#### Â NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.3 Math

Page No: 18

**Exercise 1.3**

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

**Answer**

The functions f: {1, 3, 4} â†’ {1, 2, 5} and g: {1, 2, 5} â†’ {1, 3} are defined as f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.

gof(1) = g[f(1)] = g(2) = 3 Â Â [as f(1) = 2 and g(2) = 3]

gof(3) = g[f(3)] = g(5) = 1 Â Â [as f(3) = 5 and g(5) = 1]

gof(4) = g[f(4)] = g(1) = 3 Â Â [as f(4) = 1 and g(1) = 3]

âˆ´ 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 + gohÂ

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

**Answer**

To prove: (f + g)oh = foh + goh

LHS = [(f + g)oh](x)

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

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

= (foh)(x) + (goh)(x)

= {(foh)(x) + (goh)}(x) = RHS

âˆ´ {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x âˆˆR

Hence, (f + g)oh = foh + goh

To Prove: (f.g)oh = (foh).(goh)

LHS = [(f.g)oh](x)

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

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

= (foh)(x) . (goh)(x)

= {(foh).(goh)}(x) = RHS

âˆ´ [(f.g)oh](x) = {(foh).(goh)}(x) Â Â for all x âˆˆR

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

3. Find gof and fog, if

(i) f(x) = | x | and g(x) = | 5x â€“ 2 |

(ii) f(x) = 8x

^{3}and g(x) = x^{1/3}.**Answer**

(i). f(x) = |x| and g(x) = |5x-2|

âˆ´gof(x) = g(f(x)) = g(|x|) = |5|x|-2|

fog(x) = f(g(x)) = f(|5x-2|) = ||5x-2|| = |5x-2|

(ii). f(x) = 8x

^{3}Â and g(x) = x

^{1/3}

âˆ´gof(x) = g(f(x)) = g(8x

^{3}) = (8x

^{3})

^{1/3}Â = 2x

fog(x) = f(g(x)) = f(x

^{1/3}) = 8(x

^{1/3})

^{3}Â = 8x

4. If f(x) = (4x+3)/(6x-4), x â‰ 2/3, show that fof(x) = x, for all x â‰ 2/3. What is the inverse of f ?

**Answer**

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

5. State with reason whether following functions have inverse

(i) f : {1, 2, 3, 4} â†’ {10} with

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

(ii) g : {5, 6, 7, 8} â†’ {1, 2, 3, 4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h : {2, 3, 4, 5} â†’ {7, 9, 11, 13} with

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

**Answer**

(i) f: {1, 2, 3, 4} â†’ {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

âˆ´f is not one â€“ one.

Hence, function f does not have an inverse.

(ii) g: {5, 6, 7, 8} â†’ {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.

âˆ´ g is not one â€“ one.

Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} â†’ {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.

âˆ´ Function h is one â€“ 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 â€“ one and onto function.

Hence, h has an inverse.

6. Show that f : [â€“1, 1] â†’ R, given by f (x) = Â x/(x+2)Â is one-one. Find the inverse
of the function f : [â€“1, 1] â†’ Range f.

[Hint: For y âˆˆ Range f, y = f(x) = Â x/(x+2),Â for some x in [â€“1, 1], i.e., x = 2y/(2y-1)]

**Answer**

f : [â€“1, 1] â†’ R, given by f (x) = Â x/(x+2)

For one-one,

Now,

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

**Answer**

f: R â†’ R is given by, f(x) = 4x + 3

For one â€“ one

Let f(x) = f(y)

â‡’ 4x + 3 = 4y + 3

â‡’ 4x = 4y

â‡’ x = y

âˆ´ f is a one â€“ one function.

For onto

For y âˆˆ R, let y = 4x + 3.

â‡’ x = y-3/4 âˆˆ R

Therefore, for any y âˆˆ R, there exists x = y-3/4 âˆˆ R, such that

8.Â 8. Consider f : R+ â†’ [4, âˆž) given by f(x) = x

^{2}Â + 4. Show that f is invertible with the inverse f^{-1}Â of f given by f^{-1}(y) = âˆš(y-4) , where R+ is the set of all non-negative real numbers.**Answer**

f: R+ â†’ [4, âˆž) is given as f(x) = x

^{2}Â + 4.
For one â€“ one

Let f(x) = f(y)

â‡’ x

^{2}Â + 4 = y^{2}Â + 4
â‡’ x

^{2}Â = y^{2}
â‡’ x = y [as x = y âˆˆ R+]

âˆ´ f is a oneâ€“one function.

For onto

Page No. 19

9. Consider f : R+ â†’ [â€“ 5, âˆž) given by f (x) = 9x

^{2}Â + 6x - 5. Show that f is invertible withÂ**Answer**

f: R+ â†’ [âˆ’5, âˆž) is given as f(x) = 9x

^{2}Â + 6x âˆ’ 5.
Let y be an arbitrary element of [âˆ’5, âˆž).

Let y = 9x

^{2}Â + 6x â€“ 5
â‡’ y = (3x+1)

^{2}-1 - 5 = (3x+1)^{2}-6
â‡’ y + 6 = (3x + 1)

^{2}
â‡’ 3x + 1 = âˆš(y + 6)

10. Let f : X â†’ Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g

_{1}and g_{2}Â are two inverses of f. Then for all y âˆˆ Y, fog_{1}(y) = I_{Y}(y) = fog_{2}Â (y). Use one-one ness of f).**Answer**

Let f: X â†’ Y be an invertible function.

Also, suppose f has two inverses (say g

_{1}Â and g_{2})
Then, for all yâˆˆY, we have

fog

_{1}(y) = I_{Y}(y) = fog_{2}(y)
â‡’ f(g

_{1}(y)) = f(g_{2}(y))
â‡’ g

_{1}(y) = g_{2}(y) [as f is invertible â‡’ f is oneâ€“one]
â‡’ g

_{1}Â = g_{2}Â [as g is oneâ€“one]
Hence, f has a unique inverse.

11. Consider f:{1, 2, 3} â†’ {a, b, c} given by f(1) = a,f(2) = b and f(3) = c. Find f

^{-1}Â and show that (f^{-1})^{-1}Â = f.**Answer**

Function f: {1, 2, 3} â†’ {a, b, c} is given by f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c} â†’ {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3.

We have,

(fog)(a) = f(g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f(g(c)) = f(3) = c

and

(gof)(1) = g(f(1)) = f(a) = 1

(gof)(2) = g(f(2)) = f(b) = 2

(gof)(3) = g(f(3)) = f(c) = 3

âˆ´ gof = I

_{X}Â and fog =Â Â I_{Y}, where X = {1, 2, 3} and Y= {a, b, c}.
Thus, the inverse of f exists and f

^{-1}Â = g.
âˆ´ f

^{-1}:{a, b, c} â†’ {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}Â i.e., find the inverse of g.
If we define h: {1, 2, 3} â†’ {a, b, c} as h(1) = a, h(2) = b, h(3) = c

We have

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

and

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

âˆ´ goh = I

_{X}Â and hog = I_{Y}, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g

^{-1}Â = h â‡’ (f^{-1})^{-1}Â = h.
It can be noted that h = f.

Hence, (f

^{-1})^{-1}Â = f.
12. Let f: X â†’ Y be an invertible function. Show that the inverse of f

^{-1}Â is f, i.e., (f^{-1})^{-1}Â = f**Answer**

Let f: X â†’ Y be an invertible function.

Then, there exists a function g: Y â†’ X such that gof = I

_{X}Â and fog = I

_{Y}.

Here, f

^{-1}Â = g.

Now, gof = I

_{X}Â and fog = I

_{Y}

â‡’ f

^{-1}

*o*f = I

_{X}Â and fof

^{-1}Â = I

_{Y}

Hence, f

^{-1}: Y â†’ X is invertible and f is the inverse of f

^{-1}

i.e., (f

^{-1})

^{-1}Â = f.

13. 13. If f: R â†’ R be given by f(x) = ((3 - x

^{3}))

^{1/3}, then fof(x) is

(a) x)

^{1/3}

(b) x

^{3}

(c) x

(d) (3 - x

^{3})

**Answer**

**Answer**

The correct option is B.

Exercise 1.1

Exercise 1.2

Exercise 1.4

Miscellaneous Exercise

Class 12 Math NCERT Solutions