Chapter 1 Relations and Functions Ex 1.3 Solutions
Question - 1 : - Letf: {1,3,4} –> {1,2, 5} and g : {1, 2,5} –> {1,3} be given by f = {(1, 2),(3,5), (4,1) and g = {(1,3), (2,3), (5,1)}. Write down g of.
Answer - 1 : -
f={(1,2),(3,5),(4,1)}
g= {(1,3),(2,3),(5,1)}
f(1) = 2, g(2) = 3 => gof(1) = 3
f(3) = 5, g(5)= 1 =>gof(3) = 1
f(4) = 1, g(1) = 3 => gof(4) = 3
=> gof= {(1,3), (3,1), (4,3)}
Question - 2 : - Letf, g and h be functions from R to R. Show that (f + g) oh = foh + goh, (f • g)oh = (foh) • (goh)
Answer - 2 : -
f+ R –> R, g: R –> R, h: R –> R
(i) (f+g)oh(x)=(f+g)[h(x)]
= f[h(x)]+g[h(x)]
={foh} (x)+ {goh} (x)
=>(f + g) oh = foh + goh
(ii) (f • g) oh (x) = (f • g) [h (x)]
= f[h (x)] • g [h (x)]
= {foh} (x) • {goh} (x)
=> (f • g) oh = (foh) • (goh)
Question - 3 : - Findgof and fog, if
(i) f (x) = |x| and g (x) = |5x – 2|
(ii) f (x) = 8x³ and g (x) = 
.
Answer - 3 : -
(i)f(x) = |x|, g(x) = |5x – 2|
(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|
(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|
(ii) f(x) = 8x³ and g(x) =
(a) gof(x) = g(f(x)) = g(8x³) = 
=2x
(b) fog (x) = f(g (x))=f() = 8.()³ = 8x
Question - 4 : - If 
, show that fof (x) = x, for all 
. What is the inverse of f?
Answer - 4 : - It is given that
(a) fof (x) = f(f(x)) = 
Question - 5 : - Statewith reason whether following functions have inverse
Answer - 5 : - (i) f: {1,2,3,4}–>{10} with f = {(1,10), (2,10), (3,10), (4,10)}
(ii) g: {5,6,7,8}–>{1,2,3,4} with g = {(5,4), (6,3), (7,4), (8,2)}
(iii) h: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}
Solution
(i) f:{1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From thegiven definition of f, we can seethat f is a many one functionas: f(1) = f(2)= f(3) = f(4)= 10
∴f is notone-one.
Hence,function f does not have aninverse.
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From thegiven definition of g, it is seenthat g is a many one functionas: g(5) = g(7)= 4.
∴g is notone-one,
Hence,function g does not have an inverse.
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It isseen that all distinct elements of the set {2, 3, 4, 5} have distinct imagesunder h.
∴Function h isone-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, thereexists an element x in the set{2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function.Hence, h has an inverse.
Question - 6 : - Showthat f: [-1,1] –> R, given by f(x) = 
is one-one. Find the inverse of the functionf: [-1,1] –> Range f.
Hint – For y ∈ Range f, y = f (x) = for some x in [- 1,1], i.e., x = 
Answer - 6 : - f: [−1, 1] → R is given as
Let f(x) = f(y).
∴ f is a one-onefunction.
It isclear that f: [−1, 1] → Range f is onto.
∴ f:[−1, 1] → Range f is one-oneand onto and therefore, the inverse of the function:
f: [−1, 1] → Range f exists.
Let g: Range f →[−1, 1] be the inverse of f.
Let y be an arbitrary element ofrange f.
Since f: [−1, 1] → Range f isonto, we have:
Now, let us define g: Range f →[−1, 1] as
∴gof =
and fog = 
f−1 = g
⇒
Answer - 7 : -
f:R—>R given by f(x) = 4x + 3
f (x1) = 4x1 + 3, f (x2) = 4x2 +3
If f(x1) = f(x2), then 4x1 + 3 = 4x2 +3
or 4x1 = 4x2 or x1 = x2
f is one-one
Also let y = 4x + 3, or 4x = y – 3
∴ 
For each value of y ∈ R and belonging to co-domain of y has apre-image in its domain.
∴ f is onto i.e. f is one-one and onto
f is invertible and f-1 (y) = g (y) = 
Question - 8 : - Considerf: R+ –> [4, ∞] given by f (x) = x² + 4. Show that f isinvertible with the inverse f-1 of f given by f-1 (y)= √y-4 , where R+ is the set of all non-negative real numbers.
Answer - 8 : -
f(x1)= x12 + 4 and f(x2) = x22 +4
f(x1) = f(x2) => x12 + 4 =x22 + 4
or x12 = x22 => x1 =x2 As x ∈ R
∴ x>0, x12 = x22 =>x1 = x2 =>f is one-one
Let y = x² + 4 or x² = y – 4 or x = ±√y-4
x being > 0, -ve sign not to be taken
x = √y – 4
∴ f-1 (y) = g(y) = √y-4 ,y ≥ 4
For every y ≥ 4, g (y) has real positive value.
∴ The inverse of f is f-1 (y)= √y-4
Question - 9 : - Considerf: R+ –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that fis invertible with
Answer - 9 : -
f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6x − 5.∴f isonto, thereby range f = [−5,∞).
Letus define g: [−5, ∞) → R+ as
We now have:
∴
and
Hence, f isinvertible and the inverse of f isgiven by
Hint – suppose g1 and g2 aretwo inverses of f. Then for all y∈Y, fog1(y)=Iy(y)=fog2(y).Useone-one ness of f.
Answer - 10 : -
Iff is invertible gof (x) = Ix and fog (y) = Iy
∴ f is one-one and onto.
Let there be two inverse g1 and g2
fog1 (y) = Iy, fog2 (y) = Iy
Iy being unique for a given function f
=> g1 (y) = g2 (y)
f is one-one and onto
f has a unique inverse.