excersie 1.3 continued

Question 6:

Show that f: [−1, 1] → R, given by is one-one. Find the inverse of the function f: [−1, 1] → Range f.

(Hint: For y ∈Range f, y =, for some x in [−1, 1], i.e.,)

solutions =

 

f: [−1, 1] → R is given as

Let f(x) = f(y).

∴ f is a one-one function.

It is clear that f: [−1, 1] → Range f is onto.

∴ f: [−1, 1] → Range f is one-one and 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 of range f.

Since f: [−1, 1] → Range f is onto, we have:

Now, let us define g: Range f → [−1, 1] as

∴gof = and fog = 

f⁻¹ = g

⇒ 

 

Question 7:

Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

solutions=

 

f: R → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

Therefore, for any y ∈ R, there exists  such that

∴ f is onto.

Thus, f is one-one and onto and therefore, f⁻¹ exists.

Let us define g: R→ R by.

∴

Hence, f is invertible and the inverse of f is given by

 

Question 8:

Consider f: R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f⁻¹ of given f by, where R₊ is the set of all non-negative real numbers.

solutions=

 

f: R₊ → [4, ∞) is given as f(x) = x² + 4.

One-one:

Let f(x) = f(y).

∴ f is a one-one function.

Onto:

For y ∈ [4, ∞), let y = x² + 4.

Therefore, for any y ∈ R, there exists  such that

.

∴ f is onto.

Thus, f is one-one and onto and therefore, f⁻¹ exists.

Let us define g: [4, ∞) → R₊ by,

∴

Hence, f is invertible and the inverse of f is given by

 

Question 9:

Consider f: R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with.

solutions=

 

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 is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R₊ as

We now have:

∴and

Hence, f is invertible and the inverse of f is given by

 

Question 10:

Let f: X → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y,

fog₁(y) = I_Y(y) = fog₂(y). Use one-one ness of f).

solutions=

 

Let f: X → Y be an invertible function.

Also, suppose f has two inverses (say_).

Then, for all y ∈Y, we have:

Hence, f has a unique inverse.

 

Question 11:

Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f⁻¹ and show that (f⁻¹)⁻¹ = f.

solutions=

 

Function f: {1, 2, 3} → {a, b, c} is given by,

f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:

∴and, where X = {1, 2, 3} and Y= {a, b, c}.

Thus, the inverse of f exists and f⁻¹ = g.

∴f⁻¹: {a, b, c} → {1, 2, 3} is given by,

f⁻¹(a) = 1, f⁻¹(b) = 2, f^-1(c) = 3

Let us now find the inverse of f⁻¹ i.e., find the inverse of g.

If we define h: {1, 2, 3} → {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we have:

∴, where X = {1, 2, 3} and Y = {a, b, c}.

Thus, the inverse of g exists and g⁻¹ = h ⇒ (f⁻¹)⁻¹ = h.

It can be noted that h = f.

Hence, (f⁻¹)⁻¹ = f.

 

Question 12:

Let f: X → Y be an invertible function. Show that the inverse of f⁻¹ is f, i.e.,

(f⁻¹)⁻¹ = f.

solutions=

 

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_Xand fog = I_Y.

Here, f⁻¹ = g.

Now, gof = I_Xand fog = I_Y

⇒ f⁻¹of = I_Xand fof⁻¹= I_Y

Hence, f⁻¹: Y → X is invertible and f is the inverse of f⁻¹

i.e., (f⁻¹)⁻¹ = f.

 

Question 13:

If f: R → R be given by, then fof(x) is

(A) (B) x³ (C) x (D) (3 − x³)

solutions=

 

f: R → R is given as.

The correct answer is C.

Question 14:

Let be a function defined as. The inverse of f is map g: Range

(A) (B)

(C) (D)                                                                                                     solutions =     

It is given that

Let y be an arbitrary element of Range f.

Then, there exists x ∈such that 

Let us define g: Rangeas

Now,

∴

Thus, g is the inverse of f i.e., f⁻¹ = g.

Hence, the inverse of f is the map g: Range, which is given by

The correct answer is B.