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

NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 introduces a few more concepts of relations and functions which includes inverse function, imposition of one function on other eg. fog(x) or gof(x) etc. Exercise 1.3 Class 12 Maths will help students to grasp the basic concepts related to finding the inverse of a function. Practicing this exercise is of utmost importance because most of the questions related to finding the inverse of a function are asked hence students can go through NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 to score well in CBSE class 12 board exam. In competitive exams also like JEE main ,some questions can be asked from Class 12 Maths chapter 1 exercise 1.3. Concepts related to inverse functions discussed in Class 12th Maths chapter 1 exercise 1.3 can be useful for NEET physics syllabus, as the problems in physics use the concepts of functions. The NCERT chapter Relations and Functions also has the following exercise for practice.

Relations and Functions Exercise 1.1

Relations and Functions Exercise 1.2

Relations and Functions Exercise 1.4

Relations and Functions Miscellaneous Exercise

## Question:1 Let $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$ and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$ be given by $f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$ . Write down $gof$ .

Given : $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$ and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$

$f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$

$gof(1) = g(f(1))=g(2) = 3$ $\left [ f(1)=2 \, and\, g(2)=3 \right ]$

$gof(3) = g(f(3))=g(5) = 1$ $\left [ f(3)=5 \, and\, g(5)=1 \right ]$

$gof(4) = g(f(4))=g(1) = 3$ $\left [ f(4)=1 \, and\, g(1)=3 \right ]$

Hence, $gof$ = $\left \{ (1,3),(3,1),(4,3) \right \}$

Question:2 Let $f$ , $g$ and $h$ be functions from $R$ to $R$ . Show that $\\(f + g) o h = foh + goh\\ (f \cdot g) o h = (foh) \cdot (goh)$

To prove : $\\(f + g) o h = foh + goh$

$((f + g) o h)(x)$

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

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

$=(f o h)(x) +(goh)(x)$

$=\left \{ (f o h) +(goh) \right \}(x)$ $x\forall R$

Hence, $\\(f + g) o h = foh + goh$

To prove: $(f \cdot g) o h = (foh) \cdot (goh)$

$((f . g) o h)(x)$

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

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

$=(f o h)(x) . (goh)(x)$

$=\left \{ (f o h) .(goh) \right \}(x)$ $x\forall R$

$\therefore$ $(f \cdot g) o h = (foh) \cdot (goh)$

Hence, $(f \cdot g) o h = (foh) \cdot (goh)$

Question:3(i) Find $gof$ and $fog$ , if

(i) $f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

$f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

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

$= g( | x |)$

$= |5 | x |-2|$

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

$=f( \left | 5x-2 \right |)$

$=\left \| 5x-2 \right \|$

$=\left | 5x-2 \right |$

Question:3(ii) Find gof and fog, if

(ii) $f (x) = 8x^{3}$ and $g(x) = x^{\frac{1}{3}}$

The solution is as follows

(ii) $f (x) = 8x^{3}$ and $g(x) = x^{\frac{1}{3}}$

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

$= g( 8x^{3})$

$= ( 8x^{3})^{\frac{1}{3}}$

$=2x$

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

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

$=8((x^{\frac{1}{3}} )^{3})$

$=8x$

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

$f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$

$fof (x) = x$

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

$=f( \frac{4x + 3}{6x - 4})$

$=\frac{4( \frac{4x + 3}{6x - 4}) +3}{6( \frac{4x + 3}{6x - 4}) -4}$

$= \frac{16x+12+18x-12}{24x+1824x+16}$

$= \frac{34x}{34}$

$\therefore fof(x) = x$ , for all $x \neq \frac{2}{3}$

$\Rightarrow fof=Ix$

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

Question:5(i) State with reason whether following functions have inverse

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$

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

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$ with
$f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

From the given definition,we have:

$f\left ( 1 \right )=f\left ( 2 \right )=f\left ( 3 \right )=f(4)=10$

$\therefore$ f is not one-one.

Hence, f do not have an inverse function.

Question:5(ii) State with reason whether following functions have inverse

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

From the definition, we can conclude :

$g(5)=g(7)=4$

$\therefore$ g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

From the definition, we can see the set $\left \{ 2,3,4,5 \right \}$ have distant values under h.

$\therefore$ h is one-one.

For every element y of set $\left \{ 7,9,11,13 \right \}$ ,there exists an element x in $\left \{ 2,3,4,5 \right \}$ such that $h(x)=y$

$\therefore$ h is onto

Thus, h is one-one and onto so h has an inverse function.

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

$f : [-1, 1] \rightarrow R$

$f(x) = \frac{x}{(x + 2)}$

One -one:

$f(x)=f(y)$

$\frac{x}{x+2}=\frac{y}{y+2}$

$x(y+2)=y(x+2)$

$xy+2x=xy+2y$

$2x=2y$

$x=y$

$\therefore$ f is one-one.

It is clear that $f : [-1, 1] \rightarrow Range f$ is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in $Range f\rightarrow [-1, 1]$

$g: Range f\rightarrow [-1, 1]$

let y be an arbitrary element of range f

Since, $f : [-1, 1] \rightarrow R$ is onto, so

$y=f(x)$ for $x \in \left [ -1,1 \right ]$

$y=\frac{x}{x+2}$

$xy+2y=x$

$2y=x-xy$

$2y=x(1-y)$

$x = \frac{2y}{1-y}$ , $y\neq 1$

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

$f^{-1}=\frac{2y}{1-y},y\neq 1$

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

$f : R \rightarrow R$ is given by $f (x) = 4x + 3$

One-one :

Let $f(x)=f(y)$

$4x + 3 = 4y+3$

$4x=4y$

$x=y$

$\therefore$ f is one-one function.

Onto:

$y=4x+3\, \, \, , y \in R$

$\Rightarrow x=\frac{y-3}{4} \in R$

So, for $y \in R$ there is $x=\frac{y-3}{4} \in R$ ,such that

$f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3$

$= y-3+3$

$= y$

$\therefore$ f is onto.

Thus, f is one-one and onto so $f^{-1}$ exists.

Let, $g:R\rightarrow R$ by $g(x)=\frac{y-3}{4}$

Now,

$(gof)(x)= g(f(x))= g(4x+3)$

$=\frac{(4x+3)-3}{4}$

$=\frac{4x}{4}$

$=x$

$(fog)(x)= f(g(x))= f(\frac{y-3}{4})$

$= 4\times \frac{y-3}{4}+3$

$= y-3+3$

$= y$

$(gof)(x)= x$ and $(fog)(x)= y$

Hence, function f is invertible and inverse of f is $g(y)=\frac{y-3}{4}$ .

Question: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 $f^{-1}(y)= \sqrt{y-4}$ , where R+ is the set of all non-negative real numbers.

It is given that
$f : R^+ \rightarrow [4,\infty)$ , $f(x) = x^2+4$ and

Now, Let f(x) = f(y)

⇒ x 2 + 4 = y 2 + 4

⇒ x 2 = y 2

⇒ x = y

⇒ f is one-one function.

Now, for y $\epsilon$ [4, ∞), let y = x 2 + 4.

⇒ x 2 = y -4 ≥ 0

⇒ for any y $\epsilon$ R, there exists x = $\epsilon$ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

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

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I R .

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

f-1(y) = g(y) =

Question:9 Consider $f : R_+ \rightarrow [- 5, \infty)$ given by $f (x) = 9x^2 + 6x - 5$ . Show that $f$ is invertible with $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

$f : R_+ \rightarrow [- 5, \infty)$

$f (x) = 9x^2 + 6x - 5$

One- one:

Let $f(x)=f(y)$ $for \, \, x,y\in R$

$9x^{2}+6x-5=9y^{2}+6y-5$

$9x^{2}+6x=9y^{2}+6y$

$\Rightarrow$ $9(x^{2}-y^{2})=6(y-x)$

$9(x+y)(x-y)+6(x-y)= 0$

$(x-y)(9(x+y)+6)=0$

Since, x and y are positive.

$(9(x+y)+6)> 0$

$\therefore x=y$

$\therefore$ f is one-one.

Onto:

Let for $y \in [-5,\infty)$ , $y=9x^{2}+6x-5$

$\Rightarrow$ $y=(3x+1)^{2}-1-5$

$\Rightarrow$ $y=(3x+1)^{2}-6$

$\Rightarrow$ $y+6=(3x+1)^{2}$

$(3x+1)=\sqrt{y+6}$

$x = \frac{\sqrt{y+6}-1}{3}$

$\therefore$ f is onto and range is $y \in [-5,\infty)$ .

Since f is one-one and onto so it is invertible.

Let $g : [-5,\infty)\rightarrow R_+$ by $g(y) = \frac{\sqrt{y+6}-1}{3}$

$(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{{(3x+1)^{2}}-6+6} -1$

$(gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x$

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

$=[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6$

$=(\sqrt{y+6})^{2}-6$

$=y+6-6$

$=y$

$\therefore gof=fog=I_R$

Hence, $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Question:10 Let $f : X \rightarrow Y$ be an invertible function. Show that f has a unique inverse. (Hint: suppose $g_1$ and $g_2$ are two inverses of $f$ . Then for all $y \in Y$ ,
$fog_1 (y) = I_Y (y) = fog_2 (y)$ . Use one-one ness of f).

Let $f : X \rightarrow Y$ be an invertible function

Also, suppose f has two inverse $g_1 and g_2$

For $y \in Y$ , we have

$fog_1(y) = I_y(y)=fog_2(y)$

$\Rightarrow$ $f(g_1(y))=f(g_2(y))$

$\Rightarrow$ $g_1(y)=g_2(y)$ [f is invertible implies f is one - one]

$\Rightarrow$ $g_1=g_2$ [g is one-one]

Thus,f has a unique inverse.

Question:11 Consider $f : \{1, 2, 3\} \rightarrow \{a, b, c\}$ 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
$f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$

$f(1) = a, f(2) = b \ and \ f(3) = c$

Now,, lets define a function g :
$\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ such that

$g(a) = 1, g(b) = 2 \ and \ g(c) = 3$
Now,

$(fog)(a) = f(g(a)) = f(1) = a$
Similarly,

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

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

And

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

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

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

Hence, $gof = I_X$ and $fog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, the inverse of f exists and $f^{-1} = g$

Now,
$f^{-1} : \left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ is given by

$f^{-1}(a) = 1, f^{-1}(b) = 2 \ and \ f^{-1}(c) = 3$

Now, we need to find the inverse of $f^{-1}$ ,

Therefore, lets define $h: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$ such that

$h(1) = a, h(2) = b \ and \ h(3) = c$

Now,

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

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

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

Similarly,

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

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

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

Hence, $goh = I_X$ and $hog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$

$\Rightarrow h = f$

Therefore, $(f^{-1})^{-1} = f$

Hence proved

Question:12 Let $f : X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$ , i.e., $(f^{-1})^{-1} = f$

$f : X \rightarrow Y$

To prove: $(f^{-1})^{-1} = f$

Let $f:X\rightarrow Y$ be a invertible function.

Then there is $g:Y\rightarrow X$ such that $gof =I_x$ and $fog=I_y$

Also, $f^{-1}= g$

$gof =I_x$ and $fog=I_y$

$\Rightarrow$ $f^{-1}of = I_x$ and $fof^{-1} = I_y$

Hence, $f^{-1}:Y\rightarrow X$ is invertible function and f is inverse of $f^{-1}$ .

i.e. $(f^{-1})^{-1} = f$

Question:13 If $f : R \rightarrow R$ be given by $f(x) = (3 - x^3)^{\frac{1}{3}}$ , then $fof(x)$ is

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) $x$

(D) $(3 - x^3)$

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

$fof(x)$ $=f(f(x))=f((3-x^{3})^{\frac{1}{3}})$

$=[3- ((3-x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}$

$=([3- ((3-x^{3})]^{\frac{1}{3}})$

$= (x^{3})^{\frac{1}{3}}$

$=x$

Thus, $fof(x)$ is x.

Hence, option c is correct answer.

Question:14 Let $f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x + 4}$ . The inverse of $f$ is the map $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$ given by

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(C) $g(y) = \frac{4y}{3 -4y}$

(D) $g(y) = \frac{3y}{3 -4y}$

$f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$

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

Let f inverse $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$

Let y be the element of range f.

Then there is $x \in R - \left\{-\frac{4}{3}\right\}$ such that

$y=f(x)$

$y=\frac{4x}{3x+4}$

$y(3x+4)=4x$

$3xy+4y=4x$

$3xy-4x+4y=0$

$x(3y-4)+4y=0$

$x= \frac{-4y}{3y-4}$

$x= \frac{4y}{4-3y}$

Now , define $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$ as $g(y)= \frac{4y}{4-3y}$

$gof(x)= g(f(x))= g(\frac{4x}{3x+4})$

$= \frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}$

$=\frac{16x}{12x+16-12x}$

$=\frac{16x}{16}$

$=x$

$fog(y)=f(g(y))=f(\frac{4y}{4-3y})$

$= \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$=y$

Hence, g is inverse of f and $f^{-1}=g$

The inverse of f is given by $g(y)= \frac{4y}{4-3y}$ .

The correct option is B.

More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3

The NCERT class 12 maths chapter Relations and Functions consists of a total of 5 exercises including miscellaneous exercise. Exercise 1.3 Class 12 Maths covers solutions to 14 main questions and their sub-questions. Most questions are related to finding out the inverse of a function. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 can be referred by students in case of any doubt.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

Benefits of NCERT Solutions for class 12 maths chapter 1 exercise 1.3

• The Class 12th maths chapter 1 exercise is provided above in detail which is solved by subject matter experts according to the NCERT syllabus .

• Students are recommended to practice Exercise 1.3 Class 12 Maths to prepare for topics like inverse functions etc., direct questions are asked in Board exams.

• These Class 12 Maths NCERT book chapter 1 exercise 1.3 solutions can be referred by students to revise just before the exam.

• NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 can be used to prepare inverse function topics of physics also.

### Also see-

• NCERT exemplar solutions class 12 maths chapter 1

• NCERT solutions for class 12 maths chapter 1

NCERT Solutions Subject Wise

• NCERT solutions class 12 chemistry

• NCERT solutions for class 12 physics

• NCERT solutions for class 12 biology

• NCERT solutions for class 12 mathematics

Subject wise NCERT Exemplar solutions

• NCERT Exemplar Class 12th Maths
• NCERT Exemplar Class 12th Physics

• NCERT Exemplar Class 12th Chemistry

• NCERT Exemplar Class 12th Biology

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