# 10th Maths Book Back Relations and Functions Ex 1.5

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 1 – Relations and Functions Ex 1.5 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 1 Relation and Functions consist of 6 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 1.5 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 1 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

### Exercise 1.5 Relations and Functions

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) = 2x– 1
(iii) f(x) = x+63g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4
(v) f(x) = 4x2– 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) = 4x2 – 1, g(x) = 1 + x
fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1
= 4(1 + x2 + 2x) – 1 = 4 + 4x2 + 8x – 1
= 4x2 + 8x + 3 ……………. (1)
gof(x) = g(f(x)) = g(4x2 – 1)
= 1 + 4x2 – 1 = 4x2 …………….. (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
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
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) = 2x2 + 1
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)2
= 4x2 + 4x + 1
Range of fog is
{y/y = 2x2 + 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 – 2x2 + 1 – 1
= x4 – 2x2

(ii) fofof = fof[f(x)]
= fof (x2 – 1)
= f(x2 – 1)2 – 1
= f(x4 – 2x2 + 1 – 1)
= f (x4 – 2x2)
fofof = (x4 – 2x2)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 = 3x2 …..(1)
goh (x) = g[h(x)]
= g(x2)
= 3(x2) + 1
= 3x2 +1
fo(goh) x = f [goh(x)]
= f[3x2 + 1]
= 3x2 + 1 – 1
= 3x2 ….(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
= 4x2
(fog) oh (x) = fog [h(x)]
= fog (x + 4)
= 4(x + 4)2
= 4[x2 + 8x + 16]
= 4x2 + 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
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 ….(1)
goh (x) = g[h(x)]
= g(3x – 5)
= (3x – 5)2
= 9x2 + 25 – 30x
fo(goh)x = f[goh(x)]
= f[9x2 – 30x + 25]
= 9x2 – 30x + 25 – 4
= 9x2 – 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)