{x)]

= f(x2 – 1)

= (x2 – 1)2 – 1

= x4 – 2×2 + 1 – 1

= x4 – 2×2

(ii) fofof = fof[f(x)]

= fof (x2 – 1)

= f(x2 – 1)2 – 1

= f(x4 – 2×2 + 1 – 1)

= f (x4 – 2×2)

fofof = (x4 – 2×2)2 – 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?

**Solu.:
**f(x) = x5

g(x) = x4

fog = fog(x) = f(g(x)) = f(x4)

= (x4)5 = x20

f is one-one, g is not one-one.

∵ g(1) = 14 = 1

g(-1) = ( -1)4 = 1

Different elements have same images

fog is not one-one. [∵ fog (1) = fog (-1) = 1]

### 8. Consider the functions f(x), g(x), h(x) as given below. Show that

(f o g) o h = f o(g o h) 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

**Solu.:**

**(i) f(x) = x – 1, g (x) = 3x + 1, h(x) = x2
**fog (x) = f[g(x)]

= f(3x + 1)

= 3x + 1 – 1

fog = 3x

(fog) o h(x) = fog [h(x)] ,

= fog (x2)

= 3(x2)

(fog) oh = 3×2 …..(1)

goh (x) = g[h(x)]

= g(x2)

= 3(x2) + 1

= 3×2 +1

fo(goh) x = f [goh(x)]

= f[3×2 + 1]

= 3×2 + 1 – 1

= 3×2 ….(2)

From (1) and (2) we get

(fog) oh = fo (goh)

Hence it is verified

**(ii) f(x) = x2 ; g (x) = 2x and h(x) = x + 4
**(fog) x = f[g(x)]

= f (2x)

= (2x)2

= 4×2

(fog) oh (x) = fog [h(x)]

= fog (x + 4)

= 4(x + 4)2

= 4[x2 + 8x + 16]

= 4×2 + 32x + 64 …. (1)

goh (x) = g[h(x)]

= g(x + 4)

= 2(x + 4)

= 2x + 8

fo(goh) x = fo [goh(x)]

= f[2x + 8]

= (2x + 8)2

= 4×2 + 32x + 64 …. (2)

From (1) and (2) we get

(fog) oh = fo(goh)

**(iii) f(x) = x – 4 ; g (x) = x2; h(x) = 3x – 5
**fog (x) = f[g(x)]

= f(x2)

= x2 – 4

(fog) oh (x) = fog [h(x)]

= fog (3x – 5)

= (3x – 5)2 – 4

= 9×2 – 30x + 25 – 4

= 9×2 – 30x + 21 ….(1)

goh (x) = g[h(x)]

= g(3x – 5)

= (3x – 5)2

= 9×2 + 25 – 30x

fo(goh)x = f[goh(x)]

= f[9×2 – 30x + 25]

= 9×2 – 30x + 25 – 4

= 9×2 – 30x + 21 ….(2)

From (1) and (2) we get

(fog) oh = fo(goh)

### 9. Let f ={(-1, 3),(0, -1),(2, -9)} be a linear function from Z into Z . Find f(x).

**Solu.:
**f ={(-1, 3), (0, -1), 2, -9)

f(x) = (ax) + b ………… (1)

is the equation of all linear functions.

∴ f(-1) = 3

f(0) = -1

f(2) = -9

f(x) = ax + b

f(-1) = -a + b = 3 …………… (2)

f(0) = b = -1

-a – 1 = 3 [∵ substituting b = – 1 in (2)]

-a = 4

a = -4

The linear function is -4x – 1. [From (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) = 31 is linear.

**Solu.:**

Given C(t) = 3t

C(at1) = 3at1 …. (1)

C(bt2) = 3 bt2 …. (2)

Add (1) and (2)

C(at1) + C(bt2) = 3at1 + 3bt2

C(at1 + bt2) = 3at1 + 3bt2

= Cat1 + Cbt2 [from (1) and (2)]

∴ C(at1 + bt2) = C(at1 + bt2)

The superposition principle is satisfied.

∴ C(t) = 3t is a linear function.