exercise 1.3 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.3 || ex 1.3 class 12 maths ncert solutions|| relations and functions class 12 ncert solutions (English Medium)
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NCERT Solutions for Class 12 Maths – Chapter 1 Exercise 1.3 (Relations and Functions)
Exercise 1.3 of Class 12 Maths NCERT Solutions explores advanced concepts from the chapter Relations and Functions, focusing specifically on the composition of functions and invertible functions. This section is essential for understanding how to combine functions and find their inverses, which are widely used in higher mathematics. The Class 12 Maths Exercise 1.3 NCERT Solutions provide step-by-step guidance and clear explanations to help students tackle each question with confidence. These solutions are prepared in accordance with the latest CBSE syllabus and are ideal for English Medium students. Practicing regularly from the Class 12 Maths NCERT Solutions Chapter 1 Exercise 1.3 helps develop a deeper understanding of function operations and their properties. The Exercise 1.3 Class 12 Maths NCERT Solutions are a valuable resource for thorough exam preparation and strengthening conceptual clarity in relations and functions.

relations and functions class 12 ncert solutions || ex 1.3 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.3 || exercise 1.3 class 12 maths ncert solutions
Exercise 1.3
\( \mathrm{f}=\{(1,2),(3,5),(4,1)\} \) and \( \mathrm{g}=\{(1,3),(2,3),(5,1)\} \).
Then, \( g \circ f(1)= \) \( \mathrm{g}(\mathrm{f}(1))=\mathrm{g}(2)=3 g \circ f (3) \)
\( =\mathrm{g}(\mathrm{f}(3))=\mathrm{g}(5)=1 \operatorname{g \circ f}(4)=\mathrm{g}(\mathrm{f}(4))=\mathrm{g}(1)=3 \)
Therefore, \( g \circ f=\{(1,3),(3,1),(4,3)\} \)
relations and functions class 12 ncert solutions || ex 1.3 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.3 || exercise 1.3 class 12 maths ncert solutions
2.
\( (\mathrm{f}+\mathrm{g}) \circ h = f \circ h + g \circ h\)
Let us consider \( ((\mathrm{f}+\mathrm{g}) \circ h )(x)=(\mathrm{f}+\mathrm{g})(\mathrm{h}(x)) \)
\(
=\mathrm{f}(\mathrm{h}(x))+\mathrm{g}(\mathrm{h}(x))\)
\(=(f\circ h)(x)+(g \circ h)(x)\)
\(=\{(f\circ h)+(g \circ h)\}(x)
\)
Then, \( ((\mathrm{f}+\mathrm{g}) \circ h )(x)=\{( f\circ h )+(g \circ h)\}(x) \forall x \in \mathrm{R} \)
Therefore, \( (\mathrm{f}+\mathrm{g}) \circ h = f\circ h + g\circ h.\)
\((f. g)\circ h = (f\circ h). (g\circ h)\)
Let us consider \( (({f} . {g}) \circ h )({x})=({f} . {g})({h}({x})) \)
\( ={f}({h}({x})) \cdot {g}({h}({x}))\)
\(={f}({h}({x})) \cdot {g}({h}({x}))\)
\(=( f \circ g )({x}) \cdot({g\circ h})({x})\)
\(=\{({f\circ g}) \cdot({g\circ h })\}({x})\)
Then, \( ((f. g) \circ h) (x) =\{( f \circ g ) (g\circ h) \} (x) \forall {x} \in {R} \)
Therefore, \((f. g) \circ h = (f\circ g). (g\circ h)\)
(i) \( \mathrm{f}(x)=|x| \) and \( \mathrm{g}(x)=|5 x-2| \)
Then, \( (g \circ f)(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}(|x|)=|5| x|-2| \)
\( (f \circ g)(x)=\mathrm{f}(\mathrm{g}(x))=\mathrm{f}(|5 x-2|)=\|5 x-2 \|=| 5 x-2|\)
\( \mathrm{f}(x)=8 x^{3} \) and \( \mathrm{g}(x)=x^{\frac{1}{3}} \)
Then, \( ( g \circ f )(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}(8 x^{3})=\left(8 x^{3}\right)^{\frac{1}{3}}=2 \)
\( (f \circ g)(x)=\mathrm{f}(\mathrm{g}(x))=\mathrm{f}\left(x^{\frac{1}{3}}\right)=8\left(x^{\frac{1}{3}}\right)^{3}=8 x \)
relations and functions class 12 ncert solutions || ex 1.3 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.3 || exercise 1.3 class 12 maths ncert solutions
\( ( f\circ f )(x)=\mathrm{f}(\mathrm{f}(x))=\mathrm{f} \frac{(4 x+3)}{(6 x-4)}=\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}=\frac{16 x+12+18 x-12}{24 x-18-24 x+16}=\frac{34 x}{34}=x \)
Therefore, \( f \circ f (x)=x \), for all \( x \neq \frac{2}{3} \)
\(= > f\circ f =1\)
Therefore, the given function f is invertible and the inverse of f itself.
\(\mathrm{f}:\{1,2,3,4\} \rightarrow\{10\} \) with
\(\mathrm{f}=\{(1,10),(2,10),(3,10),(4,10)\}\)
\(\mathrm{f}=\{(1,10),(2,10),(3,10),(4,10)\}\)
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.
Therefore, function \(f\) does not have an inverse.
\( g:\{5,6,7,8\} \rightarrow\{1,2,3,4\} \) with
\(g=\{(5,4),(6,3),(7,4),(8,2)\}\)
\(
g=\{(5,4),(6,3),(7,4),(8,2)\}
\)
Necessary Condition for a function to have inverse: Function should be one-one and onto.
From the given definition,
We can see that f is a many one function as:
\(G(5)=g(7)=4\)
[As at two points function have same values, the function is not one-one] i.e. g is not one- one.
Therefore, function g does not have an inverse.
\( \mathrm{h} :\{2,3,4,5\} \rightarrow\{7,9,11,13\} \) with
\(
\mathrm{h}=\{(2,7),(3,9),(4,11),(5,13)\}
\)
\(
\mathrm{h}=\{(2,7),(3,9),(4,11),(5,13)\}
\)
We can see that all distinct elements of the set \( (2,3,4,5) \) have distinct images under \( h \).
\( \Rightarrow \mathrm{h} \) is one-one.
Also, h is onto since for every element of the set \( \{7,9,11,13\} \), there exists an element \(x\) in the set \( \{2,3,4,5\} \) such that \( \mathrm{h}(x)=\mathrm{y} \).
Therefore, h is a one-one and onto function.
Therefore, \( h \) has an inverse.
Now, Let \( \mathrm{f}(x)=\mathrm{f}(\mathrm{y}) \)
\(
\Rightarrow \frac{x}{x+2}=\frac{y}{y+2}\)
\(\Rightarrow x y+2 x=x y+2 y\)
\(\Rightarrow 2 x=2 y\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is a one- one function.
Now, Let \( \mathrm{y}=\frac{x}{x+2}, \mathrm{xy}=x+2 \mathrm{y} \) so \( x=\frac{y}{1-y} \)
So, for every \( y \) in the range there exists \( x \) in the domain such that \( f(x)= \) y
\( \Rightarrow \mathrm{f} \) is onto function.
\( \Rightarrow \mathrm{f}:[-1,1] \rightarrow \) Range f is one-one and onto
\( \Rightarrow \) the inverse of the function: \( \mathrm{f}:[-1,1] \rightarrow \) Range f exists.
Let \( g \) : Range \( f \rightarrow[-1,1] \) be the inverse of range \( f \).
Let \( y \) be an arbitrary element of range \( f \).
Since, \( \mathrm{f}:[-1,1] \rightarrow \) Range f is onto, we get:
\( \mathrm{y}=\mathrm{f}(x) \) for same \( x \in[-1,1] \)
\(\Rightarrow \mathrm{y}=\frac{x}{x+2}\)
\(\Rightarrow x\mathrm{y}+2 \mathrm{y}=x\)
\(\Rightarrow x(1-\mathrm{y})=2 \mathrm{y}\)
\(\Rightarrow x=\frac{2 y}{1-y}, \mathrm{y} \neq 1
\)
Now, let us define \(g :\) Range \( \mathrm{f} \rightarrow[-1,1] \)
\(
\mathrm{g}(\mathrm{y})=\frac{2 y}{1-y}, \mathrm{y} \neq 1
\)
Now, \( (g \circ f)(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(\frac{x}{x+2}\right)=\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}=\frac{2 x}{x+2-x}=\frac{2 x}{2} \)
\(
(f \circ g)(\mathrm{y})=\mathrm{f}(\mathrm{g}(\mathrm{y}))=\mathrm{f}\left(\frac{2 y}{1-y}\right)=\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}=\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}
\)
Thus, \(g \circ f =\mathrm{I}[-1,1] \) and \( f \circ g =\mathrm{I}_{\text {Range } \mathrm{f}} \)
\(
\Rightarrow \mathrm{f}^{-1}=\mathrm{g}
\)
Therefore, \( \mathrm{f}-1(\mathrm{y})=\frac{2 y}{1-y}, \mathrm{y} \neq 1 \)
Let \( f(x)=f(y) \)
\( \Rightarrow 4 x+3=4 \mathrm{y}+3 \)
\( \Rightarrow 4 x=4 y \)
\( \Rightarrow x=\mathrm{y} \)
\( \Rightarrow \mathrm{f} \) is one- one function.
Now, for \( \mathrm{y} \in \mathrm{R} \), Let \( \mathrm{y}=4 x+3 \)
\( =x=\frac{y-3}{4} \in \mathrm{R} \)
\( \Rightarrow \) for any \( \mathrm{y} \in \mathrm{R} \), there exists \( x=\frac{Y-3}{4} \in \mathrm{R} \)
such that, \( \mathrm{f}(x)=\mathrm{f}\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=\mathrm{y} \)
\( \Rightarrow \mathrm{F} \) is onto function.
Since, f is one-one and onto
\( \Rightarrow \mathrm{f}-1 \) exists.
Let us define \( \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} \) by \( \mathrm{g}(x)=\frac{x-3}{4} \)
Now, \( (g \circ f)(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}(4 x+3)=\frac{(4 x+3)-3}{4}=x \)
\( (f \circ g)(y)=f(g(x y))=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y \)
Therefore, \(g \circ f = f \circ g =I R \)
Therefore, f is invertible and the inverse of f is given by \( \mathrm{f}^{-1}(\mathrm{y})=\mathrm{g}(\mathrm{y})=\frac{y-3}{4} \).
Now, Let \( \mathrm{f}(x)=\mathrm{f}(\mathrm{y}) \)
\(\Rightarrow x^{2}+4=y^{2}+4\)
\(\Rightarrow x^{2}=y^{2}\)
\(\Rightarrow x=\mathrm{y}
\)
\( \Rightarrow \mathrm{f} \) is one-one function.
Now, for \( \mathrm{y} \in[4, \infty) \), let \( \mathrm{y}=x^{2}+4 \).
\(
\Rightarrow x^{2}=y-4 \geq 0\)
\(=x=\sqrt{y-4} \geq 0
\)
\( \Rightarrow \) for any \( \mathrm{y} \in \mathrm{R} \), there exists \( x=\sqrt{y-4} \in \mathrm{R} \) such that
\(
=\mathrm{f}(x)=\mathrm{f}(\sqrt{y-4})=(\sqrt{y-4})^{2} y-4+4=\mathrm{y}
\)
\( \Rightarrow \mathrm{f} \) is onto function.
Therefore, f is one-one and onto function, so \( \mathrm{f}-1 \) exists.
Now, let us define \( g:[4, \infty) \rightarrow \mathrm{R}+ \) by,
\(
\mathrm{g}(\mathrm{y})=\sqrt{y-4}
\)
Now, \( g \circ f(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(x^{2}+4\right)=\sqrt{\left(x^{2}+4\right)+4}=\sqrt{x^{2}}=x \)
And, \( f \circ g(\mathrm{y})=\mathrm{f}(\mathrm{g}(\mathrm{y}))=\mathrm{f}(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=(\mathrm{y}-4)+4=\mathrm{y} \)
Therefore, \(g \circ f = g \circ f = IR\).
Therefore, \( f \) is invertible and the inverse of \( f \) is given by \( \mathrm{f}-1(\mathrm{y})=\mathrm{g}(\mathrm{y})=\sqrt{y-4} \)
Let \( y \) be any element of \( [-5, \infty) \)
Now, let \( y=9 x^{2}+6 x-5 \)
\(\Rightarrow y=(3 x+1)^{2}-1-5=(3 x+1)^{2}-6\)
\(\Rightarrow 3 x+1=\sqrt{y+6}\)
\(=x=\frac{\sqrt{y+6}-1}{3}
\)
\( \Rightarrow \mathrm{f} \) is onto and its range is \( \mathrm{f}=[-5, \infty) \)
Now, let us define \( \mathrm{g}:[-5, \infty) \rightarrow \mathrm{R}+ \) as \( \mathrm{g}(\mathrm{y})=\frac{\sqrt{y+6}-1}{3} \)
Now, we have:
\(
(g \circ f)(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(9 x^{2}+6 x-5\right)\)
\(=\mathrm{g}((3 x+1) 2-6)\)
\(=\frac{\sqrt{(3 x+1)^{2}-6+6}-1}{3}\)
\(=\frac{3 x+1-1}{3}=x
\)
And, \( (f \circ g)(y)=f(g(y))=f\left(\frac{\sqrt{y+6}-1}{3}\right) \)
\(
=\left[3\left(\frac{\sqrt{y+6}-1}{3}\right)+1\right]^{2}-6\)
\(=(\sqrt{y+6})^{2}-6=y+6-6=y
\)
Thus, \( g \circ f =\mathrm{IR} \) and \( f \circ g=\mathrm{I}(-5, \infty) \)
Therefore, f is invertible and the inverse of f is given by
\(
f-1(y)=g(y)=\frac{\sqrt{y+6}-1}{3}
\)
relations and functions class 12 ncert solutions || ex 1.3 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.3 || exercise 1.3 class 12 maths ncert solutions
Also, suppose \(f\) has two inverse
Then, for all \( \mathrm{y} \in\ Y \), we get:
\(f \circ g_{1}(y)=I_{1}(y)=f \circ g_{2}(y)\)
\(= > f\left(g_{1}(y)\right)=f\left(g_{2}(y)\right)\)
\(= > g_{1}(y)=g_{2}(y)\)
\(= > g_{1}=g_{2}\)
Therefore, f has a unique inverse.
\( \mathrm{f}(1)=\mathrm{a}, \mathrm{f}(2)=\mathrm{b} \) and \( \mathrm{f}(3)=\mathrm{c} \)
So, if we define g :
\( \{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \rightarrow\{1,2,3\} \) as
\( \mathrm{g}(\mathrm{a})=1, \mathrm{~g}(\mathrm{b})=2, \mathrm{g}(\mathrm{c})=3 \), then we get:
\( (f \circ g)(a)=f(g(a))=f(1)=a \)
\( (f \circ g)(b)=f(g(b))=f(2)=b \)
\( (f \circ g)(\mathrm{c})=\mathrm{f}(\mathrm{g}(\mathrm{c}))=\mathrm{f}(3)=\mathrm{c} \)
And
\( (g \circ f) (1)=\mathrm{g}((1))=\mathrm{g}(\mathrm{a})=1 \)
\( (g \circ f) (2)=g(f(2))=g(b)=2 \)
\( (g \circ f)(3)=\mathrm{g}(\mathrm{f}(3))=\mathrm{g}(\mathrm{c})=3 \)
Therefore, \(g \circ f =\mathrm{IX} \) and \( f \circ g =\mathrm{IY} \), where \( X=\{1,2,3\} \) and \( \mathrm{Y}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \)
Thus, the inverse of f exists and \( \mathrm{f}-1=\mathrm{g} \).
Then, \( \mathrm{f}^{-1}:\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \rightarrow\{1,2,3\} \) is given by
\(
\mathrm{f}^{-1}(\mathrm{a})=1, \mathrm{f}^{-1}(\mathrm{b})=2, \mathrm{f}^{-1}(\mathrm{c})=3
\)
Let us now find the inverse of \( \mathrm{f}^{-1} \),
So, if we define \( H_\{1,2,3\} \rightarrow\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \) as
\( h(1)=a, h(2)=b, h(3)=c \), then we get:
\((g \circ h) (1)=\mathrm{g}(\mathrm{h}(1))=\mathrm{g}(\mathrm{a})=1 \)
\( (g \circ h) (2)=\mathrm{g}(\mathrm{h}(2))=\mathrm{g}(\mathrm{b})=2 \)
\( (g \circ h)(3)=\mathrm{g}(\mathrm{h}(3))=\mathrm{g}(\mathrm{c})=3 \)
And,
\((h\circ g)(\mathrm{a})=\mathrm{h}(\mathrm{g}(\mathrm{a}))=\mathrm{h}(1)=\mathrm{a}\)
\((\operatorname{hog})(\mathrm{b})=\mathrm{h}(\mathrm{g}(\mathrm{b}))=\mathrm{h}(2)=\mathrm{b}\)
\((\operatorname{hog})(\mathrm{c})=\mathrm{h}(\mathrm{g}(\mathrm{c}))=\mathrm{h}(3)=\mathrm{c}
\)
\( \Rightarrow g \circ h =\mathrm{I}_{X} \) and hog \( =\mathrm{I}_{\mathrm{Y}} \), where \( x=\{1,2,3\} \) and \( \mathrm{Y}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \).
\( \Rightarrow \) The inverse of g exists and \( \mathrm{g}^{-1}=\mathrm{h} \)
\(\Rightarrow\left(\mathrm{f}^{-1}\right)^{-1}=\mathrm{h}\)
\(\Rightarrow \mathrm{h}=\mathrm{f}\)
\(\Rightarrow\left(\mathrm{f}^{-1}\right)^{-1}=\mathrm{f}
\)
Then, there exists a function \( \mathrm{g}: \mathrm{Y} \rightarrow X \) such that \( g \circ f=\mathrm{Ix} \) and \( f \circ g=\mathrm{Iy} \).
Then, \( \mathrm{f}^{-1}=\mathrm{g} \).
Now, \(g \circ f = Ix\) and \(f \circ g = Iy\)
\( \Rightarrow \mathrm{f}^{-1} \circ f=\mathrm{Ix} \) and \(f \circ f- 1=\mathrm{Iy} \)
Thus, \( \mathrm{f}^{-1}: \mathrm{Y} \rightarrow X \) is invertible and \(f\) is the inverse of \( \mathrm{f}^{-1} \).
Therefore, \( \left(\mathrm{f}^{-1}\right)^{-1}=\mathrm{f} \).
A. \( x^{\frac{1}{3}} \) B. \( x^{3} \) C. \(x\) D. \( \left(3-x^{3}\right) \)
Then, \( f\circ f (x)=f(f(x))=f\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)=\left[3-\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{3}} \)
\( =3\left[3-\left(3-x^{3}\right)\right]^{\frac{1}{3}}=\left(x^{3}\right)^{\frac{1}{3}}=x \)
\( \Rightarrow f\circ f(x)=x \)
A. \( \mathrm{g}(\mathrm{y})=\frac{3 y}{3-4 y} \) B. \( \mathrm{g}(\mathrm{y})=\frac{4 y}{4-3 y} \) C. \( \mathrm{g}(\mathrm{y})=\frac{3 y}{4-3 y} \) D. \( \mathrm{g}(\mathrm{y})=\frac{4 y}{3-4 y} \)
Let \( y \) be any element of Range \(f \).
Then, there exists \( x \in R-\left\{-\frac{4}{3}\right\} \) such that \( \mathrm{y}=f(x) \)
\(=\mathrm{y}=\frac{4 x}{3 x+4}\)
\(\Rightarrow 3 x\mathrm{y}+4 \mathrm{y}=4 x\)
\(\Rightarrow x(4-3 \mathrm{y})=4 \mathrm{y}\)
\(\Rightarrow x=\frac{4 y}{4-3 y}
\)
Let us define \(g :\) Range \( \mathrm{f} \rightarrow \mathrm{R}-\left\{-\frac{4}{3}\right\} \) as \( \mathrm{g}(\mathrm{y})=\frac{4 y}{4-3 y} \)
\( \operatorname{Now},(\operatorname{g \circ f})(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(\frac{4 y}{4-3 y}\right)=\frac{4\left(\frac{4 x}{3 x+4}\right)}{4-3\left(\frac{4 y}{4-3 y}\right)}=\frac{16 x}{12 x+16-12 x}=\frac{16 x}{16} \)
And, \( ( f \circ g )(\mathrm{y})=\mathrm{f}(\mathrm{g}(\mathrm{y}))=\mathrm{f}\left(\frac{4 y}{4-3 y}\right)=\frac{4\left(\frac{4 y}{4-3 y}\right)}{3\left(\frac{4 y}{4-3 y}\right)+4}=\frac{16 y}{12 y+16-12 y}=\frac{16 y}{16}=\mathrm{y} \)
Therefore, \( g \circ f ={ }^{\mathrm{I}} \mathrm{R}-\left\{-\frac{4}{3}\right\} \) and \(f \circ g =\mathrm{I}_{\text {Range } \mathrm{f}} \)
Thus, g is the inverse of f
Therefore, the inverse of f is the map
\( : \) Range \( \mathrm{f} \rightarrow \mathrm{R}-\left\{-\frac{4}{3}\right\} \), which is given by \( \mathrm{g}(\mathrm{y})=\frac{4 y}{4-3 y} \)