RBSE Class 12 Maths Chapter 1 – Composite Functions Important questions and solutions are given here. All these questions and solutions have detailed explanations, which would be useful for the students to understand clearly. The RBSE Class 12 important questions and solutions, available at BYJU’S, will help in scoring maximum marks in the exams.
Chapter 1 of RBSE Class 12 contains three exercises, such that they cover several important concepts of functions, namely composition of a function, inverse function. Various questions on types of functions, i.e. constant, identity and equal functions are covered here. Besides, different types of properties are included in this chapter, such as domain, co-domain, range of a function.
RBSE Maths Chapter 1: Exercise 1.1 Textbook Important Questions and Solutions
Question 1: If f: R → R and g: R → R are the two functions defined below, then find (f∘g)(x) and (g∘f)(x).
(i) f(x) 2x + 3, g(x) = x2 + 5
(ii) f(x) = x, g(x) = |x|
Solution:
(i) Given,
f(x) = 2x + 3 and g(x) = x2 + 5
(f∘g)(x) = f(g(x))
= f(x2 + 5)
= 2(x2 + 5) + 3
= 2x2 + 10 + 3
= 2x2 + 13
(g∘f)(x) = g(f(x))
= g(2x + 3)
= (2x + 3)2 + 5
= 4x2 + 9 + 12x + 5
= 4x2 + 12x + 14
(ii) Given,
f(x) = x and g(x) = |x|
(f∘g)(x) = f(g(x))
= f(|x|) = |x|
(g∘f)(x)=g(f(x))
= g(x) = |x|
Question 2: If A = {a, b, c}, B = {u, v. w} and f: A → B and g: B → A are defined as f = {(a, v), (b, u), (c, w)}; g = {(u, b), (v, a), (w, c)}, then find (f∘g) and (g∘f).
Solution:
Given,
f= {(a, v), (b, u), (c, w)}
g= {(u, b), (v, a), (w, c)}
Thus,
f(a)= v and g(u) = b
f(b)= u and g(v) = a
f(c)= w and g(w) = C
Now,
(f∘g)(x) = f(g(x)]
(f∘g)(u) = f(g(u)] = f(b) = u
(f∘g)(v) = f(g(v)] = f(a) = v
(f∘g)(w)= f[g(w)] = f(c) = w
Therefore, (f∘g) = {(u, u), (v, v), (w, w)}
(g∘f)(a) = g[f(a)] = g(v) = a
(g∘f)(b) = g[fb)] = g(u) = b
(g∘f)(c) = g[(c)] = g(w) = c
Therefore, (g∘f) = {(a, a), (b, b), (c, c)}
Question 3: If f: R+ → R+ and g: R+ → R+ are defined as f(x) = x2 and g(x) = √x, then find (g∘f) and (f∘g). Are they equal?
Solution:
Given,
f : R+ → R+, f(x) = x2
g : R+ → R+, g(x) = √x
(gof)(x) = g[f(x)] = g(x2) = √(x2) = x
(fog)(x) = f[g(x)] = f(√x) = (√x)2 = x
Thus, (fog)(x) = (gof)(x) = x, ∀ x ∈ R+
Hence, (fog) and (gof) are identity functions.
Question 4: If f: R → R and g: R → R are two functions such that f(x) = 3x + 4 and g(x) = 1/3(x – 4), then find (f∘g)(x) and (g∘f)(x). Also, find (g∘g)(1).
Solution:
Given,
f: R → R; f(x) = 3x + 4
g: R → R; g(x) = ⅓(x – 4)
(f∘g)(x) = f[g(x)]
= f[(x – 4)/3]
= 3[(x – 4)/3] + 4
= x – 4 + 4
= x
(g∘f)(x) = g[f(x)]
= g(3x + 4)
= (3x + 4 – 4)/3
= 3x/3
= x
Thus, (f∘g)(x) = (g∘f)(x) = x
Now,
(g∘g)(x) = g[g(x)]
(g∘g)(1) = g[g(1)]
= g[(1 – 4)/3]
= g(-3/3)
= g(-1)
= (-1 – 4)/3
= -5/3
Therefore, (g∘g)(1) = -5/3
Question 5: If three functions f, g, h defined from R to R in such a way that f(x) = x2, g(x) = cos x and h(x) = 2x + 3, then find the value of [h∘(g∘f)]√2π.
Solution:
Given,
f(x) = x2, g(x) = cos x, h(x) = 2x + 3
[h∘(g∘f)](x) = (h∘g)[f(x)]= h[g{f(x)}]
= h[g(x2)]
= h(cos x2)
= 2 cos x2 + 3
[h∘(g∘f)]√2π = 2 cos (√2π)2 + 3= 2 cos 2π + 3
= 2 (1) + 3
= 2 + 3
= 5
Question 6: If A = {1, 2, 3, 4}, f: R → R, f(x) = x2 + 3x + 1, g: R → R, g(x) = 2x – 3, then find
(i) (f∘g)(x)
(ii) (g∘f)(x)
(iii) (f∘f)(x)
(iv) (g∘g)(x)
Solution:
Given,
f : R→ R, f(x) = x2 + 3x + 1
g : R → R, g(x) = 2x – 3
(i) (f∘g)(x) = f[g(x)]
= f(2x – 3)
= (2x – 3)2 + 3(2x – 3) + 1
= 4x2 – 12x + 9 + 6x – 9 + 1
= 4x2 – 6x + 1
(ii) (g∘f)(x) = g[f(x)]
= g(x2 + 3x + 1)
= 2(x2 + 3x + 1) – 3
= 2x2 + 6x + 2 – 3
= 2x2 + 6x – 1
(iii) (f∘f)(x) = f[f(x)]
= f(x2 + 3x + 1)
= (x2 + 3x + 1)2 + 3(x2 + 3x + 1)+1
= x4 + 9x2 + 1 + 6x3 + 6x + 2x2 + 3x2 + 9x + 3 + 1
= x4 +6x3 + 14x2 + 15x + 5
(iv) (g∘g)(x) = g[g(x)]
= g(2x – 3)
= 2(2x – 3) – 3
= 4x – 6 – 3
= 4x – 9
RBSE Maths Chapter 1: Exercise 1.2 Textbook Important Questions and Solutions
Question 7: If A = {1, 2, 3, 4}, B = {a, b, c}, then find four one-one onto functions from A to B and also find their inverse function.
Solution:
Given
A = {1, 2, 3, 4), B = {a, b, c, d}
(a) f1 = {(1, a),(2, b), (3, c), (4, d)}
f1-1 = {(a, 1), (b, 2), (c, 3), (d, 4)}
(b) f2 = {(1, a), (2, c), (3, b), (4, d)}
f2-1 = {(a, 1), (c, 2), (b, 3), (d, 4)}
(c) f3 = {(1, d), (3, b), (2, a), (4, c)}
f3-1 = {(d, 1), (b, 3), (a, 2), (c, 4)}
(d) f4 = {(1, a), (3, d), (2, b),(4, c)}
f4-1 = {(a, 1), (d, 3), (b, 2), (c, 4)}
Question 8: If f: R → R, f(x) = x3 – 3, then prove that f-1 exists and find the formula of f-1 and the values of f-1(24), f-1(5).
Solution:
Given,
f : R → R, f(x) = x3 – 3
One-one/many-one:
Let a, b ∈ R
∴ f(a) = f(b)
⇒ a3 – 3 = b3 – 3
⇒ a3 = b3
⇒ a = b
Therefore, f(a) = f(b) ⇒ a = b
Hence, f is a one-one function.
Onto/into:
Consider, y ∈ R (co-domain)
f(x) = y
⇒ x3 – 3 = y
⇒ x = (y + 3)1/3 ∈ R, ∀ y ∈ R
Thus, for each value of y, x exists in domain R.
Therefore, the range of f = co-domain of f.
⇒ f is onto function.
‘f is one-one onto function.
Hence, f-1: R → R exists.
f-1(y) = x ⇒ f(x) = y ….(i)
f(x)= x3 – 3
∴ x3 = 3 = y [From (i)]
⇒ x3 = y + 3
⇒ x = (y + 3)1/3
⇒ f-1(y) = (x + 3)1/3
⇒ f-1(x) = (x + 3)1/3, ∀ x ∈ R
Substituting x = 24 in f-1,
f-1(24)= (24 + 3)1/3
= (27)1/3
= 3(3 x 1/3)
= 3
Substituting x = 5 in f-1,
f-1(5) = (5 + 3)1/3
= (8)1/3
= 2(3 x 1/3)
= 2
Question 9: If f: R → R is defined as f(x) = x3 + 5, then prove that f is bijective and also find f-1.
Solution:
Given,
f : R → R, f(x) = x3 + 5
One-one/onto:
Let a, b ∈ R
f(a) = f(b)
⇒ a3 + 5 = b3 + 5
⇒ a3 = b3
a = b
Thus, f(a) = f(b) ⇒ a = b, ∀ a, b ∈ R
∴ f is one-one function
Onto/into:
Let us take y ∈ R (co-domain)
f(x) = y
⇒ x3 + 5 = y
⇒ x3 = y – 5
⇒ x = (y – 5)1/3 ∈ R, ∀ x ∈ R
Here, pre-image for each value of y exists in the domain R.
Thus, range of f = co-domain of f
Therefore, the function ‘f’ is onto.
Hence, f is one-one onto function.
Now,
f-1: R → R defined as f-1(y)= x ⇔ f(x) = y ….(i)
⇒ f(x) = y
⇒ x3 + 5 = y
⇒ x3 = y – 5
⇒ x = ( y – 5)1/3
⇒ f-1(y)= (y – 5)1/3 [From (i)]
⇒ f-1(x) = (x – 5)1/3
Question 10: If A = {1, 2, 3, 4}, B = {3, 5, 7, 9}, C = {7, 23, 47, 79} and f: A → B, f(x) = 2x + 1, g: B → C, g(x) = x2 – 2, then write (g∘f)-1 and (f-1∘g-1) in the form of ordered pair.
Solution:
Given,
A = {1, 2, 3, 4}, B = {3, 5, 7, 9}, C = {7, 23, 47, 79}
f : A → B, f (x) = 2x + 1
g : B → C, g(x) = x2 – 2
(g∘f)(x) = g[f(x)]
= g(2x + 1)
= (2x + 1)2 – 2
= 4x2 + 4x + 1 – 2
= 4x2 + 4x – 1
∴ (g∘f)(x) = 4x2 + 4x – 1
Substituting x = 1, 2, 3, 4:
(g∘f) = {(1,7), (2, 23), (3,47),(4, 79)}
∵ (g∘f) is a bijection function.
Therefore, the inverse of (g∘f) is possible.
By the theorem, (g∘f)-1 = f-1∘g-1
⇒ (g∘f)-1 = {(7, 1), (23, 2), (47,3), (79,4)}
⇒ f-1∘g-1 = {(7,1), (23, 2) (47, 3), (79,4)}
Question 11: If f : R → R, f(x) = ax + b, a ≠ 0 is defined, then prove that f is a bijection and also find the formula of f-1.
Solution:
Given,
f : R → R and ax + b, a ≠ 0
f-1 exists if f : R → R be a bijection function.
One-one/many-one:
Let p, q ∈ R
f(p) = f(q)
⇒ ap + b = aq + b
⇒ ap = aq
⇒ p = q
So, f(p) = f(q), ∀ p, q ∈ R
∴ f is a bijection function.
Onto/into:
Consider, f(x) = y, y ∈ R
ax + b = y
⇒ x = (y – b)/a ∈ R
Thus, the preimage of every value of y exists in the domain R.
Therefore, ‘f’ is onto.
Hence, we can say that range of f = co-domain of f.
Also, f is a bijection function.
Therefore, f-1 exists.
Consider, if y ∈ R and f-1(y)= x, then f(x) = y.
⇒ ax + b = y
⇒ x = (y – b)/a
⇒ f-1(y) = (y – b)/a
⇒ f-1(x) = (x – b)/a, ∀ x ∈ R
Therefore, f-1(x) = (x – b)/a
Question 12: If f : R → R, f(x) = cos(x + 2), then does f-1 exist?
Solution:
Given,
f : R → R, f(x) = cos (x + 2)
Substituting x = 2π in f(x),
f(2π) = cos (2π+ 2)
= cos (2)
Substituting x = 0 in f(x),
f(0) = cos (0 + 2) = cos 2
Here, only one image of f(x) is obtained for 0 and 2π.
Thus, ‘f’ is not one-one.
Also, ‘f’ is not one-one onto.
Therefore, f-1: R → R does not exist.
RBSE Maths Chapter 1: Exercise 1.3 Textbook Important Questions and Solutions
Question 13: Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) a*b = a, on N
(ii) a*b = a + 3b, on N
(iii) a*b = a/b, on Q
(iv) a*b = a – b, on R
Solution:
(i) a*b = a, on N
Here, * is a binary operation because a, b ∈ N
a*b = a ∈ N
Here, a*b ∈ N
Hence, * is a binary operation.
(ii) a*b = a + 3b ∈ N
Here, * is a binary operation because 1 ∈ N, 2 ∈ N.
Thus, 1 + 3 × 2 = 1 + 6 = 7 ∈ N
(iii) (iv) a*b = a/b ∈ Q
Here, a*b is not a binary operation, this can be shown as:
Let a = 22 ∈ Q and b= 7 ∈ Q
a/b = 22/7 = π ∉ Q
Thus, 7, 22 ∈ Q but 22/7 ∉ Q
(iv) a*b = a – b ∈ R
Here, * is a binary operation because,
a ∈ R, b ∈ R
⇒ a – b ∈ R, ∀ a, b ∈ R
Question 14: For each binary operation * defined below, determine whether it is commutative or associative?
(i) * on N where a*b = 2ab
(ii) * on N where a*b = a + b + a2b
(iii) * on Z where a*b = a – b
(iv) * on Q where a*b = ab + 1
(v) * on R where a*b = a + b – 7
Solution:
(i) Given a*b = 2ab
Commutativity: Let a, b ∈ N
a*b = 2ab
= 2b.a
= b*a
So, a*b = b*a
∴ * is a commutative operation.
Associativity: Let a, b, c ∈ N
(a*b)*c = 2(ab)*2c = 2ab + c
= 2c*2(ab)
= 2c +ab
a*(b*c) = 2a*2(bc) = 2a + bc
2ab + c ≠ 2a + bc
It is clear that (a*b)*c ≠ a*(b*c)
So, (a*b)*c is not an associative operation.
Hence, a*b = 2ab is commutative but not associative.
(ii) Given a*b= a + b + a2b
Commutativity: Let a, b ∈ N
a*b = a + b + a2b
b*a = b + a + b2a
a*b ≠ b*a .
So, * is not a commutative operation.
Associativity: Let a, b, c ∈ N
(a*b)*c = (a + b + a2b)*c
a*(b*c) = a*(b + c + b2c)
t is clear that (a*b)*c ≠ a*(b*c)
So * is not an associative operation.
Hence, a*b = a + b + a2b is neither commutative nor associative.
(iii) Given, a*b = a – b
Commutativity:
a*b = a – b, (a, b ∈ Z)
b*a = b – a, (a, b ∈ Z)
a*b* ≠ b*a
So * is not a commutative operation.
Associativity :
(a*b)*c = (a – b)*c
= a – b – c
a*(b*c) = a*(b-c)
= a – b + c
∵ (a*b)*c ≠ a*(b*c)
So, it is not an associative operation.
It is clear that
a*b = a – b is neither commutative nor associative.
(iv) Given, a*b = ab + 1
Commutativity: Let a, b ∈ Q
a*b = ab + 1 and : b*a = ba + 1
⇒ a*b = b*a
∴ It is commutative.
∴ Addition and multiplication of rational numbers is commutative.
Associativity: Let a, b, c ∈ Q
(a*b)*c = (ab + 1)*c
= ab + 1 + c
(b*c)*a = (bc + 1) +a
= (a*b)*c ≠ (b*c)*a
So, * is not associative.
It is clear from above that a*b = ab + 1 is commutative but not associative.
(v) Given, a*b = a + b – 7
Commutativity: In R,
a*b = a + b – 7
= b + a – 7
= b*a’
Associativity :
(a*b)*c = (a + b – 7)*c
= (a + b – 7) + c – 7
= a + b + c – 14
a*(b*c) = a*(b + c – 7)
= a + (b + c – 7) – 7
= a + b + c – 14
So, (a*b)*c = a*(b*c)
Hence, it is clear that a*b = a + b – 7 are commutative and associative.
Question 15: If in a set of integers Z is an operation * is defined as *, a*b = a + b + 1, ∀ a, b ∈ Z, then prove that * is commutative and associative. Also, find its identity element. Find the inverse of any integer.
Solution:
Given,
a*b = a + b + 1, ∀ a, b ∈ Z
Commutativity:
a*b = a + b + 1
a*b = b + a + 1
= b*a
∴ a*b = b*a
∴ * is a commutative operation.
Associativity:
(a*b) * c = (a + b + 1)*c
= a + b + 1 + c +1
a + b + c + 2
Again a*(b*c) = a*(b + c + 1)
= a + b + c + 1 + 1
= a + b + c + 2
a*(b*c) = (a*b)*c
∴ * is associative operation
Identity:
If e is identity element, then a*e = a.
⇒ a + e + 1 = a
⇒ e = -1
Thus, – 1 ∈ Z is an identity element.
Inverse:
Let x be the inverse of a.
By definition,
a*x = -1 [∵ – 1 is identity]
⇒ a + x + 1 = -1
⇒ x = -(a + 2) ∈ Z
Inverse element a-1 = -(a + 2).
Question 16: A binary operation defined on a set R-{1} is as follows:
a*b = a + b – ab, ∀ a, b ∈ R – {1}
Prove that * is commutative and associative. Also, find its identity element and find inverse of any element a.
Solution:
Given,
a, b ∈ R – {1}
a*b = a + b – ab
= b + a – ba
= b*a
∴ * is a binary operation.
Again, (a*b)*c = (a + b – ab)*c
= (a + b – ab) + c – (a + b – ab)c
= a + b – ab + c – ac – bc + abc
= a + b + c – ab – bc – ac + abc….(i)
a*(b*c) = a*(b + c – bc)
= a + (b + c – bc) – a (b + c – bc)
= a + b + c – bc – ab – ac + abc
= a + b + c – ab – bc – ac + abc….(ii)
From (i) and (ii),
(a*b)*c = a*(b*c)
∴ * is an associative operation.
Let e be the identity of *, then for a ∈ R,
a*e = a (from definition of identity)
⇒ a + e – ae = a
⇒ e(1 – a)= 0
⇒ e = 0 ∈ R – {1}
∵ 1 – a ≠ 0
∴ 0 is identity of *.
Let b be the inverse of a.
a*b = e
a + b – ab = 0.e
b – ab = -a
b(1 – a) = -a
⇒ b = -a/)1 – a)
or
b = a/(a – 1)
Therefore, the inverse of a is b, then b = a-1 = a/(a – 1).
RBSE Maths Chapter 1: Additional Important Questions and Solutions
Question 1: If f : R → R, f(x) = 2x – 3; g : R → R, g(x) = x3 + 5, then the value of (fog)-1(x) is:
(a) [(x + 7)/2]1/3
(b) [x – (7/2)]1/3
(c) [(x – 2)/7]1/3
(d) [(x – 7)/2]1/3
Solution:
Correct answer: (d)
Given,
f: R → R, f(x) = 2x – 3
g: R → R, g(x)= x3 + 5
Now, (f∘g)(x) = f[g(x)]
= f(x3 +5)
= 2(x3 + 5) – 3
= 2x3 + 10 – 3
= 2x3 + 7
Let y = (f∘g)(x) = 2x3 + 7
∴ (f∘g)-1(y) = x = [(y – 7)/2]1/3
∴ (f∘g)-1(x) = [(x – 7)/2]1/3
Question 2: If f(x) = x/(1 – x) = 1/y, then the value of f(y) is
(a) x
(b) x – 1
(c) x + 1
(d) (1 – x)/(2x – 1)
Solution:
Correct answer: (d)
Given,
f(x) = x/(1 – x) = 1/y
⇒ x/(1 – x) = 1/y
⇒ y = (1 – x)/x
f(y) = y/(1 – y) = [(1 – x)/x] / [1 – (1 – x)/x]
= [(1 – x)/x] / [(x – 1 + x)/x]
= (1 – x)/(2x – 1)
Question 3: If f(x) = (x – 3)/(x + 1), then the value of f[f{f(x)}] is equal to
(a) x
(b) 1/x
(c) -x
(d) -1/x
Solution:
Correct answer: (a)
Given,
f(x) = (x – 3)/(x + 1)
f[f{f(x)] = f[f{(x – 3)/(x + 1)}]
= f{[(x – 3)/(x + 1) – 3] / [(x – 3)/(x + 1) + 1]}
= f{[(x – 3 – 3x – 3)/(x + 1)] / [(x – 3 + x + 1)/(x + 1)]
= f[(-2x – 6)/(2x – 2)]
= f[(-x – 3)/(x – 1)]
= [(-x – 3)/(x – 1) – ] / [(-x – 3)/(x – 1) + 1]
= [(-x – 3 – 3x + 3)/(x – 1)] / [(-x – 3 + x – 1)/(x – 1)]
= -4x/(04)
= x
Question 4: If f(x) = cos(log x), then the value of f(x).f(y) – (1/2) [f(x/y) + f(x.y)] is
(a) -1
(b) 0
(c) 1/2
(d) -2
Solution:
Correct answer: (b)
Given,
f(x) = cos(log x)
f(x).f(y) – (1/2) [f(x/y) + f(x.y)]
= cos(log x).cos(log y) – (1/2) [cos(log x/y) + cos(log xy)]
= cos(log x).cos(log y) – (1/2) [cos(log x – log y) + cos(log x + log y)]
= cos(log x).cos(log y) – (1/2) [2 cos(log x) cos(log y)]
= cos(log x).cos(log y) – cos(log x) cos(log y)
= 0
Question 5: If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to
(a) 2
(b) 1
(c) -1
(d) 0
Solution:
Correct answer: (b)
Given,
f(x) = 2x + 1, g(x) = x3
Let (g∘f)-1(27) = x
⇒ (g∘f)(x) = 27
g[f(x)] = 27
g(2x + 1) = 27
(2x + 1)3 = 27
2x + 1 = (27)1/3
2x + 1 = 3(3 x 1/3)
⇒ 2x + 1 = 3
2x = 3 – 1 = 2
Therefore, x = 1
Question 6: If f : R → R and g : R → R where f(x) = 2x + 3 and g(x) = x2 + 1, then the value of (gof)(2) is
(a) 38
(b) 42
(c) 46
(d) 50
Solution:
Correct answer: (d)
Given,
f(x) = 2x + 3 and g(x) = x2 + 1
(g∘f)(2) = g[f(2)]
= g(2 x 2 + 3)
= g(7)
= (7)2 + 1
= 49 + 1
= 50
Question 7: If an operation * defined on Q0, as *, a*b = ab/2, ∀ a, b ∈ Q0, then the identity element is
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
Correct answer: (c)
Given,
a*b = ab/2, ∀ a, b ∈ Q0
Let e be an identity element.
Now, a ∈ Q0
a*e = a
⇒ ae/2 = a
e = 2
Question 8: A binary operation defined on R as a*b = 1 + ab, ∀ a, b ∈ R, then * is
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) commutative and associative
Solution:
Correct answer: (a)
Given,
a*b = 1 + ab, ∀ a, b ∈ R
Commutativity:
a*b = 1 + ab
= 1 + b.a
= b*a
We know that the set of real numbers is commutative.
⇒ a*b = b*a
∴ It is commutative.
Associativity:
(a*b)*c = (1 + ab)*c
= (1 + ab)c
= 2 + abc
a*(b*c) = a*(1 + bc) = 1 + a*(1 + bc)
= 1 + a + abc
Thus, (a*b)*c ≠ a*(b*c)
Hence, * operation is not associative.
Question 9: For the given three functions justify the associativity of composite function operation.
f: N → Z0, f(x) = 2x; g: Z0 → Q, g(x) = 1/x; h: Q → R, h(x) = ex
Solution:
Given,
f: N→ Z0
g: Z0 → Q
h: Q → R
Thus, h∘(g∘f): N → R
and (h∘g)∘f : N → R
Hence, the domain and co-domain of h∘(g∘f) and g∘(h∘f) are the same because both the functions are defined from N to R.
That means, we have to prove
[h∘(g∘f)](x) = [(h∘g)∘f](x), ∀ X ∈ N [h∘(g∘f)](x)= h[(g∘f)(x)]= h[g{f(x)}]
= h[g(2x)]
= h(1/2x)
= e(1/2x)
[h∘(g∘f)](x) = e(1/2x) ….(i)Now, [(h∘g)∘f](x) = (h∘g)f(x) = (h∘g)(2x) = h[g(2x)] = h(1/2x)
[(h∘g)∘f](x) = e(1/2x) ….(ii)From (i) and (ii),
(h∘g)∘f = h∘(g∘f)
Therefore, the associativity of f, g, h is proved.
Question 10: If f: R → R, f(x) = sin x and g: R → R, g(x) = x2, then find (g∘f)(x).
Solution:
Given,
f: R → R, f(x) = sin x
g: R → R, g(x) = x2
(g∘f)(x) = g[f(x)]
= g(sin x)
= (sin x)2
= sin2x
Therefore, (g∘f)(x) = sin2x
Question 11: If f: R → R, f(x) = x2 – 5x + 7, then find the value of f-1(1).
Solution:
Given,
f: R → R, f(x) = x2 – 5x + 7
Let f-1(1) = x
⇒ f(x) = 1
⇒ x2 – 5x + 7 = 1
⇒ x2 – 5x + 6 = 0
⇒ x2 – 2x – 3x + 6 = 0
⇒ x(x – 2) – 3(x – 2) = 0
⇒ (x – 2)(x – 3) = 0
⇒ x – 2 = 0, x – 3 = 0
⇒ x = 2, x = 3
Therefore, f-1(1) = {2, 3}
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