Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5

Tamilnadu State Board Class 10 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) = \(\frac { 2 }{ x } \), g(x) = 2x²– 1
(iii) f(x) = \(\frac { x+6 }{ 3 } \)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)

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

(iii) f(x) = \(\frac { x+6 }{ 3 } \),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 fog = gof
(i) f(x) = 3x + 2, g(x) = 6x – k
(ii) f(x) = 2x – k, g(x) = 4x + 5
Solution:
(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)

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)

3k = -5
k = \(\frac { -5 }{ 3 } \)

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

Question 4.
(i) If f (x) = x² – 1, g(x) = x – 2 find a, if gof(a) = 1.
(ii) Find k, if f(k) = 2k – 1 and fof (k) = 5.
Solution:
(i) 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
(ii) f(k) = 2k – 1
fo f(k) = 5
f(f(k)m = f(2k – 1) = 5
⇒ 2(2k – 1) – 1 = 5
4 k – 2 – 1 = 5 ⇒ 4k = 8
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.
If f(x) = x² – 1. Find (i)f(x) = x² – 1, (ii)fofof
Solution:
(i) f(x) = x² – 1
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

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 (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
Solution:
(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) = 3×2 …(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.

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), (o,-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 = -A
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.
Solution:
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] = 3 at₁ + 3 bt₂
= a(3t₁) + b(3t₂) = a[C(t₁) + b(Ct₆₂)]
∴ Superposition principle is satisfied.
Hence C(t) = 3t is linear function.

Samacheer Kalvi 10th Maths Book Answers