# Plus Two Math's Solution Ex 1.3 Chapter1 Relations and Function

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 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 1 Exercise Ex 1.3 Chapter Name Relations and Function Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  1 Relation and Function Exercises 1.3

### Chapter  1  Relations and Function Solution

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter  1  Relations and Function Exercise   1.3

Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Write down gof.

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Let fg, and h be functions from to R. Show that To prove:     Find goand fog, if

(i) (ii) (i)  (ii)  If , show that f f(x) = x, for all . What is the inverse of f?

It is given that . Hence, the given function f is invertible and the inverse of f is f itself.

State with reason whether following functions have inverse

(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: {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) f: {1, 2, 3, 4} → {10}defined as:

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10

is not one-one.

Hence, function does not have an inverse.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.

is not one-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 is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

∴Function h is one-one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists 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.

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 fy = , for some x in [−1, 1], i.e., )

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 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 fo  f−1 = g

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

fR → 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 ∈ R, there exists such that ∴ f is onto.

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

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

fR+ → [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 ∈ 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, ∞) → Rby,   Hence, f is invertible and the inverse of f is given by Consider fR+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with .

fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x2 + 6− 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 Let fX → 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) = IY(y) = fog2(y). Use one-one ness of f).

Let fX → 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.

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

Function f: {1, 2, 3} → {abc} is given by,

f(1) = af(2) = b, and f(3) = c

If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:  and , where X = {1, 2, 3} and Y= {abc}.

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

f−1: {abc} → {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} → {abc} as

h(1) = ah(2) = bh(3) = c, then we have:  , where X = {1, 2, 3} and Y = {abc}.

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

It can be noted that h = f.

Hence, (f−1)−1 = f.

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

(f−1)−1 = f.

Let fX → Y be an invertible function.

Then, there exists a function gY → X such that gof = IXand fo= IY.

Here, f−1 = g.

Now, gof = IXand fo= IY

⇒ f−1of = IXand fof−1= IY

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

i.e., (f−1)−1 = f.

If f→ be given by , then fof(x) is

(A) (B) x3 (C) x (D) (3 − x3)

fR → R is given as . Let be a function defined as . The inverse of f is map g: Range (A) (B) (C) (D) It is given that Let y be an arbitrary element of Range f.

Then, there exists x ∈ such that  Let us define g: Range as 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 #### Chapter 1: Relations and Function EX 1.3 Solution- Preview

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