Class 12: Relations and Functions – Miscellaneous Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1:} \text{Find whether the following functions are one-one or not: }$
$\text{(i) } f : R \rightarrow R \text{ given by } f(x)= x^3 + 2 \text{ for all } x \in R$
$\text{(ii) } f : Z \rightarrow Z \text{ given by } f(x)= x^2 + 1 \text{ for all } x \in Z$
$\displaystyle \text{Answer:}$
$\text{(i) Let } x, y \text{ be two arbitrary elements of } R (\text{domain of } f) \\ \text{ such that } f (x) = f (y). \text{ Then, }$
$f(x) = f( y) \Rightarrow x^3+ 2 = y^3 + 2 \Rightarrow x^3 = y^3 \Rightarrow x = y$
Hence, $f$ is a one-one function from $R$ to itself.
$\text{(ii) Let } x, y \text{ be two arbitrary elements of } Z \text{ such that } f (x) = f(y). \text{ Then, }$
$f(x) = f( y) \Rightarrow x^2 + 1 = y^2 + 1 \Rightarrow x^2 = y^2 \Rightarrow x = \pm y$
Here, $f(x) = f(y)$ does not provide the unique solution $x =y$ but it provides $x =\pm y .$ Hence, $f$ is not a one-one function.
$\text{Infact, } f (2) = 2^2 + 1= 5 \text{ and } f(- 2) = (- 2)^2 + 1=5. \text{ Hence, } 2 \text{ and, } 1 \\ \text{ are two distinct elements having the same image. }$
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Question 2: Show that the function $f : Z \rightarrow Z$ defined by $f(x)= x^2 + x$ for all $x \in Z$ is a many-one function.
$\displaystyle \text{Answer:}$
$\text{Let } x , y \in Z. \text{Then, } f(x)=f(y)$
$\Rightarrow x^2 + x = y^2 + y$
$\Rightarrow (x^2-y^2) + ( x-y) = 0$
$\Rightarrow (x-y)(x+y+1) = 0$
$\Rightarrow x = y \text{ or } y = - x - 1$
$\text{Since, } f(x) = f(y) \text{ does not provide the unique solution } x=y \\ \\ \text{but it also provides } y= - x -1. \\ \\ \text{This means that } x \neq y \text{ but } f(x)=f(y) \text{ when } y =-x-1.$
$\text{For example, if we put } x =1 \text{ in } y = - x - 1 \text{ we obtain } y = - 2.$
$\text{This shows that } 1 \text{ and } -2 \text{ have the same image under } f. \\ \\ \text{Hence, } f \text{ is a many-one function. }$
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Question 3: Discuss the surjectivity of the following functions:
$\text{(i) } f : R \rightarrow R \text{ given by } f(x) = x^3 + 2 \text{ for all } x \in R.$
$\text{(ii) } f :R \rightarrow R \text{ given by } f(x) : x^2 + 2 \text{ for all } x \in R.$
$\text{(iii) } f : Z \rightarrow Z \text{ given by } f(x) = 3x + 2 \text{ for all } x \in Z.$
$\displaystyle \text{Answer:}$
(i) Let $y$ be an arbitrary element of $R$ . Then,
$\displaystyle f(x) = y \Rightarrow x^3 + 2 = y \Rightarrow x = ( y-2)^{\frac{1}{3}}$
$\displaystyle \text{Clearly, for all } y \in R, ( y-2)^{\frac{1}{3}} \text{ for all } x \in R.$
Hence, negative real numbers in R(co-domain) do not have their pre-images in R(domain). Hence, f is not an onto function.
$\text{(ii) Clearly, } f(x) = x^2 + 2 \geq 2 \text{ for all } x \in R.$
Hence, negative real numbers in $R$ (co-domain) do not have their pre-images in $R$ (domain). Hence, $f$ is not an onto function.
(iii) Let y be an arbitrary element of $Z$ (co-domain). Then,
$\displaystyle f(x) = y \Rightarrow 3x + 2 = y \Rightarrow x = \frac{y-2}{3}$
$\displaystyle \text{Clearly, if } y = 0 , \text{ then } x = \frac{-2}{3} \notin Z.$
Thus, $y =0 \in Z$ does not have its pre-image in Z(domain). Hence, $f$ is not an onto function.
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Question 4: Show that the function $f:N \rightarrow N \text{ given by } f(1)=f(2)=1$ and $f (x)=x-1$ for every $x \geq 2$ , is onto but not one-one.
$\displaystyle \text{Answer:}$
It is given that
$\displaystyle f(x) = \Bigg\{ \begin{array}{ll} 1, & x = 1, 2 \\ \\ x-1, & x \geq 2\end{array}$
$\text{Clearly, } f (1) = f (2)=1 \text{ i.e. } 1 \text{ and } 2 \text{ have the same image. }$
$\text{Hence, } f : N \rightarrow N \text{is a many-one function }$
Let $y$ be an arbitrary element in $N$ (co-domain). Then,
$f(x) = y \Rightarrow x - 1 = y \Rightarrow x = y +$
$\text{Clearly, } y +1 \in N \text{(domain) } \text{ for all } y \in N \text{(Co-domain). }$
$\text{Thus, for each } y \in N \text{(co-domain) there exists } y + 1 \in N \\ \text{(domain) } \text{ such that } f (y +1) =y + 1 - 1 =y.$
$\text{Hence, } f : N \rightarrow N \text{ is an onto function. }$
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Question 5: Show that the Signum function $f : R \rightarrow R$ , given by
$\displaystyle f(x) = \Bigg\{ \begin{array}{ll} 1, & \text{ if } x > 0 \\ 0, & \text{ if } x = 0 \\ -1, & \text{ if } x < 0 \end{array}$
is neither one-one nor onto.
$\displaystyle \text{Answer:}$
Clearly, all positive real numbers have the same image equal to 1.
Hence, $f$ is a many-one function.
We observe that the range of $f \text{ is } \{-1,0,1 \}$ which is not equal to the co-domain of $f$ .
Hence, $f$ is not onto.
Hence, $f$ is neither one-one nor onto
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$\text{Question 6: Prove that the function } f : Q \rightarrow Q \text{ given by } f (x) = 2x - 3 \text{ for all } x \in Q \text{ is a bijection. }$
$\displaystyle \text{Answer:}$
We observe the following properties of $f$ .
Injectivity: Let $x, y$ be two arbitrary elements in $Q$ . Then,
$f(x) = f(y)\Rightarrow 2x-3 = 2y - 3 \Rightarrow 2x = 2y \Rightarrow x = y$
$\text{Thus, } f(x) = f(y)\Rightarrow x = y \text{ for all } x,y \in Q.$
Hence, $f$ is an injective map.
Surjectivity: Let $y$ be an arbitrary element of $Q$ . Then,
$\displaystyle f(x) = y \Rightarrow 2x-3 = y \Rightarrow x = \frac{y+3}{2}$
$\displaystyle \text{Clearly, for all } y \in Q, x = \frac{y+3}{2} \in Q.$
Thus, for all $y \in Q$ (co-domain) there exists $x \in Q$ (domain).
$\displaystyle \text{Given by } x = \frac{y+3}{2} \text{ such that } f(x) = f (\frac{y+3}{2}) = 2(\frac{y+3}{2}) - 3 = y$
That is every element in the co-domain has its pre-image in $x$ .
Hence, $f$ is a surjection.
$\text{Hence } f : Q \rightarrow Q \text{ is a bijection. }$
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$\text{Question 7: Show that the function } f : R \rightarrow R \\ \\ \text{ defined by } f (x) : 3x^3 + 5 \text{ for all } x \in R \text{ is a bijection. }$
$\displaystyle \text{Answer:}$
We observe the following properties of $f.$
Injectivity: Let $x, y$ be any two elements of $R$ (domain). Then,
$f(x) = f(y)= 3x^3 +5 = 3y^3 +5 \Rightarrow x^3 = y^3 \Rightarrow x = x$
$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all } x,y \in R.$
Hence, $f$ is an injective map.
Surjectivity: Let $y$ be an arbitrary element of $R$ (co-domain). Then,
$\displaystyle f(x) = y \Rightarrow 3x^3 + 5 = y \Rightarrow x^3 = \frac{y-5}{3} \Rightarrow x = \Big( \frac{y-5}{3} \Big)^{1/3}$
$\displaystyle \text{Thus, we find that for all } y \in R \text{(co-domain) there exists } \\ \\ x = \Big( \frac{y-5}{3} \Big)^{1/3} \in R \text{(domain) such that }$
$\displaystyle f(x) = f \Bigg( \Big( \frac{y-5}{3} \Big)^{1/3} \Bigg) = 3\Bigg( \Big( \frac{y-5}{3} \Big)^{1/3} \Bigg)^3 + 5 = y - 5+5 = y$
This shows that every element in the co-domain has its pre-image in the domain. Hence, $f$ is a surjection.
Hence, $f$ is a bijection.
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Question 8: Let $A =\{ x \in R : -1 \leq x \leq 1 \} = B$ . Show that $f : A \rightarrow B$ given by $f(x) = x |x|$ is a bijection.
$\displaystyle \text{Answer:}$
We observe the following properties of $f$ .
Injectivity: Let $x, y$ be any two elements in $A$ . Then,
$x \neq y \Rightarrow x |x| \neq y|y| \Rightarrow f(x) \neq f(y)$
Hence, $f : A \rightarrow B$ is an injective map.
Surjectivity: We have
$\displaystyle f(x) = x |x| = \Bigg\{ \begin{array}{rr} x^2, & \text{ if } x \geq 0 \\ \\ -x^2, & \text{ if } x < 0 \end{array}$
If $0 \leq x \leq 1, \text{ then } f (x)=x^2$ takes all values between 0 and 1 including these two points.
Also, if $-1 \leq x < 0, \text{ then } f(x)=-x^2$ takes all values between -1 and 0 including -1. Therefore, $f(x)$ takes every value between – 1 and 1 including – 1 and 1. Hence, range of $f$ is same as its co-domain.
$\text{Hence, } f: A \rightarrow B \text{ is an onto function. }$
$\text{Thus, } f : A \rightarrow B \text{ is both one-one and onto. }$
Hence, it is a bijection.
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Question 9: Let A be the set of all 50 students of class XII in a central school. Let $f : A\rightarrow N$ be a function defined by $f (x) =$ Roll number of student $x$ . Show that $f$ is one-one but not onto.
$\displaystyle \text{Answer:}$
Here, $f$ associates each students to his (her) roll number. Since no two different students of the class can have the same roll number.
Therefore, $f$ is one-one.
We observe that $f (A)= \text{ Range of } f =\{1,2,3,...,50)\} \neq N$ i.e. range of $f$ is not same as its co-domain. Hence, $f$ is not onto.
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Question 10: Prove that $f : R \rightarrow R, \text{ given by } f (x) = 2x,$ is one-one and onto
$\displaystyle \text{Answer:}$
We observe the following properties of $f$ .
$\text{Injectivity: Let } x_1, x_2 \in R \text{ such that } f(x_1)=f(x_2). \text{ Then, }$
$f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_ 2$
Hence, $f: R \rightarrow R$ is one-one.
Surjectivity: Let $y$ be any real number in $R$ (co-domain). Then,
$\displaystyle f(x) = y \Rightarrow 2x = y \Rightarrow x = \frac{y}{2}$
$\displaystyle \text{Clearly, } \frac{y}{2} \in R \text{ for any } y \in R \text{ such that } f \Big(\frac{y}{2}\Big) = 2\Big(\frac{y}{2}\Big) = y.$
Thus, for each $y \in R$ (co-domain) has its pre-image in domain.
This means that each element in co-domain has its pre-image in domain.
Hence, $f: R \rightarrow R$ is onto.
Hence, $f : R \rightarrow R$ is a bijection.
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Question 11: Show that the function $f :R \rightarrow R, \text{ defined as } f (x)=x^2$ , is neither one-one nor onto.
$\displaystyle \text{Answer:}$
We observe that $1 \text{ and } -1 \in R \text{ such that } f (- 1) = f (1)$ i.e. there are two distinct elements in $R$ which have the same image. Hence, f is not one-one.
Since $f (x )$ assumes only non-negative values. Hence, no negative real number in $R$ (co-domain) has its pre-image in domain of $f \text{ i.e. } R$ . Consequently $f$ is not onto.
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Question 12: Show that $f : R \rightarrow R, \text{ defined as } f (x)=- x^3$ , is a bijection.
$\displaystyle \text{Answer:}$
We observe the following properties of $f$ .
Injectivity: Let $x,y \in R$ such that $f(x) = f (y).$ Then,
$f(x)=f(y) \Rightarrow x^3 =y^3 \Rightarrow x=y$
Hence, $f: R \rightarrow R$ is one-one.
Surjectivity: Let $y \in R$ (co-domain). Then,
$f(x)=y \Rightarrow x^3=y \Rightarrow x - y^{1/3}$
Clearly, $y^{1/3} \in R$ (domain) for all $y \in R$ (co-domain).
Thus, for each $y \in R$ (co-domain) there exists $x=y^{1/3} \in R$ (domain) such that $f(x)=x^3 =y.$
Hence, $f: R \rightarrow R$ is onto.
Hence, $f :R \rightarrow R$ is a bijection.
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Question 13: Show that the function $f : R_0 \rightarrow R_0, \text{ defined as } f (x) = \frac{1}{x},$ is one-one onto, where $R_0$ is the set of all non-zero real numbers. ls the result true, if the domain $R_0$ is replaced by $N$ with co-domain being same as $R_0$ ?
$\displaystyle \text{Answer:}$
We observe the following properties of $f$ .
Injectivity: Let $x, y \in R_0$ such that $f (x) = f (y ). Then,$ latex \displaystyle f(x) = f(y) \Rightarrow \frac{1}{x} = \frac{1}{y} \Rightarrow x = y $
Hence, $f: R_0 \rightarrow R_0$ is one-one
Surjectivity: Let $y$ be an arbitrary element of $R_0$ (co-domain) such that $f(x) = y$ . Then,
$\displaystyle f(x) = y \Rightarrow \frac{1}{x} = y \Rightarrow x = \frac{1}{y}$
$\displaystyle \text{Clearly, } x = \frac{1}{y} \in R_0 \text{(domain) for all } y \in R_0 \text{(co-domain). }$
$\displaystyle \text{Thus, for each } y \in R_0 \text{(co-domain) there exists } x = \frac{1}{y} \in R_0 \\ \text{ (domain) such that } f(x) \frac{1}{x} = y.$
Hence, $f: R_0 \rightarrow R_0$ is onto.
Hence, $f:R_0 \rightarrow R_0$ is one-one onto.
$\text{Let us now consider } f:N \rightarrow R_0 \text{ given by } f(x) = \frac{1}{x}$
For any $x, y \in N$ , we find that
$\displaystyle f(x) = f(y) \Rightarrow \frac{1}{x} = \frac{1}{y} \Rightarrow x = y$
Hence, $f: N \rightarrow R_0$ is one-one.
$\displaystyle \text{We find that } \frac{2}{3}, \frac{3}{5} \text{ etc. in co-domain } R_0 \text{ do not have their pre-image in domain } N.$
$\text{Hence, } f:N \rightarrow R_0 \text{ is not onto. }$
$\text{Thus, } f:N \rightarrow R_0 \text{ is one-one but not onto. }$
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Question 14: Prove that the greatest integer function $f : R \rightarrow R$ , given by $f (x) =[x]$ , is neither one-one nor, where $[x]$ denotes the greatest integer less than or equal to $x$ .
$\displaystyle \text{Answer:}$
$\text{We observe that } f (x) =0 \text{ for all } x \in [0, 1) \text{ Hence, } f :R \rightarrow R \text{ is not one-one }.$
Also, $f: R \rightarrow R$ does not attain non-integral values. Therefore, non-integer points in $R$ (co-domain) do not have their pre-images in the domain. Hence, $f : R \rightarrow R$ is not onto.
Hence, $f: R \rightarrow R$ is neither one-one nor onto.
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Question 15: show that the modulus function $f : R \rightarrow R, \text{ given by } f (x) =|x|$ is neither one-one nor onto.
$\displaystyle \text{Answer:}$
We observe that $f (- 2) = f (2)$ . Hence, f is not one-one.
Also, $f(x) = |x|$ assumes only non-negative values. Hence, negative real numbers in $R$ (co-domain) do not have their pre-images in $R$ (domain).
Hence, $f$ is neither one-one nor onto
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Question 16: Let $C$ and $R$ denote the set of all complex numbers and all real numbers respectively. Then show that $f : C \rightarrow R$ given by $f(z) = |z|$ for all $z \in C$ is neither one-one nor onto.
$\displaystyle \text{Answer:}$
Injectivity: We find that $z_1=1 -i$ and $z_2 = 1 + i$ are two distinct complex numbers in $C$ such that $|z_1| = |z_2| \text{ i.e. } z_1 \neq z_2 \text{ but } f (z_1) = f (z_2).$
This shows that different elements in $C$ may have the same image. Hence, $f$ is not an injection.
Surjectivity: $f$ is not a surjection, because negative real numbers in $R$ do not have their pre-images in $C$ . In other words, for every negative real number a there is no complex number $z \in C$ such that $f (z) =|z| = a.$
Hence, $f$ is not a surjection.
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Question 17: Show that the function $f:R \rightarrow R \text{ given by } f(x)=ax+b, \text{ where } a,b \in R, a \neq 0$ is a bijection. [CBSE 2010]
$\displaystyle \text{Answer:}$
Injectivity: Let $x, y$ be any two real numbers. Then,
$f(x) = f(y) \Rightarrow ax+b = ay+b \Rightarrow ax = ay \Rightarrow x = y$
$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all. } x,y \in R \text{(domain). }$
Hence, $f$ is an injection.
Surjectivity: Let $y$ be an arbitrary element of $R$ (co-domain). Then,
$\displaystyle f(x) = y \Rightarrow ax+b = y \Rightarrow x = \frac{y-b}{a}$
$\displaystyle \text{Clearly, } x = \frac{y-b}{a} \in R \text{(domain) for all } y \in R \text{(co-domain). }$
$\displaystyle \text{Thus, for all } y \in R \text{(co-domain) there exists } x = \frac{y-b}{a} \in R \text{(domain) such that}.$
$\displaystyle f(x) = f (\frac{y-b}{a}) = a(\frac{y-b}{a}) + b = y$
This shows that every element in co-domain has its pre-image in domain. Hence, f is a surjection.
Hence, $f$ is a bijection.
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Question 18: Show that the function $f : R \rightarrow R \text{ given by } f(x) = \cos x \text{ for all } x \in R$ , is neither one-one nor onto.
$\displaystyle \text{Answer:}$
Injectivity: We know that $f(0) = \cos 0 = 1, \text{ and } f (2\pi) = \cos 2\pi =1.$
hence, different elements in $R$ may have the same image. Hence, $f$ is not an injection.
Surjectivity: Since the values of $cos x$ lie between $-1$ and $1$ , it follows that the range of $f(x)$ is not equal to its co-domain. Hence, f is not a surjection.
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$\displaystyle \text{Question 19: Let } A = R- \{ 2 \} \text{ and } B = R - \{ 1 \}. \text{ If } f :A \rightarrow B \text{ is a mapping} \\ \\ \text{defined by } f(x) = \frac{x-1}{x-2} , \text{ show that f is bijective. }$
$\displaystyle \text{Answer:}$
Injectivity: Let $x, y$ be any two elements of $A$ . Then,
$f(x) = f(y)$
$\displaystyle \Rightarrow \frac{x-1}{x-2} = \frac{y-1}{y-2} \\ \\ \Rightarrow (x-1)(y-2) = (x-2)(y-1) \\ \\ \Rightarrow xy - y - 2x + 2 = xy - x - 2y + 2 \\ \\ \Rightarrow x = y$
$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all } x , y \in A.$
Hence, $f$ is an injective.
Surjectivity: Let $y$ be an arbitrary element of $B$ . Then,
$\displaystyle f(x) = y \Rightarrow \frac{x-1}{x-2} = y \Rightarrow (x-1) = y(x-2) \Rightarrow x = \frac{1-2y}{1-y}$
$\displaystyle \text{Clearly, } x = \frac{1-2y}{1-y} \text{ is a real number for all } y \neq 1.$
$\displaystyle \text{Also, } \frac{1-2y}{1-y} \neq 2 \text{ for any } y , \text{ for, it we take } \frac{1-2y}{1-y} = 2,\text{ then we get } \\ \\ 1 = 2 \text{ which is not possible. }$
$\displaystyle \text{Thus, every element } y \text{ in } B \text{ has its pre-image } x \text{ in } A \text{ given by } x = \frac{1-2y}{1-y} .$
Hence, $f$ is a surjective.
Hence, $f$ is a bijective.
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Question 20: Let $A$ and $B$ be two sets. Show that $f :A \times B \rightarrow B \times A$ defined by $f (a,b)=f(b,a)$ is a bijection.
$\displaystyle \text{Answer:}$
Injectivity: Let $(a_1, b_1)$ and $(a_2, b_2) \in A \times B$ such that
$f (a_1, b_1) = f (a_2,b_2)$
$\Rightarrow (b_1, a_1) = (b_2, a_2)$
$\Rightarrow b_1 = b_2 \text{ and } a_1 = a_2$
$\Rightarrow (a_1, b_1) = (a_2, b_2)$
Thus, $f(a_1, b_1) = f( a_2, b_2) \Rightarrow (a_1, b_1) = (a_2, b_2) \text{ for all } ( a_1, b_1), (a_2, b_2) \in A \times B.$
Hence, $f$ is an injective map.
Surjectivity: Let $(b, a)$ be an arbitrary element of $B \times A$ . Then,
$b \in B \text{ and } a \in A\Rightarrow (a,b) \in A \times B.$
Thus, for all, $(b, a) \in B \times A$ there exists $(a,b) \in A \times B$ such that $f (a, b)=(b, a).$
Hence, $f : A \times B \rightarrow B \times A$ is an onto function.
Hence, $f$ is a bijection.
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Question 21: Let $A$ be any non-empty set. The, prove that the identity function on set $A$ is a bijection.
$\displaystyle \text{Answer:}$
The identity function $I_A \rightarrow A$ is defined as
$I_A(x) = x \text{ for all } x \in A.$
Injectivity: Let $x, y$ be any two elements of $A$ . Then,
$I_A(x) =I_A(y) \Rightarrow x=y$
Hence, $I_A$ is an injective map.
Surjectivity: Let $y \in A$ . Then, there exists $x =y \in A$ such that
$I_A(x) = x = y$
Hence, $I_A$ is a surjective map.
Hence, $I_A: A \rightarrow A$ is a bijection.
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Question 22: Let $A = \{1, 2 \}$ . Find all one-to-one functions from $A$ to $A$ .
$\displaystyle \text{Answer:}$
Let $f : A \rightarrow A$ be a one-one function.
Then, $f (1 )$ has two choices, namely, 1 or 2.
$\text{Hence, } f(1) :1 \text{ or } f (1) = 2.$
Case 1: When $f(1) = 1:$
As $f : A \rightarrow A$ is one-one. Therefore, $f (2) =2.$
Thus, we have $f(1) =1 \text{ and } f (2) =2.$
Case 2: When $f (1) =2:$
Since $f : A \rightarrow A$ is one-one. Therefore, $f (2) =1.$
Thus, in this case, we have $f (1)=2 \text{ and } f (2)=1$
Hence, there are two one-one functions say $f \text{ and } g$ from $A \text{ to } A$ given by
$f (1)=1, f (2) =2 \text{ and, } g (1) = 2, g (2) =1.$
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Question 23: Consider the identity function $I_N: N \rightarrow N$ defined as, $I_N(x) =x$ for all $x \in N$ . Show that although $I_N$ is onto but $I_N+I_N : N \rightarrow N$ defined as $(I_N +I_N)(x) =I_N(x) + I_N(x) = x + x = 2x$
$\displaystyle \text{Answer:}$
We know that the identity function on a given set is always a bijection. Therefore, $I_N : N \rightarrow N$ is onto. We have,
$(I_N +I_N)(x) = 2x \text{ for all } x \in N$
This means that under $I_N+I_N,$ images of natural numbers are even natural numbers.
Hence, odd natural numbers in $N$ (co-domain) do not have their pre-images in domain $N$ .
For example,1,3,5 etc. do not have their pre-images. Hence, $I_N + I_N : N \rightarrow N$ is not onto.
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$\displaystyle \text{Question 24: Consider the function } f : \Big[ 0, \frac{\pi}{2} \Big] \rightarrow R \text{ given by } f(x) = \sin x \\ \text{ and } g : \Big[0, \frac{\pi}{2} \Big] \rightarrow R \text{ given by } g(x) = \cos x . \text{ Show that } f \text{ and } g \\ \text{ are one-one, but } f + g \text{ is not one-one. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We observe that for any two distinct elements } x_1 \text{ and } x_2 \text{ in } \Big[ 0, \frac{\pi}{2} \Big]$
$\sin x_1 \neq \sin x_2 \text{ and } \cos x_1 \neq \cos x_2$
$\Rightarrow f(x_1) \neq f(x_2) \text{ and } g(x_1) \neq g(x_2)$
$\Rightarrow f \text{ and } g \text{ are one-one. }$
$\text{We have } (f+g) (x) = f(x) + g( x) = \sin x + \cos x$
$\displaystyle \Rightarrow (f+g) (0) = \sin 0 + \cos 0 = 1 \text{ and } (f+g) \Big(\frac{\pi}{2}\Big) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1$
$\displaystyle \text{Thus, } 0 \neq \frac{\pi}{2} \text{ but } (f+g) (0) = (f+g) \Big(\frac{\pi}{2}\Big).$
Hence, $f+g$ is not one-one.
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Question 25: Let $f:X \rightarrow Y$ be a function. Define a relation $R$ on $X$ given by $R= \{ (a, b) :f (a) = f(b) \}$ .
$\displaystyle \text{Answer:}$
Show that $R$ is an equivalence relation on $X.$
We observe the following properties of relation R.
Reflexivity: For any $a \in X$ , we have
$f(a)=f(a) \Rightarrow (a,a) \in R \Rightarrow R \text{ is reflexive.}$
Symmetry: Let $a, b \in X$ be such that $(a, b) \in R.$ Then,
$(a, b) \in R \Rightarrow f (a)=f (b) \Rightarrow f (b)=f (a) \Rightarrow (b,a) \in R$
Hence, $R$ is symmetric.
Transitivity: Let $a, b, c \in X$ be such that $(a, b) \in R \text{ and } (b, c) \in R.$ Then,
$(a, b) \in R \text{ and } (b, c) \in R$
$\Rightarrow f (a) = f (b) \text{ and } f (b)=f (c)$
$\Rightarrow f (a) = f (c)$
$\Rightarrow (a, c) \in R$
Hence, $R$ is transitive.
Hence, $R$ is an equivalence relation.
$\\$
$\displaystyle \text{Question 26: Show that the function } f:R \rightarrow \{ x \in R : -1 < x < 1 \} \\ \\ \text{ defined by } f(x) = \frac{x}{1+|x|}, x \in R \text{ is one-one onto function. }$
$\displaystyle \text{Answer:}$
We have,
$\displaystyle f(x) = \frac{x}{1+|x|} = \Bigg\{ \begin{array}{rr} \frac{x}{1+x}, & \text{ if } x \geq 0 \\ \\ \frac{x}{1-x}, & \text{ if } x < 0 \end{array}$
$\displaystyle \text{Case 1: When } x \geq 0$
$\displaystyle \text{In this case we have } f(x) = \frac{x}{1+x}$
$\displaystyle \text{Injectivity: Let } x, y \in R \text{ such that } x \geq 0, y \geq 0. \text{ Then, }$
$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y + xy \Rightarrow x = y$
Hence, $f$ is an injective map.
Surjectivity: When $x \geq 0$ , we have
$\displaystyle \text{In this case we have } f(x) = \frac{x}{1+x} \geq 0 \text{ and } f(x) < 1$
Let $y \in [0, 1)$ be any real number. Then,
$\displaystyle f(x) = y \Rightarrow \frac{x}{1+x} = y \Rightarrow x = \frac{y}{1-y}$
$\displaystyle \text{Clearly, } x \geq 0 \text{ for all } y \in [0,1). \text{ Thus, for each } y \in [0, 1) \text{ there exists } \\ \\ x = \frac{y}{1-y} \geq \text{ such that } f(x) = y.$
Hence, $f$ is an onto function from $[0, 1) \text{ to } [0, 1)$
$\displaystyle \text{Case 2: When } x < 0$
$\displaystyle \text{In this case we have } f(x) = \frac{x}{1-x}$
$\displaystyle \text{Injectivity: Let } x, y \in R \text{ such that } x < 0, y < 0. \text{ Then, }$
$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{1-x} = \frac{y}{1-y} \Rightarrow x - xy = y - xy \Rightarrow x = y$
Hence, $f$ is an injective map.
Surjectivity: When $x < 0$ , we have
$\displaystyle \text{In this case we have } f(x) = \frac{x}{1-x} < 0$
$\displaystyle \text{Also, } f(x) = \frac{x}{1-x} = -1 + \frac{1}{1-x} > - 1$
$\displaystyle \therefore -1 < f(x) < 0$
Let $y \in (- 1, 0)$ be an arbitrary real number such that $f (x) = y$ . Then,
$\displaystyle f(x) = y \Rightarrow \frac{x}{1-x} = y \Rightarrow x = \frac{y}{1+y}$
$\displaystyle \text{Clearly, } x < 0 \text{ for all } y \in (-1,0). \text{ Thus, for each } y \in (-1,0) \text{ there exists } \\ \\ x = \frac{y}{1+y} < 0 \text{ such that } f(x) = y.$
Hence, $f$ is an onto function from $(-1, 1) \text{ to } (-1, 1)$
Hence, $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ is a one-one onto function.
$\\$
Question 27: Show that the function $f :R \rightarrow R \text{ given by } f (x): x^3 + x$ is a bijection.
$\displaystyle \text{Answer:}$
Injectivity: Let $x, y \in R$ such that
$f(x) = f(y)$
$\Rightarrow x^3+x = y^3 + y$
$\Rightarrow x^3 - y^3 +(x-y) = 0$
$\Rightarrow (x-y)(x^2 + xy + y^2 + 1) = 0$
$\because x^2 + xy + y^2 \geq 0 \text{ for all } x, y \in R \\ \\ \therefore x^2 + xy + y^2+ 1 \geq 1 \text{ for all } x, y \in R$
$\Rightarrow x-y = 0$
$\Rightarrow x = y$
$\text{Thus, } f(x) =f(y) \Rightarrow x=y \text{ for all } x, y \in R$
Hence, $f$ is an injective map.
Surjective: Let $y$ be an arbitrary element of $R$ . Then,
$f(x) = y \Rightarrow x^3 +x =y \Rightarrow x^3+ x-y = 0$
We know w that an odd degree equation has at least one real root. Therefore, for every real value of $y$ the equation $x^3 + x - y = 0$ has a real root $\alpha$ such that
$\alpha^3 + \alpha - y = 0 \Rightarrow \alpha^3 + \alpha = y \Rightarrow f(\alpha) = y$
Thus, for every $y \in R$ there exists $\alpha \in R$ such that $f(\alpha) =y$ . Hence, $f$ is a surjective map.
Hence, $f:R \rightarrow R$ is a bijection.
$\\$
Question 28: Show that $f:N \rightarrow N$ defined by
$\displaystyle f(n) = \Bigg\{ \begin{array}{rr} \frac{n+1}{2}, & \text{ if } n \text{ is odd} \\ \\ \frac{n}{2}, & \text{ if } n \text{ is even} \end{array}$
is many-one onto function. [CBSE 2009]
$\displaystyle \text{Answer:}$
We observe that
$\displaystyle f(1) = \frac{1+1}{2} = 1 \text{ and } f(2) = \frac{1}{2} = 1$
Thus $1, 2 \in N \text{ such that } 1 \neq 2 \text{ but } f (1)=f (2)$ . Hence, f is a many-one function.
Surjectivity: Let $n$ be an arbitrary element of $N$ .
If n is an odd natural number, then $2n-1$ is also an odd natural number such that
$\displaystyle f(2n-1) = \frac{2n-1+1}{2} = n$
If n is an even natural number, then $2n$ is also an even natural number such that
$\displaystyle f(2n) = \frac{2n}{2} = n$
Thus, for every $n \in N$ (whether even or odd) there exists its pre-image in $N$ . Hence, f is a surjection.
Hence, $f$ is a many-one onto function.
$\\$
$\text{Question 29: Show that the function } f : N \rightarrow N \text{ given by, } \\ \\ f (n) = n - (- 1)^n \text{ for all } n \in N \text{ is a bijection. }$
$\displaystyle \text{Answer:}$
$\text{We have, } f (n) = n-(-1)^n \text{ for all } n \in N$
$\displaystyle f(x) = \Bigg\{ \begin{array}{rr} n-1, & \text{ if } n \text{ is even } \\ \\ n+1, & \text{ if } n \text{ is odd } \end{array}$
lnjectivity: Let $n, m$ be any two even natural numbers. Then,
$f (n) = f (m) \Rightarrow n-1=m-1 \Rightarrow n = m$
If $n, m$ are any two odd natural numbers. Then,
$f (n) = f (m) \Rightarrow n+1=m+1 \Rightarrow n = m.$
Thus in both the cases, $f(n) = f (m) \Rightarrow n = m.$
$\text{If } n \text{ is even and } m \text{ is odd, then } n \neq m . \text{ Also } f (n) \text{ is odd and } f (m) \text{ is even.} \\ \\ \text{Hence, } f (n) \neq f (m).$
$\text{Thus, } n \neq m \Rightarrow f (n) \neq f (m).$
Hence, $f$ is an injective map.
Surjectivity: Let $N$ he an arbitrary natural number.
If $n$ is an odd natural number, then there exists an even natural number $n + 1$ such that
$f(n+1) = n+1-1 = n$
If $n$ is an even natural number, then there exists an odd natural number $(n - 1)$ such that
$f (n-1) = n-1+1 = n$
Thus, every $m \in N$ has its pre-image in $N.$ Hence, $f: N \rightarrow N$ is a surjection.
Hence, $f: N \rightarrow N$ is a bijection.
$\\$
$\text{Question 30: Let } f:N \cup \{ 0 \} \rightarrow N \cup \{ 0 \} \text{ be defined by }$
$\displaystyle f(x) = \Bigg\{ \begin{array}{rr} n+1, & \text{ if } n \text{ is even } \\ \\ n-1, & \text{ if } n \text{ is odd } \end{array}$
Show that $f$ is a bijection.
$\displaystyle \text{Answer:}$
$f$ is an injection : Let $n, m \in N \cup \{ 0 \}$
If $n$ and $m$ are even, then
$f (n) = f (m) \Rightarrow n+1=m+1\Rightarrow n = m$
If $n$ and $m$ are odd, then
$f (n) = f (m) \Rightarrow n-1=m-1 \Rightarrow n = m$
Thus, in both case, we have
$f(n)=f(m) \Rightarrow n=m.$
If $n$ is odd and $m$ is even then $f (n) = n-1$ is even and $f (m) = m + 1$ is odd. Therefore,
$n \neq m \Rightarrow f(n) \neq f(m).$
Similarly, if $n$ is even and $m$ is odd, then
$n \neq m \Rightarrow f(n) \neq f(m).$
Hence, $f$ is an injection.
$f$ is a surjection: Let $n$ be an arbitrary element of $N \cup \{ 0 \}$
If $n$ is an odd natural number, there exist an even natural number $n - 1 \in N \cup \{ 0 \}$ (domain) such that $f (n-1) =n-1+1 =n.$
If $n$ is an even natural number, then there exists an odd natural number $n + 1 \in N \cup \{ 0 \}$ (domain) such that $f(n + 1) = n + 1 -1 =n.$
Also, $f (1) = 0.$
Thus, every element of $N \cup \{ 0 \}$ (co-domain) has its pre-image in $N \cup \{ 0 \}$ (domain). Hence, $f$ is an onto function.
Question 31: Let $A$ be a finite set. If $f : A \rightarrow A$ is a one-one function, show that $f$ is onto also.
$\displaystyle \text{Answer:}$
$\text{Let } A = \{ a_1, a_2, a_3, \ldots a_n \}.$
In order to prove that f is onto function, we will have to show that every element in A (co-domain) has its pre-image in the domain $A$ . In other words, range of $f = A.$
Since $f: A \rightarrow A$ is a one-one function. Therefore, $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of set $A$ .
But, $A$ has only $n$ elements. Therefore, $A = \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \}$ i.e. Co-domain=Range.
Hence, $f : A \rightarrow A$ is onto.
$\\$
Question 32: Let $A$ be a finite set. If $f :A \rightarrow A$ is an onto function, show that $f$ is one-one also.
$\displaystyle \text{Answer:}$
Let $A = \{ a_1, a_2, a_3, \ldots a_n \}.$
In order to prove that $f$ is a one-one function, we will have to show that $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of $A.$
Clearly, Range of $f = \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \}$
Since, $f:A \rightarrow A$ is an onto function. Therefore,
Range of $f= A \Rightarrow \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \} = A$
But $A$ is a a finite set consisting of $n$ elements. Therefore, $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of $A$ . Hence, $f:A \rightarrow A$ is one-one.
$\\$
Question 33: Let $R$ be the set of real numbers. If $f : R \rightarrow R ; f(x) = x^2 \text{ and } g:R \rightarrow R; g(x) = 2x + 1.$ Then, find $fog \text{ and } gof.$ Also show that $fog \neq gof.$
$\displaystyle \text{Answer:}$
Clearly; range of $f$ is a subset of domain of $g$ and range of $g$ is a subset of domain of $f$ .
$\text{Hence, } fog \text{ and } gof \text{ both exist. }$
$\text{Now, } (gof) (x) = g(f(x)) = g(x^2) = 2 (x^2) + 1 = 2x^2 + 1$
$\text{and, } (fog)(x) = f(g(x)) = f(2x+1) = (2x+1)^2$
$\because 2x^2 + 1 \neq ( 2x+1)^2$
$\therefore gof \neq fog$
$\\$
$\text{Question 34: Let } f : \{ 2,3,4,5 \} \rightarrow \{ 3,4,5,9 \} \text{ and } g: \{ 3,4,5,9 \} \rightarrow \{ 7,11,15 \} \\ \\ \text{ be functions defined as } f (2) =3, f (3)=4, f (4) =f(5) =5 \\ \\ \text{ and, } g (3) =g (4) =7 \text{ and } g(5) =g(9) =11. \text{ Find gof. }$
$\displaystyle \text{Answer:}$
We have, Range of $f = \{3,4,5 \}$
Clearly, it is a subset of domain of $g$ . Hence, $gof$ exists and $gof : \{ 2, 3, 4,5 \} \rightarrow \{ 7, 11, 15 \}$ such that
$gof (2)=g(f(2)) = g(3) = 7; gof(3) = g(f(3)) = g(4) = 7$
$gof (4) = g(f(4)) = g(5) = 11 \text{ and } gof(5) = g(f(5)) = g(5) = 11$
$\text{Hence, } \\ \\ gof : \{ 2, 3,4,5 \} \rightarrow \{7,11,15 \} \text{ such that } gof = \{ (2,7),(3,7),(4,11), (5,11) \}$
$\\$
$\text{Question 35: Find } gof \text{ and fog, if } f :R \rightarrow R \text{ and } g:R \rightarrow R \text{ are given } \\ \\ f(x)= |x| \text{ and } g(x) = |5x-2|.$
$\displaystyle \text{Answer:}$
Clearly,
$\displaystyle gof(x) = g(f(x)) = g(|x|) = \Big| 5|x|-2 \Big| = \Bigg\{ \begin{array}{rr} |5x-2|, & \text{ if } x \geq 0 \\ \\ |-5x-2|, & \text{ if } x < 0 \end{array}$
$\text{and } fog(x) = f(g(x)) = f(|5x-2|) = \Big| |5x-2|-2 \Big| = |5x-2|$
$\\$
$\displaystyle \text{Question 36: If the function } f:R \rightarrow R \text{ be given by } f(x)= x^2 + 2 \text{ and } \\ \\ g:R-\{1\} \rightarrow R \text{ be given by } g(x) = \frac{x}{x-1}. \text{ Find } fog \text{ and } gof.$ [CBSE 2014]
$\displaystyle \text{Answer:}$
$\text{Clearly, range } f = \text{ domain } g \text{ and, range } g = \text{ domain } f. \\ \\ \text{ Hence, } fog \text{ and } gof \text{ both exist. }$
$\displaystyle \text{Now, } (fog)(x) = f(g(x)) = f \Big( \frac{x}{x-1} \Big) = \Big( \frac{x}{x-1} \Big)^2 + 2 = \frac{x^2}{(x-1)^2}+2$
$\displaystyle \text{and, } (gof) (x) = g(f(x)) = g( x^2 + 2) = \frac{x^2+2}{(x^2+2)- 1} = \frac{x^2+2}{x^2+1}$
$\text{Hence, } gof:R \rightarrow R \text{ and } fog:R - \{1\} \rightarrow R \text{ are given by: }$
$\displaystyle (fog)(x)= \frac{x^2}{(x-1)^2}+2 \text{ and } (gof) (x)= \frac{x^2+2}{x^2+1}$
$\\$
$\displaystyle \text{Question 37: If } f:R - \Big\{ \frac{7}{5} \Big\} \rightarrow R - \Big\{ \frac{3}{5} \Big\} \text{ be defined as } f(x) = \frac{3x+4}{5x-7} \\ \\ \text{ and } g:R - \Big\{ \frac{3}{5} \Big\} \rightarrow R - \Big\{ \frac{7}{5} \Big\} \text{ be defined as } g(x) = \frac{7x+4}{5x-3}. \text{ Show that } \\ \\ gof = I_A \text{ and } fog = I_B, \text{ where } B = R - \Big\{ \frac{3}{5} \Big\} \text{ and } A = R - \Big\{ \frac{7}{5} \Big\}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that } f:A \rightarrow A \text{ and } g:B \rightarrow A. \\ \\ \text{Therefore , } gof:A \rightarrow A \text{ and } fog : B \rightarrow B.$
$\displaystyle gof(x) = g(f(x)) = g \Big( \frac{3x+4}{5x-7} \Big) = \frac{7( \frac{3x+4}{5x-7} )+4 }{5( \frac{3x+4}{5x-7} )-3 } \\ \\ \\ { \hspace{5.5cm} = \frac{21x+28+20x-28}{15x+20-15x+21} = \frac{41x}{41} = x }$
$\displaystyle \text{Therefore } gof: A \rightarrow A \text{ is such that } gof (x) = x \text{ for all } x \in A. \\ \\ \text{ Hence, } gof = I_A$
$\displaystyle fog(x) = f(g(x)) = f \Big( \frac{7x+4}{5x-3} \Big) = \frac{3( \frac{7x+4}{5x-3} )+4 }{5( \frac{7x+4}{5x-3} )-7 } \\ \\ \\ { \hspace{5.5cm} = \frac{21x+12+20x-12}{35x+20-35x+21} = \frac{41x}{41} = x }$
$\displaystyle \text{Therefore } fog: B \rightarrow B \text{ is such that }fog (x) = x \text{ for all } x \in B. \\ \\ \text{ Hence, } fog = I_B$
$\\$
Question 38: If $f ,g : R \rightarrow R$ are defined respectively by $f(x)=x^2 + 3x+1, g(x) = 2x-3$ , find $\text{(i)} \ fog \ \text{(ii)} \ gof \ \text{(iii)} \ fof \ \text{(iv)} \ gog.$
$\displaystyle \text{Answer:}$
$\text{Clearly, Range } f=\text{ Domain } g \text{ and, } \text{ Range } g =\text{ Domain } f. \\ \text{Therefore, } fog, gof, fof \text{ and } gog \text{ all exist. }$
$\text{(i) For any } x \in R, \text{ we have }$
$(fog) (x) = f (g(x)) = f (2x - 3) = (2x - 3)^2 + 3(2x- 3) +1 = 4x^2 -6x +1$
$\text{Therefore, } fog:R \rightarrow R \text{ is defined by } (fog)(x) = 4x^2 -6x+1 \text{ for all } x \in R.$
$\text{(ii) For any } x \in R, \text{ we have }$
$(gof)(x) = g(f(x)) = g( x^2 +3x+1) = 2(x^2 + 3x+1) -3 = 2x^2 +6x-1$
$\text{Therefore, } gof : R \rightarrow R \text{ is defined by } (gof)(x)= 2x^2 + 6x -1 \text{ for all } x \in R$
$\text{(iii) For any } x \in R, \text{ we have }$
$(fog)(x) = f (f (x)) = f (x^2+3x+1) = (x^2+3x+1)^2+ 3(x^2+ 3x+1)+1$
$\Rightarrow (fof)(x) = x^4 +6x^3 +14x^2 +15x+5$
$\text{Therefore, } fof : R \rightarrow R \text{ is defined by } (fof)(x)=x^4+6x^3+14x^2+15x+5 \\ \text{ for all } x \in R$
$\text{(iv) For any } x \in R, \text{ we have }$
$(gog)(x) = g(g(x)) = g(2x-3) = 2(2x-3)-3 = 4x-9$
$\text{Therefore, } gog: R \rightarrow R \text{ is defined by } (gog) (x) =4x -9.$
$\\$
$\text{Question 39: Let } f :Z \rightarrow Z \text{ be defined by } f (x) = x + 2. \\ \\ \text{ Find } g :Z \rightarrow Z \text{ such that } gof = I_Z.$
$\displaystyle \text{Answer:}$
$\text{We have, } gof = I_Z$
$\Rightarrow gof(x) =I_Z(x) \text{ for all } x \in Z$
$\Rightarrow g(f(x)) = x \text{ for all } x \in Z$
$\Rightarrow g(x+2)=x \text{ for all } x \in Z$
$\Rightarrow g(y) =y-2 \text{ for all } y \in Z, \text{ where } x+2=y$
$\Rightarrow g(x) = x -2 \text{ for all } x \in Z$
$\text{Hence, } g: Z \rightarrow Z \text{ defined by } g (x) = x - 2 \text{ for all } x \in Z, \\ \text{is the required function. }$
$\\$
$\text{Question 40: Let } f :R \rightarrow R \text{ be defined by } f (x) = 2x. \\ \\ \text{Find } g :R \rightarrow R \text{ such that } gof = I_R.$
$\displaystyle \text{Answer:}$
$\text{We have, } gof = I_R$
$\Rightarrow gof(x) =I_R(x) \text{ for all } x \in R$
$\Rightarrow g(f(x)) = x \text{ for all } x \in R$
$\Rightarrow g(2x)=x \text{ for all } x \in R$
$\displaystyle \Rightarrow g(y) =\frac{y}{2} \text{ for all } y \in R, \text{ where } 2x=y$
$\displaystyle \Rightarrow g(x) = \frac{x}{2} \text{ for all } x \in R$
$\displaystyle \text{Hence, } g: R \rightarrow R \text{ defined by } g (x) = \frac{x}{2} \text{ for all } x \in R, \\ \text{is the required function. }$
$\\$
$\text{Question 41: Let } f , g \text{ and } h \text{ be functions from } R \text{ to } R. \\ \\ \text{ Show that: } \text{ (i) } (f+ g)oh=foh+ goh \text{ (ii) } (fg) oh = (foh) (goh)$
$\displaystyle \text{Answer:}$
$\text{(i) Since } f, g \text{ and } h \text{ are functions from } R \text{ to } R. \text{ Therefore, }$
$(f+g) oh:R \rightarrow R \text{ and } foh + goh: R \rightarrow R$
$\text{For any } x \in R$
$((f + g) oh) (x) = (f + g) (h(x)) = f (h (x)) + g (h(x)) = foh (x) + goh(x)$
$\text{Therefore, } (f + g) oh=foh+ goh$
$\text{ (ii) Clearly, } (fg) oh: R \rightarrow R \text{ and } (foh) (goh) : R \rightarrow R \text{ such that }$
$\{(fg) oh \} (x) =(fg) (h (x)) = f (h (x)) g (h (x)) = (foh) (x) (goh) (x)$
$\{ (fg) oh \} (x) = \{ (foh) .(goh) \} (x) \text{ for all } x \in R$
$\text{Therefore } (fg) oh=(foh).(goh).$
$\\$
$\text{Question 42: Let } A =\{ x \in R : 0 \leq x \leq1 \}. \text{ If } f : A \rightarrow A \text{ is defined by }$
$\displaystyle f(x) = \Bigg\{ \begin{array}{rr} x, & \text{ if } x \in Q \\ \\ 1-x, & \text{ if } x \notin Q \end{array}$
$\text{then prove that } fof(x) = x \text{ for all } x \in A$
$\displaystyle \text{Answer:}$
$\text{Let } x \in A. \text{ Then, either } x \text{ is rational or } x \text{ is irrational. Hence two cases arise. }$
$\text{Case 1: When } x \in Q$
$\text{In this case, we have } f(x) = x$
$\text{Therefore } fof(x) = f(f(x)) = f(x) = x$
$\text{Case 2: When } x \notin Q$
$\text{In this case, we have } f(x) = 1- x$
$\text{Therefore } fof(x) = f(f(x))$
$\Rightarrow fof(x) = f( 1-x)$
$\Rightarrow fof(x) = 1 - ( 1-x) = x$
$\text{Therefore,} fof(x) = x \text{ whether } x \in Q \text{ or } x \notin Q$
$\text{Hence, } fof(x) = x \text{ for all } x \in A$
$\\$
$\text{Question 43: lf } f:R \rightarrow R \text{ given by }$
$\displaystyle f(x) = \sin^2 x + \sin^2 \Big(x + \frac{\pi}{3} \Big) + \cos x \cos \Big(x + \frac{\pi}{3} \Big) \text{ for all } x \in R, \text{ and } g : R \rightarrow R$
$\displaystyle \text{be such that } g \Big(\frac{5}{4} \Big) = 1, \text{ then prove that } gof : R \rightarrow R \text{ is a constant function. }$
$\displaystyle \text{Answer:}$
We have,
$\displaystyle f(x) = \sin^2 x + \sin^2 \Big(x + \frac{\pi}{3} \Big) + \cos x \cos \Big(x + \frac{\pi}{3} \Big)$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ 2 \sin^2 x + 2 \sin^2 \Big(x + \frac{\pi}{3} \Big) + 2 \cos x \cos \Big(x + \frac{\pi}{3} \Big) \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ 1 - \cos 2x + 1 - \cos \Big(2x + \frac{2\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) + \cos \frac{\pi}{3} \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \cos 2x - \cos \Big(2x + \frac{2\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \Big[ \cos 2x + \cos \Big(2x + \frac{2\pi}{3} \Big) \Big] + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - 2 \cos \Big(2x + \frac{\pi}{3} \Big) \cos \frac{\pi}{3} + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \cos \Big(2x + \frac{\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$
$\displaystyle \Rightarrow f(x) = \frac{5}{4} \text{ for all } x \in R$
Therefore, for any $x \in R,$ we have
$\displaystyle gof(x) = g(f(x)) = g \Big( \frac{5}{4} \Big) = 1$
$\text{Thus, } gof (x) =1 \text{ for all } x \in R. \text{ Hence, } gof :R \rightarrow R \text{ is a constant function/ }$
$\\$
$\text{Question 44: Let } f : Z \rightarrow Z \text{ be defined by } f (n) = 3n \text{ for all } n \in Z \\ \\ \text{ and } g: Z \rightarrow Z \text{ be defined by }$
$\displaystyle g(n) = \Bigg\{ \begin{array}{rr} \frac{n}{3}, & \text{ if } n \text{ is a multiple of } 3 \\ \\ 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array} \text{ for all } n \in Z.$
$\text{Show that } gof = I_Z \text{ and } fog \neq I_Z$
$\displaystyle \text{Answer:}$
$\text{Since } f:Z \rightarrow Z \text{ and } g: Z \rightarrow Z. \text{ Therefore, } gof :Z \rightarrow Z \text{ and } fog:Z \rightarrow Z.$
$\text{For any } n \in Z, \text{ we have }$
$gof (n) = g(f (n))$
$\Rightarrow gof (n) = g(3n)$
$\displaystyle \Rightarrow gof (n) = \frac{3n}{3} = n$
$\Rightarrow gof (n) = n \text{ for all } n \in Z$
$\Rightarrow gof = I_Z$
$\text{For any } n \in Z, \text{ we have }$
$fog(n) = f(g(n))$
$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} f(\frac{n}{3}), & \text{ if } n \text{ is a multiple of } 3 \\ \\ f(0), & \text{ if } n \text{ is not a multiple of } 3 \end{array}$
$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} 3(\frac{n}{3}), & \text{ if } n \text{ is a multiple of } 3 \\ \\ 3 \times 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array}$
$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} n, & \text{ if } n \text{ is a multiple of } 3 \\ \\ 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array}$
$\text{ Clearly, } fog(n) \neq n \text{ for all } n \in Z. \text{ In fact, } fog(n) = n \text{ only for multiple of } 3. \\ \\ \text{ Hence, } fog \neq I_Z.$
$\\$
$\text{Question 45: Let } f :R \rightarrow R \text{ be a function given by } f (x) = ax+b \\ \\ \text{ for all } x \in R. \text{ Find the constants } a \text{ and } b \text{ such that } fof = l_R.$
$\displaystyle \text{Answer:}$
We have, $fof = I_R$
$\Rightarrow fof (x) = I_R (x) \text{ for all } x \in R$
$\Rightarrow f(f(x)) = x \text{ for all } x \in R$
$\Rightarrow f(ax+b) = x \text{ for all } x \in R$
$\Rightarrow a(ax+b)+b=x \text{ for all } x \in R$
$\Rightarrow (a^2 -1) x + ab + b = 0 \text{ for all } x \in R$
$\Rightarrow a^2-1=0 \text{ and } ab +b =0$
$\Rightarrow a \pm 1 \text{ and } b(a+1) = 0$
$\text{When } a = 1$
$b(a+1)=0 \Rightarrow 2b=0 \Rightarrow b=0$
$a=-1 \text{ and } b=0.$
$\text{When } a = -1$
$b(a+1) = 0 \text{ for all } b \in R$
$\text{Therefore, } a = -1 \text{ and } b \text{ can take any real value. }$
$\text{Hence, either } a =1 \text{ and } b = 0, \text{ or } a =-1 \text{ and } b \text{ can take any real value. }$

Question 46: Let $f : A \rightarrow A$ be a function such that $fof = f$ .Show that $f$ is onto if and only if $f$ is one-one. Describe $f$ in this case.
$\displaystyle \text{Answer:}$
$\text{We have, } fof =f$
Let $f: A \rightarrow A$ be onto. Then, we have to prove that $f$ is one-one.
$\text{Let } x, y \in A. \text{ Then, as } f : A \rightarrow A \text{ is onto there exist } \alpha, \beta \in A \text{ such that. }$
$f(\alpha) = x \text{ and } f(\beta) = y$
$\text{Now, } f(x) = f(y)$
$\Rightarrow f(f(\alpha)) = f(f(\beta))$
$\Rightarrow fof(\alpha) = fof(\beta)$
$\Rightarrow f(\alpha) = f(\beta)$
$\Rightarrow x = y$
$\text{Hence, } f \text{ is one-one. }$
$\text{Therefore, } f :A \rightarrow A \text{ is onto } \Rightarrow f :A \rightarrow A \text{ is one-one. }$
$\text{Conversely, let } f : A \rightarrow A \text{ be one-one. Then, we have to prove that } f \text{ is onto. }$
$\text{Let } y \text{ be an arbitrary element in } A. \text{ Then, }$
$fof = f$
$\Rightarrow fof(y) = f(y)$
$\Rightarrow f(f(y)) = f(y)$
$\Rightarrow f(y) = y$
$\text{Therefore, for all } y \in A, \text{ there exists } y \in A \text{ such that } f (y) = y. \\ \text{Hence, } f \text{ is onto. }$
$\text{Now, } fof = f$
$\Rightarrow fof(x) = f(x) \text{ for all } x \in A$
$\Rightarrow f(f(x)) = f(x) \text{ for all } x \in A$
$\Rightarrow f(\alpha) = \alpha \text{ for all } \alpha = f(x) \in A$
$\text{Therefore, } f(x) = x \text{ for all } x \in A$
$\\$
Question 47: lf $f :R\rightarrow R \text{ and } g: R\rightarrow R$ be functions defined by $f (x)=x^2+1 \text{ and } g(x)= \sin x,$ then find $fog \text{ and } gof$
$\displaystyle \text{Answer:}$
We have,
$f (x) = x^2+1 \text{ and } g(x) = \sin x$
$\text{Now , } x^2 \geq 1 \text{ for all } x \in R$
$\Rightarrow x^2+1 \geq 1 \text{ for all } x \in R$
$\Rightarrow f (x)\geq1 \text{ for all } x \in R$
$\Rightarrow \text{ Range }(f) = [1, \infty)$
$\text{Also, } -1\leq \sin x \leq 1 \text{ for all } x \in R$
$\Rightarrow \text{ Range }(g) = [-1,1]$
$\text{Clearly, } \\ \text{Range} (f ) =[1, \infty) \subseteq \text{ Domain } (g) \text{ and, Range} (g) =[-1. 1] \subseteq \text{ Domain } (f )$
$\text{Hence, } gof : R \rightarrow R \text{ and } fog: R \rightarrow R \text{ are given by }$
$gof (x) = g(f(x)) = g(x^2 +1) = \sin (x^2+1)$
$\text{and, } fog(x) = f (g(x)) =f ( \sin x) = \sin^2 x+1 \text{ respectively. }$
$\\$
$\text{Question 48: If } f (x) = e^x \text{ and } g (x) = \log_e x (x > 0), \text{ find } fog \text{ and } gof. \\ \\ \text{ Is } fog = gof? \hspace{5.0cm} \text{[CBSE 2002] }$
$\displaystyle \text{Answer:}$
We observe that
$\text{Domain } (f) =R, \text{ Range }(f) =(0, \infty), \text{ Domain } (g) =(0,\infty) \\ \text{and, Range }(g) =R.$
$\text{Computation of } fog: \text{ We observe that }$
$\text{Range }( g) =\text{ Domain } (f)$
$\therefore fog \text{ exists and } fog: \text{ Domain } (g) \rightarrow R \text{ i.e. } fog: (0, \infty) \rightarrow R \text{ such that }$
$\displaystyle fog(x) = f (g(x)) \rightarrow f (\log_e x) = e^{\log_e x} = x$
$\text{Thus, } fog : (0, \infty) \rightarrow R \text{ is defined as } fog (x) = x.$
$\text{Computation of } gof : \text{ We have, }$
$\text{Range }(f) = (0, \infty) = \text{ Domain } (g)$
$\therefore gof \text{ exists and } gof: \text{ Domain } (f ) \rightarrow R \text{ i.e. } gof : R \rightarrow R \text{ such that }$
$\displaystyle gof (x) = g(f(x)) = g(e^x) = \log_e e^x = x \log_e e = x$
$\text{Thus, } gof : R \rightarrow R \text{ is defined as } gof (x) = x$
$\text{We observe that Domain } (gof) \neq \text{ Domain } (fog).$
$\therefore gof \neq fog.$
$\\$
$\text{Question 49: If } f (x)=\sqrt{x} (x > 0) \text{ and } g(x)=x^2 -1 \\ \\ \text{ are two real functions, find } fog \text{ and } gof. \text{ Is } fog = gof? \hspace{1.0cm} \text{[CBSE 2002] }$
$\displaystyle \text{Answer:}$
We observe that
$\text{Domain } (f ) =[0, \infty), \text{ Range } (f ) : [0, \infty), \text{ Domain } (g) = R$
$\text{and, Range } (g) = [-1, \infty)$
$\text{Computation of } gof: \text{ We observe that : Range }(f) = [0, \infty) \subseteq \text{ Domain } (g).$
$\text{Therefore, } gof \text{ exists and } gof: [0, \infty) \rightarrow R \text{ such that }$
$gof (x) = g(f(x)) =g(\sqrt{x})=( \sqrt{x})^2 - 1 = x - 1$
$\text{Thus, } gof (x) = [0, \infty) \rightarrow R \text{ is defined as } gof(x) = x - 1$
$\text{Computation of } fog: \text{ We observe that }$
$\text{Range } (g) = [-1, \infty) \nsubseteq \text{ Domain } (f)$
$\therefore \text{Domain } (fog) = \{ x: x \in \text{ Domain } (g) \text{ and } g(x) \in \text{ Domain } (f ) \}$
$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } g (x) \in [0, \infty) \}$
$\Rightarrow \text{Domain } (fog) =\{ x: x \in R \text{ and } x^2 -1 \in [0. \infty) \}$
$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } x^2 -1 \geq 0 \}$
$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } x \leq -1 \text{ or, } x \geq 1 \}$
$\Rightarrow \text{Domain } (fog) =\{ x : x \leq-1 \text{ or } x \geq 1 \}$
$\Rightarrow \text{Domain } (fog) = (-\infty, -1] \cup [1, \infty )$
$\text{Also, } fog(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1}$
$\text{Thus, } fog: ( -\infty, -1) \cup [1, \infty) \rightarrow R \text{ is defined as } fog( x) = \sqrt{x^2 -1 }$
We find that $fog$ and $gof$ have distinct domains. Also, their formulas are not same.
$\text{Hence } fog \neq gof.$
$\\$
$\displaystyle \text{Question 50: Let} f \text{ and } g \text{ be real functions defined by } f(x) = (x){x+1} \\ \\ \text{ and } g(x) = \frac {1}{x+3}. \text{ Describe the functions } gof \text{ and } fog \text{ (if they exist). }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have } f(x) = \frac{x}{x+1} \text{ and } g(x) = \frac {1}{x+3}$
$\text{Clearly, Domain } (f) = R - \{ -1 \} \text{ and, Range } (f ) = R - \{ 1 \}$
$\text{Also, Domain } (g) = R - \{ - 3 \text{ and, Range } (g) = R - \{ 0 \}.$
$\text{Computation of } gof : \text{ We observe that }$
$\text{Range } (f) \not\subset \text{ Domain } (g)$
$\text{Therefore, Domain } (gof ) = \{ x : x \in \text{ Domain } (f) \text{ and } f (x) \in \text{ Domain } (g) \}$
$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x: x \in R - \{ -1 \} \text{ and } \frac{x}{x+1} \in R - \{ -3 \} \Big\}$
$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x \in R : x \neq -1 \text{ and } \frac{x}{x+1} \neq -3 \Big\}$
$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x \in R : x \neq -1 \text{ and } x \neq - \frac{3}{4} \Big\}$
$\displaystyle \Rightarrow \text{Domain } (gof ) = R - \Big\{ -\frac{3}{4} , -1 \Big\}$
$\displaystyle \text{Hence, } gof : R - \Big\{ -\frac{3}{4} , -1 \Big\} \rightarrow R \text{ is defined as } gof(x) = \frac{x+1}{4x+3}$
Computation of $fog:$ We observe that: Range $(g) \not\subset \text{ Domain } (f )$
$\text{Therefore, Domain } (fog) = \{ x: x \in \text{ Domain } (g) \text{ and } g (x) \in \text{ Domain } (f ) \}$
$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x: x \in R - \{ -3 \} \text{ and } \frac{1}{x+3} \in R - \{ -1 \} \Big\}$
$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x : x \neq -3 \text{ and } \frac{1}{x+3} \neq -1 \Big\}$
$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x : x \neq -3 \text{ and } x \neq - 4 \Big\}$
$\displaystyle \Rightarrow \text{Domain } (fog ) = \{ x \in R : x \neq -3, -4 \}$
$\displaystyle \Rightarrow \text{Domain } (fog ) = R - \{ -3 , -4 \}$
$\displaystyle \text{Also, } fog(x) = f(g(x)) = f \Big( \frac{1}{x+3} \Big) = \frac{\frac{1}{x+3}}{\frac{1}{x+3} + 1} = \frac{1}{x+4}$
$\displaystyle \text{Hence, } fog : R - \{ -3, -4 \} \rightarrow R \text{ is defined as } fog(x) = \frac{1}{x+4}$
$\\$
$\displaystyle \text{Question 51: If } f(x) = \frac{1}{2x+1}, x \neq -\frac{1}{2}, \text{ then show that } f(f(x)) = \frac{2x+1}{2x+3}, \\ \\ \text{ provided that } x \neq -\frac{1}{2}, -\frac{3}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } f(x) = \frac{1}{2x+1}$
$\displaystyle \text{Clearly, Domain } (f) = R - \Big\{ -\frac{1}{2} \Big\}$
$\displaystyle \text{Let } y = \frac{1}{2x+1}. \text{ Then, }$
$\displaystyle y = \frac{1}{2x+1} \Rightarrow 2x + 1 = \frac{1}{y} \Rightarrow x = \frac{1-y}{2y}$
$\displaystyle \text{Since, } x \text{ is a real number distinct from } -\frac{1}{2}. \\ \\ \text{Therefore, } y \text{ can take any non-zero real value. }$
$\displaystyle \text{Hence, Range } (f) = R - \{ 0 \}$
$\displaystyle \text{We observe that Range } (f) = R - \{ 0 \} \nsubseteq \text{ Domain } (f) = R - \Big\{ -\frac{1}{2} \Big\}$
$\displaystyle \text{Therefore, Domain } (fof) = \{ x: x \in \text{ Domain } (f) \text{ and } f (x) \in \text{ Domain } (f) \}$
$\displaystyle \Rightarrow \text{Domain } (fof) = \Bigg\{ x:x \in R - \Big\{ -\frac{1}{2} \Big\} \text{ and } f(x) \in R - \Big\{ -\frac{1}{2} \Big\} \Bigg\}$
$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } f(x) \neq -\frac{1}{2} \Big\}$
$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } \frac{1}{2x+1} \neq -\frac{1}{2} \Big\}$
$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } x \neq -\frac{3}{2} \Big\} = R - \Big\{ -\frac{1}{2}, -\frac{3}{2} \Big\}$
$\displaystyle \text{Thus, } fof : R - \Big\{ -\frac{1}{2}, -\frac{3}{2} \Big\} \rightarrow \text{ is defined by } fof(x) = \frac{2x+1}{2x+3}.$
$\displaystyle \text{Hence, } f(f(x)) = \frac{2x+1}{2x+3} \text{ for all } x \in R , x \neq -\frac{1}{2}, -\frac{3}{2}.$
$\\$
$\displaystyle \text{Question 52: Let } f(x) = \frac{x}{\sqrt{1+x^2 } } . \text{ Then, show that } (fofof)(x) = \frac{x}{\sqrt{1+3x^2 } }.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } f(x) = \frac{x}{\sqrt{1+x^2 } }.$
Clearly, domain $(f ) = R.$
In order to find the range of $f$ , we proceed as follows:
Let $f(x)=y$ . Then,
$\displaystyle y = f(x) \Rightarrow \frac{x}{\sqrt{1+x^2 } } = y \Rightarrow \frac{x^2}{1+x^2} = y^2 \Rightarrow x = \pm \frac{y}{\sqrt{1-y^2}}$
Since $x$ takes real values. Therefore
$1- y^2 > 0 \Rightarrow y^2 - 1 < 0 \Rightarrow y \in ( -1 , 1).$
$\text{Hence, Range } (f) = ( - 1, 1)$
$\text{Clearly, Range } (f) \subset \text{ Domain } f. \\ \\ \text{Therefore, } fof :R \rightarrow R \text{ and } fofof :R \rightarrow R.$
$\text{Now, } (fofof)(x) = ((fof)of) (x) = ( fof) (f(x))$
$\displaystyle \Rightarrow (fofof)(x) = (fof) \Bigg( \frac{x}{\sqrt{1+x^2 } } \Bigg) = f \Bigg( f \Big( \frac{x}{\sqrt{1+x^2 } } \Big) \Bigg)$
$\displaystyle \Rightarrow (fofof)(x) = f \Bigg( \frac{\frac{x}{\sqrt{1+x^2 } }}{\sqrt{1+\frac{x^2}{1+x^2}}} \Bigg) = f \Big( \frac{x}{\sqrt{1+2x^2}} \Big) \\ \\ \\ { \hspace{3.5cm} = \frac{\frac{x}{\sqrt{1+2x^2 } }}{\sqrt{1 + \frac{x^2}{1+2x^2}}} = \frac{x}{\sqrt{1+3x^2}} }$
$\\$
Question 53: Let $f$ be a real function defined by $f (x)=\sqrt{x-1}.$ Find $(fofof)(x).$ Also, show that $fof \neq f^2.$
$\displaystyle \text{Answer:}$
$\text{We have, } f(x) = \sqrt{x-1}$
$\text{Clearly, Domain } (f) = [1, \infty) \text{ and Range } (f) = [0, \infty)$
$\text{We observe that range } (f ) \text{ is not a subset of domain of } f$
$\text{Domain } (fof) = \{ x : x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (f) \}$
$= \{ x : x \in [1, \infty) \text{ and } \sqrt{x-1} \in [1, \infty) \}$
$= \{ x : x \in [1, \infty) \text{ and } \sqrt{x-1} \geq 1 \}$
$= \{ x : x \in [1, \infty) \text{ and } x \geq 2 \}= [2, \infty)$
$\text{Clearly, Range } (f) = [ 0, \infty) \not\subset \text{ Domain } (fof)$
$\text{Therefore, Domain } ( ( fof)of) = \{ x: x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (fof) \}$
$= \{ x: x \in [1, \infty) \text{ and } f(x) \in [2, \infty) \}$
$= \{ x: x \in [1, \infty) \text{ and } \sqrt{x-1} \in [2, \infty) \}$
$= \{ x: x \geq 1 \text{ and } \sqrt{x-1} \geq 2 \}$
$= \{ x :x \geq 1 \text{ and } x-1 \geq 4 \}$
$\Rightarrow \text{Domain } ((fof)of) - \{ x \geq 1 \text{ and } x \geq 5 \} = [ 5, \infty )$
$\text{Now, } (fof)(x) = f(f(x)) = f(\sqrt{x-1} ) = \sqrt{\sqrt{x-1}-1}$
$\text{and } (fofof)(x) = ((fof)of)(x)$
$= (fof)(f(x))$
$= (fof)(\sqrt{x-1})$
$= f(f(\sqrt{x-1}) )= f(\sqrt{\sqrt{x-1}-1}) = \sqrt{\sqrt{\sqrt{x-1}-1}-1}$
$\text{Thus, } fof: [2, \infty) \rightarrow R \text{ and } fofof : [5, \infty) \text{ are defined as }$
$fof(x) = \sqrt{\sqrt{x-1}-1} \text{ and } (fofof)(x) = \sqrt{\sqrt{\sqrt{x-1}-1}-1}$
$\text{Now, } f^2(x) = [f(x)]^2 = ( \sqrt{x-1})^2 = x-1$
$\text{Therefore, } f^2:[1, \infty) \rightarrow \text{ is given by } f^2(x) = x-1$
$\text{Clearly, } fof \neq f^2$
$\\$
$\displaystyle \text{Question 54: If } f(x) = \frac{x-1}{x+1}, x \neq -1, \text{ then show that } f(f(x)) = -\frac{1}{x} \\ \\ \text{ provided that } x \neq 0, -1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } f(x) = \frac{x-1}{x+1}$
$\text{Clearly, } f(x) \text{ is defined for all } x \in R \text{ except } x+1 = 0 \text{ i.e. } x = - 1$
$\text{Therefore, Domain } (f) = R - \{ -1 \}$
$\text{Let us now find the range of } f.$
$\text{Let } y = f(x). \text{ Then, }$
$\displaystyle y = \frac{x-1}{x+1} \Rightarrow x = \frac{y+1}{1-y}$
$\text{As } x \text{ takes all real values other than } - 1. \\ \\ \text{Therefore, } y \text{ also takes all real values other than } 1.$
$\text{Therefore, Range } (f) = R - \{ 1 \}$
$\text{We observe that Range } (f) \not\subset \text{ Domain } (f)$
$\text{Therefore, Domain } (fof) = \{ x : x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (f) \}$
$\displaystyle = \Big\{ x : x \in R - \{-1 \} \text{ and } \frac{x-1}{x+1} \in R - \{ - 1 \} \Big\}$
$\displaystyle = \Big\{ x : x\neq -1 \text{ and } \frac{x-1}{x+1}\neq -1 \Big\}$
$= \{ x : x \neq -1 \text{ and } x \neq 0 \}$
$= R - \{ -1, 0 \}$
$\displaystyle \text{Now, } fof(x) = f(f(x)) = f \Big( \frac{x-1}{x+1} \Big) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{-2}{2x}= -\frac{1}{x}$
$\text{Thus, } fof : R - \{ -1, 0 \} \rightarrow \text{ is defined as }$
$\displaystyle fof(x) = -\frac{1}{x} \text{ or , } f(f(x)) = -\frac{1}{x}$
$\displaystyle \text{Hence, } f(f(x)) = - \frac{1}{x} \text{ for all } x \neq 0, -1$
$\\$
$\text{Question 55: If the function } f: R \rightarrow \text{ be defined by } f(x) = x^2 + 5x + 9 , \\ \\ \text{ find } f^{-1} ( 8) \text{ and } f^{-1}(9).$
$\displaystyle \text{Answer:}$
Let $f^{-1} (8) = x.$ Then,
$\displaystyle f(x) = 8 \Rightarrow x^2 + 5x + 9 = 8 \Rightarrow x = \frac{-5\pm\sqrt{21}}{2} \text{ which are in } R.$
$\displaystyle \therefore f^{-1}(8) = \Bigg\{ \frac{-5 +\sqrt{21}}{2}, \frac{-5- \sqrt{21}}{2} \Bigg\}$
$\text{Now, Let } f^{-1} (9) = x$
$\Rightarrow f(x) = 9$
$\Rightarrow x^2 + 5x + 9 = 9 \Rightarrow x^2 + 5x = 0 \Rightarrow x(x+5) = 0 \Rightarrow x = 0, -5, \text{ which are in } R$
$\text{Therefore, } f^{-1} (9) = \{ 0, -5 \}$
$\\$
Question 56: Let $f:R \rightarrow R$ be defined as $f(x) = x^2 + 1.$ Find:
$\displaystyle \text{Answer:}$
$(i) f^{-1}(5) \hspace{1.0cm} (i) f^{-1}(26) \hspace{1.0cm} (i) f^{-1}\{10, 37\}$
(i) Let $f^{-1} (-5) = x.$ Then,
$f(x) = - 5 \Rightarrow x^2 + 1 = - 5 \Rightarrow x^2 = - 6 \Rightarrow = \pm \sqrt{-6}, \text{ which is not in R. }$
(ii) Let $f^{-1} (26) = x.$ Then,
$f(x) = 26 \Rightarrow x^2 + 1 = 26 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5 \text{ which are real numbers. }$
(iii) $f^{-1}\{10, 37\} = \{ x\in R : f(x) = 10 \text{ or } f(x) = 37 \}$
$= \{ x \in R : x^2 + 1 = 10 \text{ or } x^2 + 1 = 37 \}$
$= \{ x \in R : x^2 = 9 \text{ or } x^2 = 36 \} = \{ 3, -3, 6, -6 \}$
$\\$
Question 57: Let $S =\{ 1,2, 3\}.$ Determine whether the function $f : S \rightarrow S$ defined as below have inverse. Find $f^{-1},$ if it exists
$\displaystyle \text{Answer:}$
$\text{(i) } f = \{ (1,1),(2,2), (3,3) \} \hspace{1.0cm} \text{(ii) } f = \{ (1,2), (2,1), (3, 1) \} \hspace{1.0cm} \\ \\ \text{(iii) } f = \{ (1, 3), (3, 2), (2, 1) \}$
(i) Clearly, $f : S \rightarrow S$ is a bijection. Hence, $f$ is invertible and its inverse is given by $f^{-1} =\{ (1,1),(2,2), (3,3) \}.$
(ii) Clearly, $f (2) = f ( 3) = 1 .$ Therefore, $f$ is many-one and hence it is not invertible.
(iii) Clearly, $f : S \rightarrow S$ is a bijection and hence invertible. The inverse of $f$ is given by $f^{-1} = \{ (3,1), (2, 3), (1, 2) \}$
$\\$
$\text{Question 58: Consider } f :\{1,2,3\} \rightarrow \{ a,b, c \} \text{ given by } \\ \\ f (1) =a, f (2)=b \text{ and } f (3) =c. \text{ Find the inverse } (f^{-1})^{-1} \text{ of } f^{-1}. \\ \\ \text{Show that } (f^{-1})^{-1}=f.$
$\displaystyle \text{Answer:}$
$\text{We have } f= \{ (1, a), (2, b), (3, c) \}$
Clearly, $f$ is a bijection and hence invertible. The inverse of $f$ is given by
$f^{-1} = \{ (a, 1), (b, 2), (c, 3) \}$
$\Rightarrow (f^{-1})^{-1} = \{ (1, a), (2, a), (3, c) \}$
$\text{Therefore } (f^{-1})^{-1}=f$
$\\$
$\text{Question 59: Let } f : N \cup \{ 0 \} \rightarrow N \cup \{0\} \text{ be defined by }$
$\displaystyle f(n) = \Bigg\{ \begin{array}{rr} n+1, & \text{ if } n \text{ is even } \\ \\ n-1, & \text{ if } n \text{ is odd } \end{array}$
Show that $f$ is invertible and $f = f^{-1}$
$\displaystyle \text{Answer:}$
$\text{In order to find } f^{-1} , \text{ let } n, m \in N \cup \{0\} \text{ such that}$
$f(n) = m$
$\Rightarrow n+1 = m, \text{ if } n \text{ is even}$
$n-1 = m, \text{ if } n \text{ is odd}$
$\displaystyle \Rightarrow n = \Bigg\{ \begin{array}{rr} m-1, & \text{ if } m \text{ is odd } \\ \\ m+1, & \text{ if } m \text{ is odd } \end{array}$
$\displaystyle \Rightarrow f^{-1}(m) = \Bigg\{ \begin{array}{rr} m-1, & \text{ if } m \text{ is odd } \\ \\ m+1, & \text{ if } m \text{ is odd } \end{array}$
$\displaystyle \text{Hence, } f^{-1}(n) = \Bigg\{ \begin{array}{rr} n+1, & \text{ if } n \text{ is even } \\ \\ n-1, & \text{ if } n \text{ is odd } \end{array}$
Therefore, $f = f^{-1}.$
$\\$
Question 60: Prove that the function $f : R \rightarrow R$ defined as $f (x) =2x - 3$ is invertible. Also, find $f^{-1}.$
$\displaystyle \text{Answer:}$
In order to prove that $f$ is invertible, it is sufficient to show that $f$ is a bijection. $f \text{ is one-one: Let } x, y \in R. \text{ Then, }$
$f(x) = f(y) \Rightarrow 2x-3 = 2y - 3 \Rightarrow x = y$
$\text{Thus, } f(x) = f(y) \Rightarrow x=y \text{ for all } x, y \in R$
Hence, $f \text{ is one-one }$
$f \text{ is onto: Let } y \text{ be an arbitrary element in } R ( \text{ co-domain of } f). \text{ Then, }$
$\displaystyle f(x) = y \Rightarrow 2x-3 = y \Rightarrow x = \frac{y+3}{2}$
$\displaystyle \text{Clearly, } x = \frac{y+3}{2} \in R \text{ (domain) for all } y \in R \text{ (co-domain). } \\ \\ \text{Thus, for each } y \in R \text{ there exists } x \in R \text{ such that } f(x) = y. \\ \\ \text{ Hence, } f \text{ is onto. }$
$\text{Since } f:R \rightarrow R \text{ is one-one and onto both. } \\ \\ \text{Hence, it is a bijection and hence invertible. }$
$fof^{-1} (x) = x$
$\Rightarrow f(f^{-1}(x)) = x$
$\Rightarrow 2f^{-1}(x) - 3 = x$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{x+3}{2}$
$\displaystyle \text{Thus } f^{-1}: R \rightarrow R \text{ is given by } f(x) =\frac{x+3}{2} \text{ for all } x \in R.$
$\\$
$\displaystyle \text{Question 61: Show that } f:R - \{ -1\} \rightarrow R - \{ 1 \} \text{ given by } f(x) = \frac{x}{x+1} \\ \\ \text{ is invertible. Also, find } f^{-1} .$
$\displaystyle \text{Answer:}$
In order to prove the invertibility of $f(x)$ , it is sufficient to show that it is a bijection. $f$ is one-one: For any $x, y \in R - \{ -1 \}.$
$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{x+1} = \frac{y}{y+1} \Rightarrow xy + x = xy + y \Rightarrow x = y$
$\text{Hence, } f \text{ is one-one. }$
$f \text{ is onto : Let } y \in R - \{ 1 \}. \text{ Then }$
$\displaystyle f(x) = y \Rightarrow \frac{x}{x+1} = y \Rightarrow x = \frac{y}{1-y}$
$\text{Clearly, } x \in R \text{ for all } y \in R - \{ 1 \}. \text{ Also, } x \neq -1. \text{ Because, }$
$\displaystyle x = - 1 \Rightarrow \frac{y}{1-y} = - 1 \Rightarrow y = -1 + y, \text{ which is not possible. }$
$\displaystyle \text{Thus, for each } y \in R - \{ 1 \} \text{ there exists } x = \frac{y}{1-y} \in R - \{ - 1 \} \text{ such that }$
$\displaystyle f(x) = \frac{x}{x+1} = \frac{ \frac{y}{1-y}}{ \frac{y}{1-y}+1} = y$
$\text{So, } f \text{ is onto. }$
$\text{Thus, } f \text{ is both one-one and onto. Consequently it is invertible. }$
Now,
$fof^{-1}(x) = x \text{ for all } x \in R - \{ 1 \}$
$\displaystyle \Rightarrow f(f^{-1}(x)) = x \Rightarrow \frac{f^{-1}(x)}{f^{-1}(x) + 1} = x \Rightarrow f^{-1}(x) = \frac{x}{1-x} \text{ for all } x \in R - \{1\}.$
$\\$
$\text{Question 62: Let } f: R \rightarrow R \text{ be defined as } f(x) = 10x + 7. \\ \\ \text{ Find the function } g: R \rightarrow R \text{ such that } gof = fog = I_R \hspace{1.0cm} \text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\text{We have } fog = I_R$
$fog(x) = I_R(x) \text{ for all } x \in R$
$\Rightarrow f(g(x)) = x \text{ for all } x \in R$
$\Rightarrow 10g(x) + 7 = x \text{ for all } x \in R$
$\displaystyle \Rightarrow g(x) = \frac{x-7}{10} \text{ for all } x \in R$
$\\$
$\text{Question 63: If the function } f:[1, \infty ) \rightarrow [1, \infty ) \text{ defined by } f(x) = 2^{x(x-1)} \\ \\ \text{ is invertible, find } f^{-1}(x).$
$\displaystyle \text{Answer:}$
It is given that f is invertible with $f^{-1}$ as its inverse.
$\displaystyle \text{Therefore } (fof^{-1}) (x) = x \text{ for all } x \in [1, \infty)$
$\Rightarrow f(f^{-1}(x)) = x$
$\displaystyle \Rightarrow 2^{f^{-1}(x)\{ f^{-1}(x)-1\}} = x$
$\displaystyle \Rightarrow f^{-1}(x) \{ f^{-1} (x)- 1 \} = \log_2 x$
$\displaystyle \Rightarrow \{ f^{-1}(x) \}^2 - f^{-1}(x) - \log_2 x = 0$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{1 \pm \sqrt{1 + 4 \log_2 x}}{2}$
$\displaystyle \because f^{-1} (x) \in [1, \infty ) \therefore f^{-1} (x) \geq 1$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{1 + \sqrt{1 + 4 \log_2 x}}{2}$
$\\$
Question 64: Let $f :N \rightarrow Y$ be a function defined as $f (x)=4x+ 3$ ,where $Y =\{ y \in N : y = 4x + 3 \text{ for some } x \in N).$ Show that f is invertible. Find its inverse
$\displaystyle \text{Answer:}$
In order to prove that $f$ is invertible, it is sufficient to show that it is a bijection. $f$ is one-one: For any $x, y \in N$ , we find that
$f(x) = f(y) \Rightarrow 4x+3 = 4y+ 3 \Rightarrow x = y$
$\text{Hence, } f:N \rightarrow Y \text{ is one-one. }$
$f$ is onto : Let $y$ be an arbitrary element of $Y$ . Then there exists $x \in N$ such that $y=4x+3$
$\Rightarrow y = f(x)$
Thus, for each $y \in N$ there exists $x \in N$ such that $f (x) = y$ . So, $f : N \rightarrow Y$ is onto.
Thus, $f : N \rightarrow Y$ is both one and onto. Consequently, it is invertible. Let $f^{-1}$ be the inverse of $f$ .
Then, $\displaystyle fof^{-1}(x) = x \text{ for all } x \in Y$
$\displaystyle \Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in Y$
$\displaystyle \Rightarrow 4 f^{-1}(x) + 3 = x \text{ for all } x \in Y$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{x-3}{4} \text{ for all } x \in Y$
$\displaystyle \text{Hence, } f^{-1} : Y \rightarrow N \text{ is given by } f^{-1} (x) = \frac{x-3}{4} \text{ for all } x \in Y.$
$\\$
Question 65: Let $Y = \{ n^2 : n \in N \} \subset N.$ Consider $f : N \rightarrow Y$ given by $f(n) = n^2$ . Show that $f$ is invertible. Find the inverse of $ff$ .
$\displaystyle \text{Answer:}$
In order to prove that $f$ is invertible, it is sufficient to show that it is a bijection.
$f \text{ is one-one: For any } n,m \in N, \text{ we find that }$
$f (n)=f (m)$
$\Rightarrow n^2 = m^2$
$\because n, m \in N$
$\Rightarrow n = m$
$\text{Hence, } f :N \rightarrow Y \text{ is one-one. }$
$f \text{ is onto: Let } y \text{ be an arbitrary element of } Y. \text{ Then there exists } \\ \\ n \in N \text{ such that } y = n^2$
$\Rightarrow y = f(n)$
$\text{Thus, for each } y \in Y \text{there exists } n \in N \text{such that } y = f (n). \\ \\ \text{Hence, } f : N \rightarrow Y \text{is onto. }$
$\text{Hence, } f: N \rightarrow Y \text{is a bijection. Consequently, it is invertible. }$
$\text{Let } f^{-1} \text{denote the inverse of } f. \text{Then, }$
$fof^{-1}(x)=x \text{for all } x \in Y$
$\Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in Y$
$\Rightarrow \{ f^{-1}(x) \}^2 = x \text{ for all } x \in Y$
$\Rightarrow f^{-1} (x) = \sqrt{x} \text{ for all } x \in Y$
$\text{Hence, } f^{-1} : Y \rightarrow N \text{ is given by } f^{-1}(x) = \sqrt{x} \text{ for all } x \in Y.$
$\\$
Question 66: Let $f:N \rightarrow R$ be a function defined as $f(x) = 4x^2 + 12 x + 15.$ Show that $f:N \rightarrow \text{ Range } (f)$ is invertible. Find the inverse of $f.$
$\displaystyle \text{Answer:}$
In order to prove that $f$ is invertible, it is sufficient to show that $f : N \rightarrow \text{ Range } (f )$ is a bijection.
$f \text{ is one-one: } \text{For any } x, y \in N, \text{ we find that } f(x)=f(y)$
$4x^2 +12x+15=4y^2 +12y +15$
$4(x^2 -y^2) +12(x -y) =0$
$(x-y)(4x+4y + 3)=0$
$x-y=0$
$x=y$
$\text{Hence, } f:N \rightarrow \text{ Range } (f) \text{ is one-one. }$
$\text{Clearly, } f:N \rightarrow \text{ Range } (f) \text{ is onto. Hence, } f:N \rightarrow \text{ Range } (f) \text{ is invertible. }$
$\text{Let } f^{-1} \text{ denote the inverse of } f. \text{ Then, }$
$fof^{-1} (x) = x \text{ for all } x \in \text{ Range } (f)$
$\Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in \text{ Range } (f)$
$\Rightarrow 4 \{ f^{-1} (x) \}^2 + 12 f^{-1} (x) + 15 = x \text{ for all } x \in \text{ Range } (f)$
$\Rightarrow 4 \{ f^{-1} (x) \}^2 + 12 f^{-1} (x) + 15 - x = 0$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{-12 \pm \sqrt{144-16(15-x)}}{8}$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{-12 \pm \sqrt{16x-96}}{8}$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{-3 \pm \sqrt{x-6}}{2}$
$\because f^{-1}(x) \in N, \therefore f^{-1}(x) > 0$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{-3 + \sqrt{x-6}}{2}$

$\displaystyle \textbf{Question 67. } \text{Statement 1 }\text{The function }f:N\rightarrow N\text{ given by } f(x)=5x$
$\displaystyle \text{ is one to one and onto.} \ \$
$\displaystyle \text{Statement 2 }\text{The function }f:R\rightarrow R\text{ given by } f(x)=5x\text{ is one to one and onto.}$
$\displaystyle \text{Which one of the following options is correct? } \ \$
$\displaystyle \text{(a) Both the statements are true.} \ \$
$\displaystyle \text{(b) Both the statements are false.} \ \$
$\displaystyle \text{(c) Statement 1 is true and Statement 2 is false.} \ \$
$\displaystyle \text{(d) Statement 1 is false and Statement 2 is true.} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Statement I } f : N \to N$
$\displaystyle f(x) = 5x$
$\displaystyle \text{Let } y = 5x \Rightarrow x = \frac{y}{5}$
$\displaystyle \text{Here, if } y = 12, \text{ then } x = \frac{12}{5} \notin N$
$\displaystyle \text{Hence, } f \text{ is not onto.}$
$\displaystyle \text{Statement II } f : R \to R$
$\displaystyle f(x) = 5x$
$\displaystyle \text{Now, } f(x_1) = f(x_2)$
$\displaystyle \Rightarrow 5x_1 = 5x_2 \Rightarrow x_1 = x_2$
$\displaystyle \therefore f \text{ is one-one.}$
$\displaystyle \text{Let } y = 5x \Rightarrow x = \frac{y}{5} \in R \text{ for each } y \in R$
$\displaystyle \therefore f \text{ is onto.}$
$\\$
$\displaystyle \textbf{Question 68. } \text{The function }f:X\rightarrow Y\text{ is not invertible. State the reason.} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, element } a \text{ and } e \text{ have same image}$
$\displaystyle \therefore f \text{ is not one-one.}$
$\displaystyle \text{and hence } f \text{ is not invertible.}$
$\\$
$\displaystyle \textbf{Question 69. } \text{If a relation }R\text{ on the set }A=\{1,2,3\}\text{ be defined by} \ \$
$\displaystyle R=\{(1,2),(2,1)\}\text{, then classify the function }R \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } A = \{1, 2, 3\}$
$\displaystyle \text{and } R = \{(1,2), (2,1)\}$
$\displaystyle (1,1) \notin R$
$\displaystyle \therefore R \text{ is not reflexive.}$
$\displaystyle (1,2) \in R \text{ and } (2,1) \in R$
$\displaystyle \therefore R \text{ is symmetric.}$
$\displaystyle (1,2) \in R \text{ and } (2,1) \in R \text{ but } (1,1) \notin R$
$\displaystyle \therefore R \text{ is not transitive.}$
$\\$
$\displaystyle \textbf{Question 70. } \text{Which of the following is not an equivalence relation } \\ \text{of }Z\text{? } \ \$
$\displaystyle \text{(a) }aRb\Leftrightarrow a+b\text{ is an even integer} \ \$ $\displaystyle \text{(b) }aRb\Leftrightarrow a-b\text{ is an even integer} \ \$
$\displaystyle \text{(c) }aRb\Leftrightarrow a<b \ \$ $\displaystyle \text{(d) }aRb\Leftrightarrow a=b \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) According to option (c), } aRb \Leftrightarrow a < b$
$\displaystyle \text{As, } a < b \text{ does not imply } b < a, \text{ which makes the relation not symmetric}$
$\displaystyle \text{and thus, it is not an equivalence relation.}$
$\\$
$\displaystyle \textbf{Question 71. } \text{Let }A\text{ be the set of all }50\text{ cards numbered from }1\text{ to }50. \ \$
$\displaystyle \text{Let }f:A\rightarrow N\text{ be a function defined by }f(x)=\text{card } \text{number of the card }x \ \$
$\displaystyle \text{. Then, the function }f\text{ is } \ \$
$\displaystyle \text{(a) one to one but not onto.} \ \$ $\displaystyle \text{(b) onto but not one to one.} \ \$
$\displaystyle \text{(c) neither one to one nor onto.} \ \$ $\displaystyle \text{(d) one to one and onto.} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ (a) } A \text{ is the set of all } 50 \text{ cards numbered from } 1 \text{ to } 50.$
$\displaystyle \text{Number on each card will be unique.}$
$\displaystyle \text{Now, } 51 \in N \text{ but } 51 \notin A$
$\displaystyle \text{So, there is no pre-image of } 51 \text{ in } A.$
$\displaystyle \text{Therefore, } f \text{ must be one-one.}$
$\\$
$\displaystyle \textbf{Question 72. } \text{The function }f:R\rightarrow R\text{ defined by} \ \$
$\displaystyle f(x)=\sin(3x+2),\ \forall x\in R\text{ is } \ \$
$\displaystyle \text{(a) One-One} \ \$ $\displaystyle \text{(b) Onto} \ \$ $\displaystyle \text{(c) Neither one-one nor onto} \ \$ $\displaystyle \text{(d) One-One but not onto} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Let } h(x) = \sin x \text{ and } g(x) = 3x + 2$
$\displaystyle \therefore f(x) = h \circ g(x)$
$\displaystyle g(x) = 3x + 2 \text{ is one-one and onto function but } h(x) \text{ is oscillatory.}$
$\displaystyle \text{So, } h(x) \text{ is neither one-one nor onto function.}$
$\displaystyle \text{Hence, } f(x) = h \circ g(x) \text{ is also neither one-one nor onto function.}$
$\\$
$\displaystyle \textbf{Question 73. } \text{If set }A\text{ contains }5\text{ elements and set }B\text{ contains 6 elements, then the } \ \$
$\displaystyle \text{number of one-one onto mapping} \text{from }A\text{ to }B\text{ is } \ \$
$\displaystyle \text{(a) }720 \ \$ $\displaystyle \text{(b) }120 \ \$ $\displaystyle \text{(c) }0 \ \$ $\displaystyle \text{(d) None of these} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ (c) } n(A) = 5 \text{ and } n(B) = 6$
$\displaystyle \text{We know that, the number of one-one and onto functions from } A \text{ to } B$
$\displaystyle \text{can only be possible when provided } n(A) \geq n(B).$
$\displaystyle \text{But in this case, } n(A) < n(B).$
$\displaystyle \text{Hence, the number of one-one onto mappings from } A \text{ to } B = 0.$
$\\$
$\displaystyle \textbf{Question 74. } \text{Let }A\text{ be the set of all students of a boy's school. } \text{Then, the relation } \ \$
$\displaystyle R\text{ in }A\text{ defined by }R=\{(a,b):a\text{ is } \text{sister of }b\}\text{ is } \ \$
$\displaystyle \text{(a) an equivalence relation} \ \$ $\displaystyle \text{(b) symmetric relation} \ \$
$\displaystyle \text{(c) an empty relation} \ \$ $\displaystyle \text{(d) a universal relation} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ (c) Since, it is a boys school, there are no girls students.}$
$\displaystyle \text{So, there cannot be sister of any student.}$
$\displaystyle \text{Hence, } R \text{ is an empty relation.}$
$\\$
$\displaystyle \textbf{Question 75. } \text{Let }R\text{ be the relation in the set }N\text{, given by} \ \$
$\displaystyle R=\{(x,y):x=y+3,\ y>5\}\text{. Choose the correct answer from the } \\ \text{following } \ \$
$\displaystyle \text{(a) }(7,4)\in R \ \$ $\displaystyle \text{(b) }(9,6)\in R \ \$ $\displaystyle \text{(c) }(4,1)\in R \ \$ $\displaystyle \text{(d) }(8,5)\in R \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, } R = \{x, y : x = y + 3, y > 5\}$
$\displaystyle \text{Here, } y > 5$
$\displaystyle \text{So, } y = 6$
$\displaystyle \text{when } y = 6, \text{ then } x = 6 + 3 = 9$
$\displaystyle \text{Hence, } (9, 6) \in R$
$\\$
$\displaystyle \textbf{Question 76. } \text{If }A=\{1,2,3\},\ B=\{x,y\}\text{, then the number of functions} \ \$
$\displaystyle \text{from }A\text{ to }B\text{ is } \ \$
$\displaystyle \text{(a) }3 \ \$ $\displaystyle \text{(b) }6 \ \$ $\displaystyle \text{(c) }8 \ \$ $\displaystyle \text{(d) }12 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Given, } A = \{1, 2, 3\}, \; B = \{x, y\}$
$\displaystyle \text{Here, } m = 3, \; n = 2$
$\displaystyle \text{So, number of functions from } A \text{ to } B = n^m = 2^3 = 8$
$\\$
$\displaystyle \textbf{Question 77. } \text{Write total number of functions from set }A\text{ to set }B\text{, where} \ \$
$\displaystyle A=\{1,2,3\},\ B=\{p,q,r,s\}\text{. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } A = \{1, 2, 3\}, \; B = \{p, q, r, s\}$
$\displaystyle \text{So, } n(A) = 3 \text{ and } n(B) = 4$
$\displaystyle \text{Number of functions from set } A \text{ to set } B = (n(B))^{n(A)}$
$\displaystyle = 4^3 = 64$
$\\$
$\displaystyle \textbf{Question 78. } \text{Let }f:R\rightarrow R\text{ defined as }f(x)=2x-3\text{. Find} \ \$
$\displaystyle \text{(i) }f^{-1}(x) \ \$
$\displaystyle \text{(ii) domain and range of }f^{-1}(x) \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, function } f(x) = 2x - 3 \text{ and } f : R \to R$
$\displaystyle \text{Let } a, b \in R$
$\displaystyle f(a) = f(b)$
$\displaystyle \Rightarrow 2a - 3 = 2b - 3$
$\displaystyle \Rightarrow 2a = 2b$
$\displaystyle \Rightarrow a = b \; \forall \; a, b \in R$
$\displaystyle \therefore f \text{ is one-one function.}$
$\displaystyle \text{Let } y \in R \text{ (codomain)}$
$\displaystyle f(x) = y \Rightarrow 2x - 3 = y$
$\displaystyle \Rightarrow x = \frac{y+3}{2} \in R \; \forall \; y \in R$
$\displaystyle \text{So, pre-image for each value of } y \text{ exists in domain } R.$
$\displaystyle \text{Thus, } f \text{ is onto function.}$
$\displaystyle \text{Therefore, } f \text{ is one-one onto function.}$
$\displaystyle \text{Hence, } f^{-1} : R \to R \text{ exists.}$
$\displaystyle \text{Let } x \in R \text{ (domain of } f) \text{ and } y \in R \text{ (codomain of } f)$
$\displaystyle \text{Let } f(x) = y, \text{ then } f^{-1}(y) = x$
$\displaystyle \Rightarrow f(x) = y$
$\displaystyle \Rightarrow 2x - 3 = y$
$\displaystyle \Rightarrow x = \frac{y+3}{2} \in R$
$\displaystyle \Rightarrow f^{-1}(y) = \frac{y+3}{2}$
$\displaystyle \Rightarrow f^{-1}(x) = \frac{x+3}{2}, \; \forall \; x \in R$
$\displaystyle \text{(ii) Domain of } f^{-1}(x) = \text{Range of } f(x)$
$\displaystyle \text{Here, } y = 2x - 3 \Rightarrow x = \frac{y+3}{2}$
$\displaystyle \text{Range of } f(x) = R$
$\displaystyle \text{and range of } f^{-1}(x) = \text{domain of } f(x)$
$\displaystyle \text{Domain of } f(x) = R$
$\\$
$\displaystyle \textbf{Question 79. } \text{Show that the function }f:N\rightarrow N\text{, defined by} \ \$
$\displaystyle f(x)=5x-3,\ \forall x\in N,\text{ is one-one function but not} \text{ onto function. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given a function } f : N \to N \text{ defined by } f(x) = 5x - 3$
$\displaystyle \text{For one-one}$
$\displaystyle \text{Let } x_1, x_2 \in N \text{ such that}$
$\displaystyle f(x_1) = f(x_2)$
$\displaystyle \Rightarrow 5x_1 - 3 = 5x_2 - 3$
$\displaystyle \Rightarrow 5x_1 = 5x_2 \Rightarrow x_1 = x_2$
$\displaystyle \text{So, } f \text{ is one-one.}$
$\displaystyle \text{For onto}$
$\displaystyle \text{Let } y \in N \text{ (codomain) be any arbitrary constant.}$
$\displaystyle \text{Then, } y = f(x)$
$\displaystyle \Rightarrow y = 5x - 3 \Rightarrow y + 3 = 5x$
$\displaystyle \Rightarrow x = \frac{y+3}{5}$
$\displaystyle \text{Now, for } y = 1, \; x = \frac{4}{5} \notin N$
$\displaystyle \text{Thus, } y = 1 \in N \text{ (codomain) does not have a pre-image in domain } (N).$
$\displaystyle \text{So, } f \text{ is not onto.}$
$\\$
$\displaystyle \textbf{Question 80. } \text{Let }R^{+}\text{ be the set of all positive real numbers and} \ \$
$\displaystyle f:R^{+}\rightarrow [4,\infty):f(x)=x^{2}+4.\text{ Show that inverse of }f \text{ exists and find } f^{-1}.$
$\displaystyle \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } f : R^{+} \to [4, \infty) \text{ is defined as } f(x) = x^2 + 4$
$\displaystyle \text{For one-one}$
$\displaystyle \text{Let } f(x_1) = f(x_2), \text{ for some } x_1, x_2 \in R^{+} \Rightarrow x_1^2 + 4 = x_2^2 + 4$
$\displaystyle \Rightarrow x_1^2 - x_2^2 = 0 \Rightarrow (x_1 + x_2)(x_1 - x_2) = 0$
$\displaystyle \Rightarrow x_1 = x_2, \; x_1 \neq -x_2; \; x_1, x_2 \in R^{+}$
$\displaystyle \text{Thus, } f(x_1) = f(x_2) \Rightarrow x_1 = x_2; \; x_1, x_2 \in R^{+}$
$\displaystyle \text{So, } f(x) \text{ is one-one function.}$
$\displaystyle \text{For onto}$
$\displaystyle \text{Let } y = x^2 + 4 \Rightarrow x = \sqrt{y-4} \qquad \ldots (i)$
$\displaystyle \text{Since, } x \in R^{+}, \; \forall \; y \in [4, \infty)$
$\displaystyle \text{So, range of } f(x) = [4, \infty)$
$\displaystyle \therefore \text{Range = Codomain}$
$\displaystyle \text{So, } f(x) \text{ is onto function.}$
$\displaystyle \text{Hence, } f(x) \text{ is bijective and inverse of } f(x) \text{ exists.}$
$\displaystyle \text{From Eq. (i), } f^{-1}(y) = \sqrt{y-4} \qquad [\because \; y = f(x) \Rightarrow f^{-1}(y) = x]$
$\displaystyle \text{or}$
$\displaystyle f^{-1}(x) = \sqrt{x-4}$
$\\$
$\displaystyle \textbf{Question 81. } \text{Consider the mapping }f:A\rightarrow B\text{ is defined by } f(x)=\frac{x-1}{x-2} \ \$
$\displaystyle\text{ such that }f(x)\text{ is one-one onto. Based on the above information, answer the} \ \$
$\displaystyle \text{following questions by choosing the correct option. } \ \$
$\displaystyle \text{(i) Domain of }f(x)\text{ is} \ \$
$\displaystyle \text{(a) }R-\{2\} \ \$ $\displaystyle \text{(b) }R \ \$ $\displaystyle \text{(c) }R-\{1,2\} \ \$ $\displaystyle \text{(d) }R-\{0\} \ \$
$\displaystyle \text{(ii) Range of }f(x)\text{ is} \ \$
$\displaystyle \text{(a) }R-\{2\} \ \$ $\displaystyle \text{(b) }R \ \$ $\displaystyle \text{(c) }R-\{1\} \ \$ $\displaystyle \text{(d) }R-\{0\} \ \$
$\displaystyle \text{(iii) If }g(x)=2f(x)-1,\text{ then }g(x)\text{ in terms of }x\text{ is} \ \$
$\displaystyle \text{(a) }\frac{x+2}{x} \ \$ $\displaystyle \text{(b) }\frac{x+1}{x-2} \ \$ $\displaystyle \text{(c) }\frac{x-2}{x} \ \$ $\displaystyle \text{(d) }\frac{x}{x-2} \ \$
$\displaystyle \textbf{(iv) } \text{A function }f(x)\text{ is said to be one-one, if} \ \$
$\displaystyle \text{(a) }f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2} \ \$ $\displaystyle \text{(b) }f(-x_{1})=f(-x_{2})\Rightarrow x_{1}=x_{2} \ \$
$\displaystyle \text{(c) }f(x_{1})=f(x_{2})\Rightarrow -x_{1}=x_{2} \ \$ $\displaystyle \text{(d) }-f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) (a) The denominator of the function } f(x) = \frac{x-1}{x-2}$
$\displaystyle \text{should not be zero.}$
$\displaystyle \text{So, } x \neq 2 \qquad [\because 2-2=0]$
$\displaystyle \text{Therefore, domain of } f(x) = R - \{2\}$
$\displaystyle \textbf{(ii) (c) Let } f(x) = y$
$\displaystyle \therefore y = \frac{x-1}{x-2}$
$\displaystyle \Rightarrow xy - 2y = x - 1$
$\displaystyle \Rightarrow xy - x = 2y - 1$
$\displaystyle \Rightarrow x(y-1) = 2y - 1$
$\displaystyle \Rightarrow x = \frac{2y-1}{y-1}$
$\displaystyle \text{Now, } y-1 \neq 0 \Rightarrow y \neq 1$
$\displaystyle \text{Hence, range of } f(x) = R - \{1\}$
$\displaystyle \textbf{(iii) (d) } g(x) = 2\left(\frac{x-1}{x-2}\right) - 1$
$\displaystyle = \frac{2x-2}{x-2} - 1$
$\displaystyle = \frac{2x-2 - (x-2)}{x-2}$
$\displaystyle = \frac{x}{x-2}$
$\displaystyle \textbf{(iv) (a) A function is said to be one-one if each element of set } A$
$\displaystyle \text{has a unique image in set } B.$
$\displaystyle \text{So, the function } f(x) \text{ will be one-one if }$
$\displaystyle f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
$\\$