# 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.