## Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.2

Question 1.

If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions.

Solution:

Given

A = {1, 2, 3, 4), B = {a, b, c, d}

(a) f_{1} = {(1, a),(2, b), (3, c), (4, d)}

f_{1}^{-1} = {(a, 1), (1, 2), (c, 3), (d, 4)}

(b) f_{2} = {(1, a), (2, c), (3, b), (4, d)}

f_{2}^{-1} = {(a, 1), (C, 2), (6, 3), (d, 4)}

(c) f_{3} = {(1, b), (2, a), (3, d), (4, b)}

f_{3}^{-1} = {(b, 1), (a, 2), (d, 3), (6,4)}

(d) f_{4} = {(1, c), (2, d), (3, a),(4, b)}

f_{4}^{-1} = {(c, 1), (d, 2), (a, 3), (b, 4)}

Question 2.

If f : R → R, such that f(x) = x^{3} – 3, then prove that f^{-1} exists and find its formula. Thus, find f^{-1}(24) and f^{-1}(5).

Solution:

Given, f : R → R, f(x) = x^{3} – 3

One-one/many-one:

Let a, b ∈ R

∴ f(a) = f(b)

⇒ a^{3} – 3 = b^{3} – 3

⇒ a^{3} = b^{3}

⇒ a = b

Hence, . f(a) = f(b) ⇒ a = b

∴ f is a one-one function.

Onto/into:

Let y ∈ R (Co-domain)

f(x) = y

⇒ x^{3} – 3 = y

⇒ x= (y + 3)^{1/3} ∈ R, ∀ y ∈ R

Here, for each value of y, x exists in domain R.

Thus, Range of F= co-domain of f.

So, ‘f’ is onto function.

It is clear that ‘f is one-one onto function.

Hence, f^{-1}: R → R exists.

f^{-1}(y) = x ⇒ f(x) = y

But f(x)= x^{3} – 3

∴ x^{3} = 3 = y

⇒ x^{3} = y + 3

⇒ x = (y + 3)^{1/3}

⇒ f^{-1}(y) = (x + 3)^{1/3}

⇒ f^{-1}(x) = (x + 3)^{1/3}, ∀ X ∈ R

For x = 24

∴ f^{-1}(24)= (24 + 3)^{1/3}

= (27)^{1/3}

= 3^{3 x 1/3} = 3

For x = 5

f^{-1}(5) = (5 + 3)^{1/3}

= (8)^{1/3}

= 2^{3 x 1/3} = 2

Question 3.

If f : R → R, defined as

(i) f(x) = 2x – 3

(ii) f (x) = x^{3} + 5

Then, prove that f is bijection in both conditions, Also find f^{-1}.

Solution:

(i) Given function,

f : R → R, f(x) = 2x – 3

One-one/many-one:

Let a, b ∈ R

f(a) = f(b)

⇒ 2a – 3 = 2b – 3

⇒ 2a = 2b

⇒ a = b

So, f(a) = f(b) ⇒ a = b, ∀ a, b ∈ R

∴ f is one-one function.

onto/into:

Let y ∈ R (co-domain)

f(x)= y

⇒ 2x – 3 = y

⇒ x = \(\frac { y+3 }{ 2 }\) ∈ R , ∀ y ∈ R

So, pre-image for each value of y exists in domain R. Thus, function ‘f’ is onto function.

It is clear that ‘f’ is one-one into function.

Hence f^{-1} : R → R exists.

Let x ∈ R (domain of f)

and y ∈ R (co-domain of f)

Let f(x) = y, then f^{-1}(y) = x

⇒ f(x) = y

⇒ 2x – 3 = y

(ii) According to question,

f : R → R, f(x) = x^{3} + 5

One-one/onto : Let a, b ∈ R

f(a) = f(b)

⇒ a^{3} + 5 = b^{3} + 5

⇒ a^{3} = b^{3}

a = b

So, f(a) = f(b)

⇒ a = b, ∀ a, b ∈ R

∴ f is one-one function

Onto/into:

Let y ∈ R (co-domain)

f(x) = y

⇒ x^{3} + 5 = y

⇒ x^{3} = y – 5

⇒ x = (y – 5)^{1/3} ∈ R, ∀ x ∈ R

So, pre-image for each value of y exists in domain R.

Thus, range of f = co-domain of f.

So, function ‘f’ is onto function.

Hence, we can say that f is one-one onto function.

So, f^{-1}: R → R defined as

f^{-1}(y)= x ⇔ f(x) = y ………(i)

⇒ f(x) = y

⇒ x^{3} + 5 = y

x^{3} = y – 5

⇒ x = ( y – 5)^{1/3}

⇒ f^{-1}(y)= (y – 5)^{1/3} [from (1)]

⇒ f^{-1}(x) = (x – 5)^{1/3}

Question 4.

If A = {1, 2, 3, 4), B = {3, 5, 7, 9), C = {7, 23, 47, 79} and f: A → B, g : B → C such that f(x) = 2x + 1 and g(x) = x^{2} – 2, then find (gof)^{-1} and f^{-1}og^{-1} in ordered form.

Solution:

A = {1, 2, 3, 4), B = {3, 5, 7,9} and

C = {7, 23, 47, 79}

f : A → B, f (x) = 2x + 1

g : B → C, g(x) = x^{2} – 2

Now, gof (x)=g{f(x)} = g(2x + 1)

= (2x + 1)^{2} – 2 = 4x^{2} + 4x – 1

∴ gof (x) = 4x^{2} + 4x – 1

On putting x = 1, 2, 3, 4

gof = {(1,7), (2, 23), (3,47),(4, 79)}

∵ gof is bijection function.

∴ Its inverse is possible

⇒ (gof)^{-1} = {(7, 1), (23, 2), (47,3), (79,4)}

⇒ f^{-1}og^{-1} = {(7,1), (23, 2) (47, 3), (79,4)}

[∵ (gof)^{-1} = f^{-1}og^{-1} by theorem.

Question 5.

If f : R → R, such that f(x) = ax + b, a ≠ 0, then prove that f is a bijection function. Also, find f^{-1}.

Solution:

Given function f : R→ R and ax + b, a ≠ 0

f^{-1} exists if f : R → R be a bijection function, before this we have to prove f 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:

Let f(x) = y, y ∈ R

ax + b = y

⇒ x = \(\frac { y-b }{ a }\) ∈ R

So, pre-image of every value of y exist in domain R,

∴ ‘f’ is onto function.

Thus, range of f = co-domain of f.

So, f is a bijection function

So, f^{-1} exists.

Let, y ∈ R and f^{-1}(y)= x,

then f(x) = y

⇒ ax +b = y

Question 6.

If f : R → R,f(x) = cos (x + 2), is f^{-1 }exists.

Solution:

Given function

f : R → R, f(x) = cos (x + 2).

Putting x = 2π

f(2π) = cos (2π+ 2)

= cos (2)

Putting x = 0

f(0) = cos (0 + 2) = cos 2

Here, only one image is obtained for 0 and 2π.

So, ‘f’ is not one-one.

Thus, ‘f’ is not one-one onto.

Hence, f^{-1} : R → R does not exist.

Question 7.

Find f^{-1} (if exists), where f: A → B, such that

(i) A = {0,-1,- 3, 2}, B = {-9, – 3,0, 6}, f(x) = 3x

(ii) A = {1, 3, 5, 7,9}, B = {0, 1, 9, 25, 49, 81), f(x) = x^{2}

(iii) A = B = R, f(x) = x^{3}

Solution:

(i) Give function

f : A → B, f(x) = 3x where,

A = {0,-1,-3, 2}

B = {-9, – 3,0, 6}

Now, f(0)= 3 x 0 = 0

f(-1) = 3 x – 1 = -3

f(-3) = 3 x (- 3) = -9

f(2)= 3 x 2 = 6

So, f= {(0,0), (- 1, -3), (- 3,-9), (2, 6)}

Since, different element of A have different image in B under f.

So ‘f is one-one function.

Here, range of f = {-9, -3,0, 6} = B (co-domain).

So, f is onto function.

It is clear that ‘f’ is one-one onto function.

So, f^{-1} : B → A exist.

f^{-1} = {(0,0), (-3,- 1), (-9, – 3), (6, 2)}

(ii) Given function

f : A → B, f(x) = x^{2}

where, A = {1, 3, 5, 7, 9}

B = {0, 1, 9, 25, 49, 81}

Now, f(1) = 1^{2} = 1

f(3) = 3^{2} = 9

f(5)= 5^{2} = 25

f (7) = 7^{2} = 49

f(9) = 9^{2} = 81

So, f= {(1, 1), (3,9), (5, 25), (7,49), (9,81)}

Since, different element of A have different image in B under f.

So ‘f’ one-one function.

Here, range of R = {1,9, 25, 49, 81} ≠ B (co-domain)

∴ ‘f’ is not onto function.

Hence, f^{-1} does not exist.

(iii) Given, A = B = R, f(x) = x^{3}

f(x)= x^{3}

Let a, b ∈ R

and f(a) = f(b)

⇒ a^{3} = b^{3}

a = b, ∀ a, b ∈ R

Let f(x)= y and y ∈ B or y ∈ R

∴ x3 = y

⇒ x = y^{1/3}

Since, different element of A have different image of B under f.

So ‘f’ is one-one function.

and range of f= co-domain of f.

So, ‘f’ is onto function.

Hence, it is clear that f is one-one function f^{-1} exists.

If y is image of x under ‘f’, then

f^{-1} : B → A is defined

as f^{-1}(y) = x ⇒ f(x) = y

⇒ x^{3} = y

⇒ x = y^{1/3}

⇒ f^{-1}(y) = y^{1/3}

⇒ f^{-1}(x) = x^{1/3}

So, f^{-1}: B → A, f^{-1}(x) = x^{1/3}, ∀ x ∈ B.

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