# NCERT Solutions for Class 12 Maths Chapter 1 - Relations and Functions - Exercise 1.3

### Access Exercises of Class 12 Maths Chapter 1 - Relations and Functions

Exercise 1.1 Solutions : 16 Questions (14 Short Answers, 2 MCQ)

Exercise 1.2 Solutions : 12 Questions (10 Short Answers, 2 MCQ)

Exercise 1.3 Solutions : 14 Questions (12 Short Answers, 2 MCQ)

Exercise 1.4 Solutions : 13 Questions (12 Short Answers, 1 MCQ)

Miscellaneous Exercise Solutions: 19 Questions (7 Long answers, 9 Short Answer Type, 3 MCQs)

** 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.**

Sol. The functions f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {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

[∴ f(1) = 2 and g(2) = 3] gof(3) = g(f(3)) = g(5) = 1

[∴ f(3) = 5 and g(5) = 1] gof(4) = g(f(4)) = g(1) = 3

[∴ f(4) = 1 and g(1) = 3]

∴ gof = {(1, 3), (3, 1), (4, 3)}

**2. Let f, g and h be function from R to R. Show that****(f + g)oh = foh + goh;****(f.g)oh = (foh).(goh)**

**Sol.** To prove (f + g)oh = foh + goh Consider

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

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

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

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

∴ ((f + g)oh)(x) = ((foh) + (goh)}(x) ∀ x ∈ R

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

To prove (f.g)oh = (foh).(goh) Consider LHS = ((f.g)oh)(x)

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

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

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

∴ ((f.g)oh(x) = {(foh).(goh)}(x) ∀ x ∈ R

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

**3. Find gof and fog, if****(i) f(x) = |x| and g(x) = |5x – 2|****(ii) f(x) = 8x ^{3} and 8(x) = x^{1/3}.**

**Sol.** **(i)** f(x) = |x| and g(x) = |5x – 2|

∴ (gof)(x) = g(f(x)) = g(|x|)

= |5|x| – 2|

and fog(x) = f(g(x)) = f(|5x – 2|)

= ||5x – 2| = |5x – 2|

**(ii)** f(x) = 8x^{3} and g(x) = x^{1/3}

∴ (gof)(x) = g(f(x))

= g(^{8x3}) = (8x^{3})^{1/3 }= 2x

and fog(x) = f(g(x)) = f(x^{1/3})

= 8(x^{1/3})^{3 }= 8x

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

$$\textbf{Sol.}\text{It is given that f(x)}=\frac{(4x+3)}{(6x-4)},x≠\frac{2}{3},$$

$$\text{(fof)(x) = f(f(x)) }=f(\frac{(4x+3)}{(6x-4)})\\=\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}\\=\frac{16x+12+18x-12}{24x+18-24x+16}\\=\frac{34x}{34}=x\\\text{Therefore, fof(x) = x, for all}\space x≠\frac{2}{3}\space\text{ fof =I}$$

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

**5. 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)}**

**Sol.** **(i)** **Function 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.

Therefore, f is not one-one.

Hence, function f does not have an inverse.

**(ii)** **Function 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. Therefore, g is not one-one.

Hence, function g does not have an inverse.

**(iii)** **Function 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.

Therefore, 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} have distinct images under h. Thus, h is one-one and onto function. Hence, h has an inverse.**Note :** Student should remember that any function has inverse exist only when f is one-one and onto.

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

$$\textbf{Sol}.\text{ Show that f : [– 1, 1] → R, given by f(x)}$$$$=\frac{x}{(x+2)},$$

Let f(x) = f(y)

$$⇒ \frac{x}{x+2}=\frac{y}{y+2}⇒ xy + 2x = xy + 2y\\⇒ 2x = 2y ⇒ x = y\\\text{Therefore, f is a one-one function.}\\\text{Let}\space y=\frac{x}{x+2}⇒x = xy + 2y\\⇒x=\frac{2y}{1-y}$$

So, for every y except 1 in the range there exists x in the domain such that f(x) = y. Hence, function f is onto.

Therefore, f : [– 1, 1] → Range f is one-one and onto and therefore, the inverse of the function f : [– 1, 1] → Range f exists.

Let y be an arbitrary element of range f.

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

y = f(x) for some x → [– 1, 1]

$$⇒y= \frac{x}{x+2}⇒xy+2y=x$$

$$⇒ x(1 – y) = 2y ⇒ x =\frac{2y}{1-y},y≠1$$

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

$$g(y) =\frac{2y}{1-y},y≠1\\Now, (gof)(x) = g(f(x)) =g(\frac{x}{x+2})$$

$$=\frac{2(\frac{x}{x+2})}{1-\frac{x}{x+2}}=\frac{2x}{x+2-x}=\frac{2x}{2}=2$$

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

$$=f(\frac{2y}{1-y})=\frac{(\frac{2y}{1-y})}{\frac{2y}{1-y}+2}\\=\frac{2y}{2y+2-2y}=\frac{2y}{y}=y$$

Therefore, gof = fog = I_{R}, Therefore, f^{ –1} = g

$$\text{Therefore,},f^{-1}(y)=\frac{2y}{1-y},y≠1$$

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

**Sol.** Here, f : R → R is given by f(x) = 4x + 3

Let x, y ∈ R, such that

f(x) = f(y) ⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y ⇒ x = y

Therefore, f is a one-one function.

Let y = 4x + 3

$$\text{⇒ There exist,}\space x=(\frac{y-3}{4})\space \varepsilon R,\forall \space y\varepsilon R$$

$$\text{Therefore, for any y ∈ R, there exist x}=\frac{y-3}{4}∈R$$

$$\text{such that}\space f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3=y$$

Therefore, f is onto function.

Thus, f is one-one and onto and therefore, f^{–1}

exists.

$$\text{Let us define g : R → R by g(x) =}\frac{x-3}{4}\\Now, (gof)(x) = g(f(x)) = g(4x + 3)\\=\frac{(4x+3)-3}{4}=x\\(fog)(y) = f(g(y)) =f(\frac{y-3}{4})\\=4(\frac{y-3}{4})+3=y-3+3=y$$

Therefore, gof = fog = IR

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

$$\text{by} f^{–1}(y) = g(y) =\frac{y-3}{4}.$$

**8. Consider f : R _{+} → [4, ∞) given by f(x) = x^{2} + 4. **

**Show that f is invertible with the inverse f**

^{–1}of**f given by f**

^{–1}(y) =√( y − 4 ), where R_{+}is the set**of all non-negative real numbers.**

**Sol.** Here, function f : R_{+} → [4, ∞) is given as f(x) = x^{2} + 4

Let x, y ∈ R_{+}, such that f(x) = f(y)

⇒ x^{2} + 4 ⇒ x^{2} = y^{2}

⇒ x = y [as x = y ∈ R_{+}]

Therefore, f is one-one function.

For y ∈ [4, ∞), let y = x^{2} + 4

⇒ x^{2} = y – 4 ≥ 0 [as y ≥ 4]

⇒ x =**√** (y − 4) ≥ 0

Therefore, for any y ∈ R_{+}, There exists

x = √(y − 4) ∈ R_{+} such that

f(x) = f (√( y − 4)) = ( √(y − 4))^{2} + 4

= y – 4 + 4 = y

Therefore, f is onto. Thus, f is one-one and onto

and therefore, f^{–1} exists.

Let us define g : [4, ∞) → R+ by g(y) =√(y-4)

Now, gof(x) = g(f(x)) = g(x^{2} + 4)

=√(x^{2}+4)-4=√x^{2}=x

and fog(y) = f(g(y)) =f(√(y-4))

=(√(y-4))^{2}+4

= (y – 4) + 4 = y

Therefore, gof = I_{R}_{+} and I + fog =I [4, ∞)

f^{–1}(y) = g(y) =√(y-4)

**9. Consider f : R+ → [–5, ∞) given by f(x) = 9x ^{2 }**

**+ 6x – 5. Show that f is invertible with f**

^{–1}(y)**$$=(\frac{(\sqrt{y+6})-1}{3})$$**

**Sol.** Here, function f : R+ → [– 5, ∞) is given as

f(x) = 9x^{2} + 6x – 5.

Let y be any arbitrary element of [–5, ∞).

Let y = 9x^{2} + 6x – 5

⇒ y = (3x + 1)^{2} – 1 – 5

= (3x + 1)^{2} – 6

⇒ (3x + 1)^{2} = y + 6

⇒ 3x + 1 = √(y + 6)

[as y ≥ – 5 ⇒ y + 6 ≥ 0]

$$⇒x=\frac{\sqrt{y+6}-1}{3}$$

Therefore, f is onto, there by range f = [– 5, ∞).

$$\text{Let us define g : [– 5, ∞) → R+ as g(y)}=\frac{\sqrt{y+6}-1}{3}$$

Now, (gof)(x) = g(f(x)) = g(9x^{2} + 6x – 5)

= g((3x + 1)^{2} – 6)

$$=\frac{\sqrt{(3x+1)^2-6+6}-1}{3}\\\frac{3x+1-1}{3}=x\\\text{and (fog)(y) = f(g(y))}\\=f(\frac{\sqrt{(y+6)-1}}{3})\\=[3(\frac{\sqrt{(y+6)-1}}{3})+1]^2-6\\=(\sqrt{y+6})^2-6\\= y + 6 – 6 = y$$

Therefore, gof = I_{R}+ and fog = I_{[– 5, ∞)}

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

by

$$f^{–1}(y) = g(y) =\frac{\sqrt{y+6}-1}{3}$$

**10. Let f : X → Y be an invertible function. Show that f has unique inverse. (Hint : suppose g _{1} and g_{2} are two inverses of f. Then for all y ∈ Y, (fog_{1})(y) = Iy(y) = fog_{2}(y). Use one-one of f).**

**Sol.** Let f : X → Y be an invertible function.

Also, suppose f has two inverses (say g_{1} and g_{2}).

Then, for all y ∈ Y, we have

⇒ fog_{1}(y) = Iy(y) = fog_{2}(y)

⇒ f(g_{1}(y)) = f(g_{2}(y))

⇒ g_{1}(y) = g_{2}(y)

[f is invertible ⇒ f is one-one]

⇒ g_{1} = g_{2} [g is one-one]

Hence, f has a 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.**

**Sol.** Here, 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

(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

Therefore, gof = I_{X} and fog = I_{Y}, where X =

{1, 2, 3} and Y = {a, b, c}

Thus, the inverse of f exists and f^{–1} = g.

Therefore, 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

(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

Therefore, goh = IX and hog = IY, 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}) = 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.**

**Sol.** Let f : X → Y be an invertible function.

Then, there exists a function g : Y → X such that gof = I_{X} and fog = I_{Y}.

Here, f^{–1} = g

Now, gof = I_{X} and fog = I_{Y}

Therefore, f^{–1}of = I_{X} and fof^{–1} = I_{Y}

Hence, f^{–1} : Y → X is invertible and f is the inverseof f^{–1}.

i.e., (f^{–1})^{–1} = f.

$$\textbf{13.\space If f} \textbf{: R}\xrightarrow{}\textbf{R}\space\textbf{be given by}\\\textbf{f(x) = (3-x}^{\textbf{3}})^{\frac{\textbf{1}}{\textbf{3}}}\textbf{,}\space \textbf{then fof(x) is :}\\\textbf{(A)\space x}^{\frac{\textbf{1}}{\textbf{3}}}\\\textbf{(B)\space x}^{\textbf{3}}\\\textbf{(C)\space} \textbf{x}\\\textbf{(D)\space}\textbf{(3-x}^{\textbf{3}})$$

**Sol.** (C) x

$$\text{Here, function}\space\text{f}: \text{R}\xrightarrow{}\text{R}\space\\\text{is given as f(x) = (3-x}^{3})^{\frac{1}{3}}\\\therefore\space\text{fof(x) = f(f(x)) = f((3-x}^{3})^{\frac{1}{3}})\\=\lbrack 3 - ((3-x^{3})^{\frac{1}{3}})^{3}\rbrack^{\frac{1}{3}}\\=\lbrack 3 - (3 - x^{3})\rbrack^{\frac{1}{3}}\\=(x^{3})^{\frac{1}{3}} = x$$

∴ fof(x) = x.

$$\textbf{14.\space Let}\space \textbf{f : R} - \begin{Bmatrix}-\frac{\textbf{4}}{\textbf{3}}\end{Bmatrix}\xrightarrow{}\textbf{R}\space$$

**be a function defined as**

$$\textbf{f(x)} =\frac{\textbf{4x}}{\textbf{3x+4}}\textbf{.}\space\textbf{The inverse of }\space\\\textbf{f is the map g : Range}\\\textbf{f}\xrightarrow{}\textbf{R} -\begin{Bmatrix}-\frac{\textbf{4}}{\textbf{3}}\end{Bmatrix}\space\textbf{is given by}\\\textbf{(A)\space g(y) =}\frac{\textbf{3y}}{\textbf{3-4y}}\\\textbf{(B)\space}\textbf{g(y) =} \frac{\textbf{4y}}{\textbf{4-3y}}\\\textbf{(C)\space}\textbf{g(y) = }\frac{\textbf{4y}}{\textbf{3-4y}}\\\textbf{(D)\space g(y) =}\frac{\textbf{3y}}{\textbf{4-3y}}\\\textbf{Sol.\space}\text{(B)} \space g(y) =\frac{4y}{4-3y} $$

$$\text{Given that}\space \text{f : R-}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}\xrightarrow{}\text{R}\space$$

is defined as

$$\text{f(x) = }\frac{4x}{3x + 4}.$$

Let y be an arbitrary element of range f.

$$\text{Then, there exists x} \epsilon \text{R −}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}$$

such that y = f(x)

$$\Rarr\space y =\frac{4x}{3x+4}\space\\\Rarr\space 3xy + 4y = 4x\\\Rarr\space x(4-3y) = 4y\\\Rarr\space x = \frac{4y}{4-3y}$$

Let us define g : Range

$$f\xrightarrow{} \text{R} -\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}\space\text{as}\space \text{g(y)} =\frac{4y}{4-3y}.$$

$$\text{Now, (gof)(x) = g(f(x)) =}\\g\bigg(\frac{4x}{3x+4}\bigg)\\=\frac{4\bigg(\frac{4x}{3x+4}\bigg)}{4-3\bigg(\frac{4x}{3x+4}\bigg)}\\=\frac{16x}{12x + 16 - 12 x} =\frac{16x}{16} = x$$

$$\text{and (fog)(y) = f(g(y))} =\\ f\bigg(\frac{4y}{4-3y}\bigg)\\=\frac{4\bigg(\frac{4y}{4-3y}\bigg)}{3\bigg(\frac{4y}{4-3y}\bigg) + 4}\\=\frac{16y}{12y +16- 12y}\\=\frac{16y}{16} = y$$

$$\text{Therefore,}\space\text{gof = I}_{\text{R -}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}}\\\text{and fog = I}_{R}$$

Thus, g is the inverse of f i.e., f^{–1} = g.

Hence, the inverse of f is the map g : Range

$$f\xrightarrow{}\text{R} -\bigg(-\frac{4}{3}\bigg),\space\\\text{which is given by g(y)} =\frac{4y}{4-3y}.$$

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