TN 10th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.5

TN 10th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.5

Question 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) = 2/x, g(x) = 2x² – 1
(iii) f(x) = x+6/3 g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4
(v) f(x) = 4x²– 1,g(x) = 1 + x
Solution:
(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+6/3 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)

Question 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 = −10/2 = -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 = −5/3
The value of k = −5/3

Question 3.
if f(x) = 2x – 1, g(x) = x+1/2, show that fog = gof = x
Solution:
f(x) = 2x – 1, g(x) = x+1/2, fog = gof = x

Question 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.
Answer:
(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 = 8/4 = 2
The value of k = 2

Question 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
Solution:
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}.

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

Question 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?
Solution:
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]

Question 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
Answer:
(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)

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

Question 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.
Answer:
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)
Superposition principle is satisfied.
∴ C(t) = 3t is a linear function.