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**CBSE mathematics class 12 chapter Relations and Functions Exercise 1.3 Solutions**. You can find the questions/answers/solutions for the**chapter 1/Exercise 1.3**of**CBSE class 12 mathematics**in this page. So is the case if you are looking for**CBSE class 12 Maths**related topic**Relations and Functions**Exercise 1.3

1. Let f:{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 gof.

It is given that f:{1, 3, 4} → {1, 2, 5} and g:{1, 2, 5} → {1, 3}

gof:{1, 3, 4} → {1, 3}

The function f is given as

f(1) = 2

f(3) = 5

f(4) = 1

The function g is given as

g(1) = 3

f(2) = 3

f(5) = 1

gof(1) = g(f(1)]) = g(2) = 3

gof(3) = g(f(3)]) = g(5) = 1

gof(4) = g(f(4)]) = g(1) = 3

∴ gof = {(1,3), (3, 1), (4, 3)}

2. Let f, g and h be functions from R to R. Show that

(f + g)oh = foh + goh

(f.g)oh = (foh).(goh)

(f + g)oh = foh + goh

(f.g)oh = (foh).(goh)

We have

((f + g)oh)(x)

= (f + g)(h(x))

= f(h(x)) + g(h(x))

= foh(x) + goh(x)

∴ (f + g)oh = foh + goh

Similarly,

((f.g)oh)(x)

= (f.g)(h(x))

= f(h(x)).g(h(x))

= foh(x).goh(x)

∴ (f.g)oh = (foh).(goh)

3. Find gof and fog, if

- f(x) = |x| and g(x) = |5x – 2|
- f(x) = 8x³ and g(x) = x^{\frac{1}{3}}

To find gof and fog when f(x) = |x| and g(x) = |5x – 2|

We have,

gof(x)

= g(f(x))

= g(|x|) (∵ f(x) = |x|)

= |5|x| – 2| (∵ g(x) = |5x – 2|)

∴ gof(x) = |5|x| – 2|

Similarly

fog(x)

= f(g(x))

= f(|5x – 2|) (∵ g(x) = |5x – 2|)

= ||5x – 2|| (∵ f(x) = |x|)

= |5x – 2|

To find gof and fog when f(x) = 8x³ and g(x) = x^{\frac{1}{3}}

We have

gof(x)

= g(f(x))

= g(8x³) (∵ f(x) = 8x³)

= \left(8x^3\right)^\frac{1}{3} (∵ g(x) = x^\frac{1}{3})

= \left(\left(2x\right)^3\right)^\frac{1}{3}

= \left(2x\right)^{3\times\frac{1}{3}} (∵ {\left(a^m\right)^n = a^{mn}})

= 2x

∴ gof(x) = 2x

Similarly

fog(x)

= f(g(x))

= f(x^\frac{1}{3}) (∵ g(x) = x^\frac{1}{3})

= 8\left(x^\frac{1}{3}\right)^3 (∵ f(x) = 8x³)

= 8x^{\frac{1}{3}\times3} (∵ {\left(a^m\right)^n = a^{mn}})

= 8x

∴ fog(x) = 8x

4. If f(x) = \frac{(4x + 3)}{(6x - 4)}, x ≠ \frac{2}{3}, show that fof = x, for all x ≠ \frac{2}{3}. What is the inverse of f ?

We have,

fof(x)

= f(f(x))

= f\left(\dfrac{4x - 3}{6x - 4}\right) (∵ f(x) = \dfrac{4x - 3}{6x - 4})

= \dfrac{4\left(\dfrac{4x - 3}{6x - 4}\right) - 3}{6\left(\dfrac{4x - 3}{6x - 4}\right) - 4} (∵ f(x) = \dfrac{4x - 3}{6x - 4})

= {\dfrac{\dfrac{4(4x - 3) - 3(6x - 4)}{6x - 4}}{\dfrac{6(4x - 3) - 4(6x - 4)}{6x - 4}}}

= \dfrac{16x - 12 - 18x + 12}{24x - 18 - 24x + 16} (After cancelling/cutting 6x – 4 in both numerator and denominator)

= \dfrac{-2x}{-2}

= x

∴ fof(x) = x

∵ fof(x) = x, the inverse of f is f itself.

∴ We can say that f-1 = f

5. State with reason whether following functions have inverse

- f:{1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
- g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
- h:{2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

As we know, a function f will have an inverse only if the function is both one-one and onto.

Let’s now check one by one whether the given functions f, g and h are both one-one and onto.

To check whether f is one-one

In the function f, we have

f(1) = 10

f(2) = 10

f(3) = 10 and

f(4) = 10

As different elements in the domain have the same image in the co-domain, f is not one-one

∴ f does not have an inverse (as f is not one-one, the check for f being an onto is not required)

To check whether g is one-one

In the function g, we have

g(5) = 4

g(6) = 3

g(7) = 4

g(8) = 2

As two different elements 5 and 7 in the domain have the same image 4 in the co-domain, g is not one-one.

∴ g does not have an inverse (as g is not one-one, the check for g being an onto is not required)

To check whether h is one-one

In the function h, we have

h(2) = 7

h(3) = 9

h(4) = 11

h(5) = 13

Clearly different elements in the domain have the different images in the co-domain.

∴ h is one-one

To check whether h is onto

As every element in the co-domain {7, 9, 11, 13} has an inverse-image in the domain, h is also onto.

∴ h is both one-one and onto.

∴ h will have an inverse and the inverse of h will be

h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

6. Show that f: [–1, 1] → R, given by f(x) = \dfrac{x}{(x + 2)} is one-one. Find the inverse of the function f:[–1, 1] → Range f.

(Hint: For y ∈ Range f, y = f(x) = \dfrac{x}{x + 2}, for some x in [-1, 1], i.e., x = \dfrac{2}{2 - y})

The function f is given by

f(x) = \dfrac{x}{2 + x}

This can also be written as

f(x) = \dfrac{x + 2 - 2}{x + 2} = 1 - \dfrac{2}{x + 2}

To Check whether f is one-one

Consider the two elements in the domain x1, x2 ∈ [-1, 1] such that

f(x1) = f(2)

⇒ 1 - \dfrac{2}{x_1 + 2} = 1 - \dfrac{2}{x_2 + 2}

⇒ -\dfrac{2}{x_1 + 2} = -\dfrac{2}{x_2 + 2} (After cutting/cancelling 1 on both sides)

⇒ \dfrac{1}{x_1 + 2} = \dfrac{1}{x_2 + 2} (After cutting/cancelling 2 on both sides)

⇒ x1 + 2 = x2 + 2 (After inversing both sides)

⇒ x1 = x2 (After cutting/cancelling 2 on both sides)

∴ f is one-one

To Check whether f is onto

We have y = f(x) = 1 – \dfrac{2}{x + 2}

⇒ \dfrac{2}{x + 2} = 1 – y

⇒ \frac{x + 2}{2} = \dfrac{1}{1 - y}

⇒ x + 2 = \dfrac{2}{1 - y}

⇒ x = \dfrac{2}{y - 1} – 2

⇒ x = \dfrac{2 - 2(1 - y)}{1 - y}

⇒ x = \dfrac{2y}{1 - y}

As the above equation is valid for all values of y except 1 (∵ when y = 1, 1 – y = 0 and division by 0 is not defined.

Also, as y is defined as y = \dfrac{x}{x + 2}, y can never be 1 (∵ x ≠ x + 2)

⇒ Range of f never includes 1.

Thus for every y ∈ Range of f, there exists an inverse image in the domain.

∴ f is onto.

As f is both one-one and onto f is invertible and is given as

f-1(y) = \dfrac{2y}{1 - y}

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

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function f is given as f(x) = 4x + 3

Consider two elements x1, x2 ∈ R such that

f(x1) = f(x2)

⇒ 4x1 + 3 = 4x2 + 3

⇒ 4x1 = 4x2 (After cutting/cancelling 3 on both sides)

⇒ x1 = x2 (After cutting/cancelling 4 on both sides)

∴ f is one-one

To check whether f is onto

The function f is given as y = f(x) = 4x + 3

⇒ y = 4x + 3

⇒ y – 3 = 4x

⇒ 4x = y – 3

⇒ x = \dfrac{y - 3}{4}

The above expression is valid for all real values of y.

In otherwords, every element y ∈ R will have an inverse image x in the domain R

∴ f is onto

As f is both one-one and onto, f is invertible and is given as

f-1 = \dfrac{y - 3}{4}

Method 2: To show that f is invertible by showing that fog = gof = IR by defining the inverse function g

Let y be an arbitrary element in the range of f.

⇒ For some x ∈ R, y = 4x + 3

⇒ y – 3 = 4x

⇒ 4x = y – 3

⇒ x = \dfrac{y - 3}{4}

Let’s now define g:R → R given as g(y) = \dfrac{y - 3}{4}

Now,

gof(x) = g(f(x))

= g(4x + 3)

= \dfrac{(4x + 3) - 3}{4}

= \dfrac{4x}{4}

= x

Also,

fog(y) = f(g(y))

= f\left(\dfrac{y - 3}{4}\right)

= 4\times\left(\dfrac{y - 3}{4}\right) + 3

= y – 3 + 3 (After cancelling/cutting 4 in the numerator and denominator)

y

Hence gof = IR and also fog = IR

⇒ f is invertible and is given as f-1 = g

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

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function f is given as f(x) = x² + 4

Consider two elements x1, x2 ∈ R+ in the domain such that

f(x1) = f(x2)

⇒ x1² + 4 = x2² + 4

⇒ x1² = x2² (After cutting/cancelling 4 on both sides)

⇒ x1 = ±x2

As it is given that the domain R+ is the set of non-nagative real numbers, we can ignore the negative values of x

∴ f is one-one

To check whether f is onto

The function f is given as y = f(x) = x² + 4

⇒ x² + 4 = y

⇒ x² = y – 4

⇒ x = \sqrt{y - 4}

The above equation is valid for all y – 4 ≥ 0 (∵ square root of negative numbers is not defined in real numbers)

or y ≥ 4

It is given that the co-domain of f is [4, ∞)

Every y ∈ [4, ∞) in the co-domain will be an image of an element x ∈ R+ in the domain.

∴ f is onto.

As f is both one-one and onto, f is invertible and the inverse of f is given as

f-1(y) = \sqrt{y - 4}

Method 2: To show that f is invertible by showing that fof-1 = f-1of = IR+ by using the inverse function f-1

Let y be an arbitrary element in the range of f.

For some x ∈ R+, y = x² + 4

⇒ y – 4 = x²

⇒ x² = y – 4

⇒ x = \sqrt{y - 4}

Let’s now define g:[4, ∞) → R+ given as g(y) = \sqrt{y - 4}

Now

gof(x) = g(f(x))

= g(x² + 4)

= \sqrt{x^2 + 4 - 4} (∵ g(y) = \sqrt{y - 4}

= \sqrt{x^2}

= x

Also,

fog(y) = f(g(y))

= f(\sqrt{y - 4})

= \left({\sqrt{y - 4}}\right)^2 + 4 (∵ f(x) = x²)

= y – 4 + 4

= y

Hence gof = IR+ and also fog = IR+

⇒ f is invertible and is given as f-1 = g

9. Consider f: R+ → [–5, ∞) given by f(x) = 9x² + 6x – 5. Show that f is invertible with f-1(y) = \left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right).

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function if is given as f(x) = 9x² + 6x – 5

This function can be written as y = f(x) = 9x² + 6x + -5

Consider two elements x1, x2 ∈ R+, in the domain R+ such that

f(x1) = f(x2)

⇒ 9x1² + 6x1 – 5 = 9x2² + 6x2 – 5

⇒ 9x1² + 6x1 = 9x2² + 6x2 (After cutting/cancelling -5 on both sides)

⇒ 9x1² – 9x2² + 6x1 – 62 = 0

⇒ 9(x1² – x2²) + 6(x1 – x2) = 0

⇒ 9(x1 + x2)(x1 – x2) + 6(x1 – x2) = 0(∵ a² – b² = (a + b)(a – b))

⇒ 3(x1 – x2)[3(x1 + x2) + 2] = 0

⇒ x1 – x2 = 0(∵ x ∈ R+ and hence 3x1 + 3x2 + 2 being sum of 2 positive real numbers and 2 can never be 0)

⇒ x1 = x2

∴ f is one-one

To check whether f is onto:

The function f is given as y = f(x) = 9x² + 6x – 5

⇒ 9x² + 6x – 5 = y

⇒ 9x² + 6x + 1 – 6 = y (so that we can use the identity a² + 2ab + b² = (a + b)²)

⇒ (3x)² + 2×(3x)×1 + 1² – 6 = y

⇒ (3x + 1)² – 6 = y (∵ a² + 2ab + b² = (a + b)²)

⇒ (3x + 1)² = y + 6

⇒ (3x + 1) = \sqrt{y + 6}

⇒ 3x = \left(\sqrt{y + 6}\right) - 1

⇒ x = \dfrac{\left(\sqrt{y + 6}\right) - 1}{3}

The above equation is valid when y + 6 ≥ 0

⇒ y ≥ -6

As the co-domain of f is given as [–5, ∞), every element in y ∈ [–5, ∞) will have an inverse image in the domain.

∴ f is onto.

As f is both one-one and onto, f is invertible and the inverse of f is given as

f-1(y) = \left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right)

Method 2: To show that f is invertible by using hte inverse function f-1

Let y be an arbitrary element in the range of f.

For some x ∈ R+, y = 9x² + 6x – 5

⇒ y = (3x)² + 2×(3x)×1 + 1² – 6

⇒ y + 6 = (3x + 1)²

⇒ (3x + 1)² = y + 6

⇒ 3x + 1 = \sqrt{y + 6}

⇒ 3x = \sqrt{y + 6} – 1

⇒ x = \dfrac{\sqrt{y + 6} - 1}{3}

Let’s now define g:[-5, ∞) → R+, given as

g(y) = \dfrac{\sqrt{y + 6} - 1}{3}

Now

gof(x) = g(f(x))

= g(9x² + 6x – 5) (∵ f(x) = 9x² + 6x – 5)

= \dfrac{\sqrt{9x² + 6x - 5 + 6} - 1}{3} (∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3})

= \dfrac{\sqrt{9x² + 6x + 1} - 1}{3}

= \dfrac{\sqrt{(3x)² + 2\times3x\times1 + 1²} - 1}{3}

= \dfrac{\sqrt{(3x + 1)²} - 1}{3}

= \dfrac{3x + 1 - 1}{3}

= \dfrac{3x}{3}

= x

Also,

fog(y) = f(g(y))

= f\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) (∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3})

= 9\left(\dfrac{\sqrt{y + 6} - 1}{3}\right)^2 + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5 (∵ f(x) = 9x² + 6x – 5)

= {9\left(\dfrac{\left(\sqrt{y + 6}\right)^2 + 2\times\sqrt{y + 6}\times1 + 1^2}{3^2}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5

= {9\left(\dfrac{y + 6 + 2\sqrt{y + 6} + 1}{9}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5

= y + 7 - 2\sqrt{y + 6} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5 (After cutting/cancelling 9 in the numerator and denominator)

= \dfrac{3y + 21 - 6\sqrt{y + 6} + 6\sqrt{y + 6} - 6 - 15}{3}

= \dfrac{3y}{3}

= y

Hence gof = IR+ and fog = I[–5, ∞)

∴ f is convertible and is given as f-1 = g

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) = 1Y(y) = fog2(y). Use one-one ness of f).

(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f).

As f is defined as f:X → Y and also as it is given that f is invertible.

⇒ f is both one-one and onto.

As f is one-one, if there exists any two elements x1, x2 ∈ X such that

f(x1) = f(x2) then x1 = x2

Let’s now assume that the invertible function f has two different inverses g1 and g2 defined as g1:Y → X and g2:Y → X

⇒ fog1(y) = IY and fog2(y) = IY

⇒ fog1(y) = fog2(y)

⇒ f(g1(y)) = f(g2(y))

⇒ g1(y) = g2(y)(∵ f(x1) = f(x2) ⇒ x1 = x2)

⇒ g1 = g2

∴ f has unique inverse.

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.

It is given that

f(1) = a ⇒ f-1(a) = 1

f(2) = b ⇒ f-1(b) = 2

f(3) = c ⇒ f-1(c) = 3

∴ f-1 is defined as f-1:{a, b, c} → {1, 2, 3} and f-1 = {(a, 1), (b, 2), (c, 3)}

As f-1(a) = 1 ⇒ (f-1)-1(1) = a

As f-1(b) = 2 ⇒ (f-1)-1(2) = b

As f-1(c) = 3 ⇒ (f-1)-1(3) = c

∴ (f-1)-1 is defined as (f-1)-1:{1, 2, 3} → {a, b, c} and (f-1)-1 = {(1, a), (2, b), (3, c)}

∴ From the above, it is clear that (f-1)-1 = f

12. Let f:X → Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1)-1 = f.

The inverse of f is f-1

It is given that f is invertible

⇒ f is both one-one and onto.

Consider an arbitrary elemnet x ∈ X, such that f(x) = y

⇒ f-1(y) = x

⇒ (f-1)-1(x) = y

But we know that f(x) = y

So (f-1)-1 = f(x)

As x is an arbitrary element in X, the above statement is true for all x ∈ X

∴ (f-1)-1 = f

13. If f:R → R be given by f(x) = \left(3 - x^3\right)^{\frac{1}{3}}, then fof(x) is

- x^\frac{1}{3}
- x³
- x
- (3 – x³)

fof(x) = f(f(x))

= f\left(\left(3 - x^3\right)^{\frac{1}{3}}\right)

= \left(3 - \left(\left(3 - x^3\right)^\frac{1}{3}\right)^3\right)^\frac{1}{3}

= \left(3 - \left(3 - x^3\right)^{3\times\frac{1}{3}}\right)^\frac{1}{3}

= \left(3 - \left(3 - x^3\right)^1\right)^\frac{1}{3}

= \left(3 - \left(3 - x^3\right)\right)^\frac{1}{3}

= \left(3 - 3 + x^3\right)^\frac{1}{3}

= \left(x^3\right)^\frac{1}{3}

= x

∴ C is the correct answer.

14. Let f:R – \{\dfrac{4}{3}\} → R be a function defined as f(x) = \dfrac{4x}{3x + 4}. The inverse of f is the map g: Range f → R – \{\dfrac{4}{3}\} given by

- g(y) = \dfrac{3y}{3 - 4y}
- g(y) = \dfrac{4y}{4 - 3y}
- g(y) = \dfrac{4y}{3 - 4y}
- g(y) = \dfrac{3y}{4 - 3y}

The given function is y = f(x)

⇒ f-1(y) = x

⇒ g(y) = x (∵ The inverse of f is given as g)

Also, it is given that y = f(x) = \dfrac{4x}{3x + 4}

⇒ \dfrac{4x}{3x + 4} = y

⇒ 4x = (3x + 4)y

⇒ 4x = 3xy + 4y

⇒ 4x – 3xy = 4y

⇒ x(4 – 3y) = 4y

⇒ x = \dfrac{4y}{4 - 3y}4y

⇒ g(y) = \dfrac{4y}{4 - 3y}4y

∴ B is the correct answer