NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

Question:2

Let D be the domain of the real-valued function f defined by $f(x) = \sqrt {25 - x^{2}}$ . Then, write D.

Answer:

$f(x) = \sqrt {25 - x^{2}}$
Therefore, the function can be defined as:
$\\25 - x^{2} \geq 0\\ $

$or, x^{2} \leq 25\\ -5 \leq x \leq 5\\$

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g:$R \rightarrow R$ be defined by f(x) = 2x + 1 and $g (x) = x^{2} - 2, \forall x \in R,$ respectively. Then, find gof.

Answer:

Here,

f(x) = 2x + 1
$g (x) = x^{2} -2, \forall x \in R\\$
Therefore,

$gof= g(f(x)) = g (2x + 1)$

$ = (2x + 1)^{2} -2 $

$= 4x^{2} + 4x + 1 -2 $

$= 4x^{2} + 4x - 1\\$

Question:4

. Let f: $R \rightarrow R$ be the function defined by $f (x) = 2x - 3, \forall x \in R$. write $f^{-1}$.

Answer:

Here, $f (x) = 2x - 3, \forall x \in R\\$
Let’s say,
$y = 2x - 3\\$

$ or, x = (y + 3)/ 2$

Therefore,
$f^{-1}(x) = (x + 3)/ 2\\$

Question:5

If $A = \{ a, b, c, d \}$ and the function $f = \{ (a, b), (b, d), (c, a), (d, c) \}$ , write $f^{-1}$.

Answer:
Here, $A = \{ a, b, c, d \}$ and the function $f = \{ (a, b), (b, d), (c, a), (d, c) \} \\$
Therefore, $f^{-1} = \{ (b, a), (d, b), (a, c), (c, d) \} \\$

Question:6

If $f:R \rightarrow R$ is defined by $f (x) = x^{2} - 3x + 2$, write $f (f (x))$.

Answer:

Here, $f (x) = x^{2} - 3x + 2$

Therefore,
$\\ f(f(x)) = f(x^{2} - 3x + 2)\\ = (x^{2} - 3x + 2)^{2} - 3(x^{2} - 3x + 2) + 2,\\$

$ = x^{4} + 9x^{2} + 4 - 6x^{3} + 4x^{2} - 12x - 3x^{2} + 9x - 6 + 2\\$

$ = x^{4} - 6x^{3} + 10x^{2} - 3x\\$
Similarly,
$\\f(f(x)) = x^{4} - 6x^{3} + 10x^{2} - 3x\\$

Question:7

Is $g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}$ a function? If g is described by $g (x) = \alpha x + \beta$ , then what value should be assigned to $\alpha\: \: and\: \: \beta$ .

Answer:

Here,$g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}$
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, $g (x) = \alpha x + \beta$
Therefore,
$\\g(1) = \alpha (1) + \beta = 1\\ \alpha + \beta = 1 \ldots \ldots .. (1)\\$
Similarly,
$g(2) = \alpha (2) + \beta = 2 \alpha + \beta = 3 \ldots \ldots .. (2)\\$
By solving (1) and (2), we get
$\alpha = 2 and \beta = -1\\$
Therefore, $g (x) = 2x - 1$

Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) $\{ (x, y): \text{x is a person, y is the mother of x} \} .\\$
(ii) $\{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\$

Answer:
(i)Here, $\{ (x, y): \text{x is a person, y is the mother of x} \} .\\$
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs makes a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, $\{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\$
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by $f = \{ (1, 2), (3, 5), (4, 1) \}$ and $g = \{ (2, 3), (5, 1), (1, 3) \}$ , write $fog.\\$

Answer:
Here,

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

$fog (5) = f(g(5)) = f(1) = 2\\$

$ fog (1) = f(g(1)) = f(3) = 5\\$
So, we can write that:
$\\fog = \{ (2, 5), (5, 2), (1, 5) \} \\$

Question:10

Let C be the set of complex numbers. Prove that the mapping $f: C \rightarrow R$ is given by $f (z) = \vert z \vert , \forall z \in C$, is neither one-one nor onto.

Answer:

Here, $f: C \rightarrow R$ is given by $f (z) = \vert z \vert , \forall z \in C$
If we assume $z = 4 + 3i$
Then,
$f(4 + 3i) = \vert 4 + 3i \vert = \sqrt (4^{2} + 3^{2})= \sqrt 25 = 5\\$
Similarly, for $z = 4 - 3i$
$f(4 - 3i) = \vert 4 - 3i \vert = \sqrt (4^{2} + 3^{2}) = \sqrt 25 = 5\\$
Therefore, it is clear that f(z) is many-one.
So, $\vert z \vert \geq 0, \forall z \in C,\\$
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function $f: R \rightarrow R$ be defined by $f (x) = cos x, \forall x \in R.$ Show that f is neither one-one nor onto.

Answer:
It is given that, $f: R \rightarrow R, f(x) = cos x, \forall x \in R\\$
So, we can write:
$f(x_{1}) = f(x_{2})\\ or, cos x_{1}= cos x_{2}\\$
Hence, $x_{1 }= 2n \pi \pm x_{2}, where n \in Z\\$
It is understandable that for any value of $x_{1 }and\: \: x_{2}$, the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let $X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \}$ . Find whether the following subsets of $X \times Y$ are functions from X to Y or not.
(i) $f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \}$
(ii) $g = \{ (1, 4), (2, 4), (3, 4) \} \\$
(iii) $h = \{ (1,4), (2, 5), (3, 5) \}$
(iv) $k = \{ (1,4), (2, 5) \} .\\$

Answer:

Here, $X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} \\$
Therefore, $X \times Y = \{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) \} \\$
(i) $f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \} \\$
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) $g = \{ (1, 4), (2, 4), (3, 4) \} \\$
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) $h = \{ (1,4), (2, 5), (3, 5) \} \\$
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) $k = \{ (1, 4), (2, 5) \} \\$
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions $f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA$, then show that f is one-one and g is onto.

Answer:

Here, it is given:
$f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA$
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let $f: R \rightarrow R$ be the function defined by $f(x) = 1/(2 - cos x) \forall x \in R$. Then, find the range off.

Answer:

Here, $f(x) = 1/(2 - cos x) \forall x \in R$
Let’s say,
$\\y = 1/(2 - cos x)\\ \therefore2y - ycos x = 1\\ Or, cos x = (2y - 1)/ y\\ Or, cos x = 2 - 1/y\\$
As the range of cos x is, $-1 \leq cos x \leq 1\\$
Therefore,
$\\-1 \leq 2 - 1/y \leq 1\\ -3 \leq - 1/y \leq -1\\ 1 \leq - 1/y \leq 3\\ 1/3 \leq y \leq 1\\$
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: $\forall a, b \in Z$, aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: $\forall a, b \in Z$, aRb if and only if a - b is divisible by n.
Here, aRa $\Rightarrow (a - a)$ is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, $aRb \Rightarrow (a - b)$ is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric, and transitive.

Answer:

Here, $A = \{ 1, 2, 3 \} .\\$
(i) Assume $R_{1}= \{ (1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3) \} \\$
Here, (1, 1), (2, 2) and (3, 3) $\in \ R_{1}. R_{1}$ is reflexive.
$(1, 2) \in R_{1}, (2, 3) \in R_{1} \Rightarrow (1, 3) \in R_{1}$
. Hence,$R_{1}$ is transitive.
Now, $(1, 2) \in R_{1 } \Rightarrow (2, 1) \notin R_{1}.$
Therefore, $R_{1}$ is not symmetric.
(ii) Let say, $R_{2}= \{ (1, 2), (2, 1) \}$
So, $(1, 2) \in R_{2}, (2, 1) \in R_{2}\\$
Therefore, $R_{2 }$ is symmetric,
$(1, 1) \notin R_{2}$. Therefore, $R_{2 }$ is not reflexive.
$(1, 2) \in R_{2}$, $(2, 1) \in R_{2 }$ but $(1, 1) \notin R_{2}$. Hence, $R_{2 }$ is not transitive.
(iii) Let $R_{3 }$ = $\{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) \} \\$
$R_{3 }$ is reflexive as $(1, 1) (2, 2) and (3, 3) \in R_{3 }\\$
$R_{3 }$ is symmetric as $(1, 2), (1, 3), (2, 3) \in R_{3 }$ $\Rightarrow (2, 1), (3, 1), (3, 2) \in R_{3 }\\$
Therefore, $R_{3 }$ is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
$R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}$ . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, $R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}$
So, the domain $D= \{ 1, 2, 3, \ldots .., 20 \} \\$
And the Range $= \{ 1, 3, 5, \ldots .., 39 \} \\$
Here, $(2, 2) \notin \: \: R\: \: as\: \: 2 \times 2 + 2 \neq 41$. Therefore, R is not reflexive.
Again, $(1, 39) \in R \: \: but\: \: (39, 1) \: \: \notin R$. So, R is not symmetric.
Again, $(11, 19) \notin R, (19, 3) \notin R; but (11, 3) \notin R$. So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

(c) a mapping from B to A.

Answer:

Here, $A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \} \\$
(i) Assume, $f: A \rightarrow B \text{ denote a mapping}$
Then, $f = \{ (x, y): y = x + 3 \} or\\$
$f = \{ (2, 5), (3, 6), (4, 7) \}$ , this is an injective mapping.
(ii) Again assume, $g: A \rightarrow B$ denotes a mapping: $g = \{ (2, 2), (3, 2), (4, 5) \}$ . Hence, it is not an injective mapping.
(iii) Again assume, h: $B \rightarrow A$ denotes a mapping: $h = \{ (2, 2), (5, 3), (6, 4), (7, 4) \}$ . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, $f: N \rightarrow N,$ be a mapping defined by $f (x) = x^{2}\\$
If, $f(x_{1}) = f (x_{2})\\$
Then, $x_{1}^{2}= x_{2}^{2}\\$
So,$x_{1 }= x_{2 }(as, x_{1 }+ x_{1 }\text{cannot be } 0)$
$f(x_{1 }) = f(x_{2 })$, hence, f(x) is one-one.
However, ‘f’ is not onto, as for $1 \in N$, therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let $f: R \rightarrow [0, \infty )$, is a mapping that is defined by $f(x) = \vert x \vert \\$
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But $\vert x \vert \geq 0$, so the range is $[0, \infty ].\\$
Therefore, f(x) is onto.
(iii) Assume, $f: R \rightarrow R$ , be a mapping which is defined by $f(x) = x^{2}\\$
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is $[0, \infty )$.
Therefore, f (x) is neither one-one nor onto.

Question:20

Let $A = R - \{ 3 \} , B = R - \{ 1 \}$ . Let $f : A \rightarrow B$ be defined by $f(x) = x - 2/ x - 3 \forall x \in A$ . Then show that f is bijective.

Answer:

Here,

$A = R - \{ 3 \} , B = R - \{ 1 \} \\$
$f : A \rightarrow B$ be defined by $f (x) = x - 2/ x - 3 \forall x \in A\\$
Hence, $f (x) = (x - 3 + 1)/ (x - 3) = 1 + 1/ (x - 3)\\$
Say $f(x1) = f (x2)\\$
$1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}$
$\\\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3} \\\\ x_{1}=x_{2}$
So, f (x) is an injective function.
$\\If, y = (x - 2)/ (x -3)\\$

$ Or, x - 2 = xy - 3y\\$

$ Or, x(1 - y) = 2 - 3y\\$

$ Or, x = (3y - 2)/ (y - 1)\\$

$ Or, y \in R - \{ 1 \} = B\\$
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) $f(x) = x/2$ (ii) $g(x) = \vert x \vert \\$
(iii) $h(x) = x \vert x \vert$ (iv) $k(x) = x^{2}\\$

Answer:

Here, A = [-1, 1]
(i) $f: [-1, 1] \rightarrow [-1, 1],$

$ f (x) = x/2\\$
If, $f(x_{1}) = f(x_{2})\\$
$x_{1}/2 = x_{2}/2\\$
So, f(x) is one-one.
Also, $x \in [-1, 1]\\$
$x/2 = f(x) = [-1/2, 1/2]\\$
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
$\\g(x) = \vert x \vert \\$

$ \text{Let} g(x_{1}) = g (x_{2})\\ $

$\vert x_{1} \vert = \vert x_{2} \vert \\ $

$x_{1}= \pm x_{2}\\$
So, g(x) is not one-one.
Also, $g(x) = \vert x \vert \geq 0, \text{for all real x}\\$
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
$\\\\h(x) = x \vert x \vert \\$

$ \text{Let } h(x_{1}) = h(x_{2})\\ x_{1} \vert x_{1} \vert = x_{2} \vert x_{2} \vert \\ $

$\text{If }x_{1}, x_{2}> 0\\ x_{1}^{2}= x_{2}^{2}\\ x_{1}^{2}- x_{2}^{2}= 0\\ (x_{1} - x_{2})(x_{1} + x_{2}) = 0\\ x_{1 }= x_{2 }(as x_{1}+ x_{2} \neq 0)\\$

$ \text{Again }, x_{1}, x_{2 }< 0, and x_{1}= x_{2}\\$

$ \text{Therefore }, x_{1 }and x_{2 }\text{of opposite sign}, x_{1 } \neq x_{2 }.\\$
Hence, f(x) is one-one.
$\\\text{For x} \in [0, 1], f(x) = x_{2} \in [0, 1]\\$

$ \text{For x} < 0, f(x) = -x_{2} \in [-1, 0)\\$
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
$\\k(x) = x^{2}\\$

$ \text{Let }k (x_{1}) = k (x_{2})\\$

$ x_{1}^{2}= x_{2}^{2}\\ $

$x_{1 }= \pm x_{2}\\$
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, $x, y \in N\\$
(ii) x + y = 10, $x, y \in N\\$
(iii) x y is square of an integer $x, y \in N\\$
(iv) x + 4y = 10 $x, y \in N\\$
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; $x, y \in N\\$
If $(x, x) \in R$, then x > x, that does not satisfy for any $x \in N$.\\
Therefore, R is not reflexive.
Say,$(x, y) \in R$
$\\Or, xRy\\ $

$\\ Or, x > y\\$
$\Rightarrow y > x \\$ For any $x, y \in N$, the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
$\\Or, x > y and y > z\\ $

$Or, x > z\\$

$ Or, xRz\\$
Hence, R is transitive.
(ii) x + y = 10;$x, y \in N$
Thus,
$\\R = \{ (x, y); x + y = 10, x, y \in N \} \\ $

$R = \{ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \} \\$
Therefore, $(1, 1) \notin R\\$
So, R is not reflexive.
Again, $(x, y) \in R$

$ \Rightarrow (y, x) \in R\\$
Therefore, R is symmetric.
And, $(1, 9) \in R, (9, 1) \in R, but (1, 1) \notin R\\$
Therefore, R is not transitive.
(iii)Here, xy is square of an integer $x, y \in N$
$R = \{ (x, y) : \text{xy is a square of an integer} x, y \in N \} \\$
So, $(x, x) \in R, \forall x \in N\\$
For any $x \in N, x^{2 }$ is an integer.
Thus, R is reflexive.
If $(x, y) \in R $

$\Rightarrow (y, x) \in R\\$
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, $xy = m^{2} \: \: and \: \: yz = n^{2} \text{for some m, n} \in Z\\$
$\\x = m^{2} /y and z = n^{2} /y\\ xz = m^{2} n^{2} / y^{2}$, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; $x, y \in N\\$
$\\R = \{ (x, y): x + 4y = 10; x, y \in N \} \\$

$ R = \{ (2, 2), (6, 1) \} \\ Here, (1, 1) \notin R\\$
Hence, R is not symmetric.
$\\(x, y) \in R$

$ \Rightarrow x + 4y = 10\\ And, (y, z) \in R $

$\Rightarrow y + 4z = 10\\$

$ Or,\ x - 16z = -30\\$

$ Or,\ (x, z) \notin R\\$
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in $A \times A$ defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in $A \times A$. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

$\\Here, A = \{ 1, 2, 3, \ldots 9 \} \: \: and \: \: (a, b) R (c, d)\: \:$

$ if a + d = b + c for (a, b), (c, d) \in A \times A.\\ Say (a, b) R (a, b)\\ $

$Therefore, a + b = b + a, \forall a, b \in A .$
This must be true for any $a, b \in A.\\$
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
$\\a + d = b + c\\ c + b = d + a\\ (c, d) R (a, b)\\$
Therefore, R is symmetric.
Let $\\(a, b) R(c, d) ,and, (c, d) R(e, f)$
$\\ a + d = b + c ,and, c + f = d + e\\ a + d = b + c ,and, d + e = c + f\\$

$ Or, (a + d) - (d + e = (b + c) - (c + f)\\ $

$Or, a - e = b - f\\ $

$Or, a + f = b + e\\ $

$Or, (a, b) R(e, f)\\$
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function $f: A \rightarrow B$ is invertible if and only if f is both one-one and onto.

Answer:

Say, $f: A \rightarrow B$ be many-one functions.
If, $f(a) = p \text{ and } f(b) = p\\$
Then $f^{-1}(p) = a \: \: and \: \: f^{-1}(p) = b\\$
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, $f: A \rightarrow B$ is not onto function.
$B = \{ p, q, r \} and \{ p, q \}$ is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then $f = X \rightarrow Y$ is invertible.

Question:25

Functions $f, g : R \rightarrow R$ are defined, respectively, by $f(x) = x^{2} + 3x + 1, g(x) = 2x - 3$, find
(i) fog (ii) gof (iii) fof (iv) gog

Answer:

Here,
$f(x) = x^{2} + 3x + 1, g (x) = 2x - 3\\$

$\\(i)fog = f(g(x))\\ = f(2x - 3)\\ = (2x - 3)^{2} + 3(2x - 3) + 1\\ $$= 4x^{2}\ + 9 - 12x + 6x - 9 + 1\\ = 4x^{2}\ - 6x + 1\\ $

$(ii) gof = g(f(x))\\ = g(x^{2}\ + 3x + 1)\\ = 2(x^{2}\ + 3x + 1) - 3\\$

$ = 2x^{2}\ + 6x - 1\\$

$ (iii) fof = f(f(x))\\ = f(x^{2}\ + 3x + 1)\\ = (x^{2}\ + 3x + 1)2 + 3(x^{2} + 3x + 1) + 1\\ $

$= x4 + 9x^{2}\ + 1 + 6x^{3} + 6x + 2x^{2}\ + 3x^{2}\ + 9x + 3 + 1\\$

$ = x4 + 6x^{3} + 14x^{2}\ + 15x + 5\\ (iv) gog = g(g(x))\\ $

$= g(2x - 3)\\ = 2(2x - 3) - 3\\ = 4x - 6 - 3\\ = 4x - 9\\$

Question:26

Let $\ast$ be the binary operation defined on Q. Find which of the following binary operations are commutative
$\\(i) a \ast b = a - b\: \: \forall a, b \in Q\\ (ii) a \ast b = a^2 + b^2 \: \: \forall a, b \in Q\\ (iii) a \ast b = a + ab \: \: \forall a, b \in Q\\ (iv) a \ast b = (a - b)^{2}\: \: \forall a, b \in Q\\$

Answer:

Here, $\ast$ is a binary operation defined on Q.
(i)
$\\a \ast b = a - b, \forall a, b \in Q \: \: and \: \: b \ast a = b - a\\ So, a \ast b \neq b \ast a\\$
Hence, $\ast$ is not commutative.
(ii)
$\\a \ast b = a^{2}+ b^{2}\\ b \ast a = b^{2}+ a^{2}\\$
Therefore, $\ast$ is commutative.
(iii)
$\\a \ast b = a + ab\\ b \ast a = b + ab\\ Hence, a + ab \neq b + ab\\$
Therefore, $\ast$ is not commutative.
(iv)
$\\a \ast b = (a - b)^{2}, \forall a, b \in Q\\ b \ast a = (b -a)^{2}\\ As, (a - b)^{2} = (b - a)^{2}\\$
Therefore, $\ast$ is commutative.

Question:27

Let $\ast$ be binary operation defined on R by $a \ast b = 1 + ab, \forall a, b \in R$. Then the operation $\ast$ is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,

$\ast$ is a binary operation defined on R by $a \ast b = 1 + ab, \forall a, b \in R$
So, $a \ast b = ab + 1 = b \ast a\\$
Therefore, $\ast$ is a commutative binary operation.
$\\Now, a \ast (b \ast c) = a \ast (1 + bc) = 1 + a (1 + bc) = 1 + a + abc\\ $

$Again, (a \ast b) \ast c = (1 + ab) \ast c = 1 + (1 + ab) c = 1 + c + abc\\$

$ Therefore, a \ast (b \ast c) \neq (a \ast b) \ast c\\$
Hence, $\ast$ is not associative.
Therefore, $\ast$ is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to $b \forall a, b \in T$. Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, $\forall a, b \in T.\\$
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, $aRb \Rightarrow a \sim b\\$
$b \sim a \Rightarrow bRa\\$
Therefore, R is symmetric.
Say, aRb and bRc
$\\a \sim b \: \: and\: \: b \sim c\\ a \sim c \Rightarrow aRc\\$
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

(C) neither symmetric nor transitive (D) both symmetric and transitive

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.

Question:30

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(A) 1 (B) 2

(C) 3 (D) 5

Answer:

(D) 5

Given, set $A = \{ 1, 2, 3 \} \\$
Now, the number of equivalence relations as follows
$\\R1 = \{ (1, 1), (2, 2), (3, 3) \} \\$

$ R2 = \{ (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) \} \\$

$ R3 = \{ (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) \} \\$

$ R4 = \{ (1, 1), (2, 2), (3, 3), (2, 3), (3, 2) \} \\$

$ R5 = \{ (1, 2, 3) \Leftrightarrow A \times A = A^2 \} \\$
Thus, the maximum number of equivalence relations is ‘5’.

Question:31

If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(A) reflexive (B) transitive

(C) symmetric (D) none of these

Answer:

(D) none of these

If, R be defined on the set {1, 2, 3} by R = {(1, 2)}

Then, we can say that R is not reflexive, transitive, and symmetric.

Question:32

Let us define a relation R in R as aRb if $a \geq b.$ Then R is

(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.

Answer:

(b).
The defined relation R in R as aRb if $a \geq b.\\$
Similarly, aRa implies $a \geq a$ which is true.
Therefore, it is reflexive.
Let $aRb \rightarrow a \geq b, but, b \ngtr a.\\$
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, $a \geq b, b \geq c$. So, $a \geq c$, and this is true.
Therefore, R is transitive.

Question:33

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.

Answer:

(a)

Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Therefore, it can be written as: 1R1, 2R2 and 3R3.

Therefore, R is reflexive.

Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.

Therefore, R is not symmetric.

Again, 1R1 and $1R2 \rightarrow 1R3\\$

Therefore, R is transitive.

Question:34

The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is

(a) 1 (b) 0 (c) 2 (d) None of these

Answer:

(c)
Here: $a \ast b = ab/2 a, b Q \sim \{ 0 \} \\$
Assume ‘e’ as the identity element.
Therefore, $a \ast e = ae/2 = a \rightarrow e = 2 \\$

Question:35

If the set A contains 5 elements and set B contains 6 elements, then the number of one-one and onto mapping from A to B is

(a) 720 (b) 120 (c) 0 (d) None of these

Answer:

Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping

from A to B is n! when m = n

and, 0 if m ≠ n

It is given that, m = 5 and n = 6.

As,5 ≠ 6 So, from A to B mapping = 0

Question:36

Let $A = \{ 1, 2, 3, ..., n \}$ and $B = \{ a, b \}$ . Then the number of surjections from A to B is
(a) $^{n}P_{2}$
(b) $2^{n} - 2$
(c)$2^{n} - 1$
(d) None of these

Answer:

(d)
It is given that, $A = \{ 1, 2, 3, ..., n \}$ and $B = \{ a, b \} \\$
Say, the number of elements in set A and B are m and n respectively.
Therefore, $^{n}C_{m} \times m!$ is the number of surjections from A to B, for $n \ngeq m \\$
Given, m=2.
Therefore, the number of surjections from A to B
$= nC_2 \times 2! = n!/[2! \times (n-2)!] = n(n - 1) = n^{2} - n \\$

Question:37

Let $f : R \rightarrow R$ be defined by $f(x) = 1/x \forall x \in R$ then f is
(a) one-one (b) onto (c) bijective (d) f is not defined

Answer:

(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.

Question:38

Let $f : R \rightarrow R$ be defined by $f(x) = 3x^{2} - 5$ and $g : R \rightarrow R$ by $g(x) = x/(x^{2}+1)$, then gof is
$\\(a) (3x^{2}-5)/(9x^{4}-30x^{2}+ 26)\\ (b) (3x^{2}-5)/(9x^{4}-6x^{2}+ 26) \\(c) 3x^{2}/(x^{4}+2x^{2}-4) \\(d) (3x^{2})/(9x^{4}-30^{2}+2)\\$

Answer:

(a)

Given, $f(x) = 3x^{2} - 5 \ and \ g(x) = x/(x^{2}+1)\\$
Therefore,
$\\gof = gof (x) = g(3x^{2}-5)$

$ = (3x^{2}-5)/[(3x^{2}-5)^{2}+1]\\$

$ = (3x^{2}-5)/(9x^{4}-30x^{2}+ 26)\\$

Question:39

Which of the following functions from Z to Z are bijections?
$\\(a) f(x) = x^{3 }\\(b) f(x) = x + 2 \\(c) f(x) = 2x + 1 \\(d) f(x) = x^{2} + 1 \\$

Answer:

(b)
Given that $f : Z \rightarrow Z \\$
Say, $x_{1}, x_{2} \in f(x) \rightarrow f(x_{1}) = x_{1}+ 2; f(x_{2}) = x_{2}+ 2;\\$
For, $f(x_{1}) = f(x_{2})\ \rightarrow x_{1}+ 2 = x_{2}+\ 2 \\$
or, $x_{1 }= x_{2}\\$
Therefore, the function f(x) is one-one.
Say, y = x + 2
or, $x = y - 2 \forall y \in Z \\$
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.

Question:40

Let $f : R \rightarrow R$ be the functions defined by $f(x) = x^{3} + 5$. Then $f^{-1}(x) is$
$\\(a) (x + 5)^{1/3}\\ (b) (x - 5)^{1/3}\\ (c) (5 - x)^{1/3}\\ (d) 5 - x \\$

Answer:

(b)Here,

$\\f(x) = x^{3} + 5\\ $

$Say, y = x^{3} + 5\\ $

$or, x^{3} = y - 5 \\ $

$or, x = (y - 5)^{1/3}\ f^{-1}1(x) = (x - 5)^{1/3} \\$

Question:41

Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions. Then$(gof)^{-1}$ is
$\\(a) f^{-1 }og^{-1} \\(b) fog \\(c) g^{-1}of^{-1}\\ (d) gof \\$
Answer:

(a)
Given that, $f : A \rightarrow B \text{ and } g : B \rightarrow C \\$
Then, $(gof)^{-1} = f^{-1}og^{-1}\\$

Question:42

Let $f: \{ R-3/5 \} \rightarrow R$ be defined by $f(x) = (3x + 2)/(5x - 3)$ then
$\\(a) f(x) = f(x); \\(b) f^{-1}(x) = - f(x) \\(c) (fof)x = - x \\(d) f^{-1}(x) = f(x)/19\\$

Answer:

(a)

Here,$f(x) = (3x + 2)/(5x - 3) \forall x \neq 3/5\\$
$\\So, y = (3x + 2)/(5x - 3)\\ $

$or, y (5x-3) = (3x + 2)\\$

$ or, 5xy-3y = 3x + 2\\ $

$or, x = (3y+2)/(5y-3)\\ $

$Therefore, f^{-1}(x) = (3x+2)/(5x-3)\\$

$ Hence, f^{-1}(x) = f(x)\\$

Question:43

Let $f : [0, 1] \rightarrow [0, 1]$ be defined by

f(x) = { x, if is rational

{ 1-x , if is irrational
Then (fof)x is

(a) constant

(b) 1 + x

(c) x

(d) None of these

Answer:

(c)
Here, $f : [0, 1] \rightarrow [0, 1]\\$
$\\f = f^{-1} \\ Therefore, (fof)x = x\\$

Question:44

Let $f : [2, \infty )$ and R be the function defined by $f(x) = x^{2} - 4x + 5$, then the range of f is
$\\(a) R\\ (b) [1, \infty ) \\(c) [4, \infty ) \\(d) [5, \infty ) \\$
Answer:

(b)

$\\Here, f(x) = x^{2} - 4x + 5 \\$

$ Assume, y = x^{2}\ - 4x + 5 \\ $

$or, x^{2}\ - 4x + 5 - y = 0 \\$

$ or, x = [-4 \pm \sqrt \{ (-4)^{2}- 4(-4)(5-y) \} ]/2x = 2 \pm \sqrt (y-1)\\ $

$\text{When x is real}, y-1 \geq 0\\$

$ Therefore, y \geq 1\\ \text{Hence the range is}[1, \infty )\\$

Question:45

Let $f : N \rightarrow R$ be the function defined by $f(x) = (2x-1)/2$ and $g : Q \rightarrow R$ be another function defined by $g(x) = x + 2$ then, gof(3/2) is
$\\(a) 1 \\(b) - 1 \\(c) 7/2 \\(d) None of these \\$

Answer:

(d)
Given that, $f(x) = (2x-1)/2 \: \: and\: \: g(x) = x + 2 \\$
Therefore, $gof(x) = g[(f(x)] = f(x) + 2 = (2x-1)/2 = (2x+3)/2\\$
Hence, $gof(3/2) = 3 \\$

Question:46

Let f : $R \rightarrow R$ be defined by

$\\ (a) 9 \\(b) 14 \\(c) 5 \\(d) None of these \\$

Answer:

(a)
$Here, f(- 1) + f(2) + f(4) = 3(- 1) + (2)^{2} + 2(4) = - 3 + 4 + 8 = 9 \\$

Question:47

If $f : R \rightarrow R$ be given by $f(x) = tan x$, then f^-1 is
$\\(a) \frac{\pi}{4} \\ (b) (n \pi + \pi /4) where\ n \in Z \\(c) \ does \ not \ exist\\ (d) None\ of \ these \\$
Answer:

(a)

$\\Here, f(x) = tan x \\$

$ Assume, f(x) = y = tan x\\ $

$Then,\ \ x = tan^{-1} (y)\\ $

$or, f^{-1}(x) = tan^{-1}\ (x) \\ $

$or, f^{-1}(1) = tan^{-1} (1)\\$

$ or,\ f^{-1}(1) = tan^{-1} tan ( \pi /4)=\frac{\pi}{4}\\$

Question:48

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............

Answer:

Here, aRb : 2a + 3b = 30

Or, 3b = 30 – 2a

Or, b = (30 - 2a)/ 3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question:49

Let the relation R be defined on the set $A = \{ 1, 2, 3, 4, 5 \}$ by $R = \{ (a, b) : \vert a^{2} - b^{2} \vert < 8 \}$. Then R is given by ............

Answer:

Here
$\\A = \{ 1, 2, 3, 4, 5 \} and R = \{ (a, b) : \vert a^{2} - b^{2} \vert < 8 \} \\$

$ \text{Therefore, we can say}, $

$R = \{ (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (4, 3) (3, 4), (4, 4), (5, 5) \} \\$

Question:50

Let f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}. Then gof = ................ and fog = ...................

Answer:

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

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

$Then, gof(1) = g[f(1)] = g(2) = 3\\$

$ Again, gof(3) = g[f(3)] = g(5) = 1 \\$

$ Again, gof(4) = g[f(4)] = g(1) = 3\\ $

$Therefore, gof = \{ (1, 3), (3, 1), (4, 3) \} \\$

$ Again, fog(2) = f[g(2)] = f(3) = 5\\ $

$Again, fog(5) = f[g(5)] = f(1) = 2\\ $

$Again, fog(1) = f[g(1)] = f(3) = 5\\$

$ Hence, fog = \{ (2, 5), (5, 2), (1, 5) \} \\$

Question:51

Let $f : R \rightarrow R$ be defined by $f(x) = x/ \sqrt (1+x^{2})$ then $(fofof)(x) = ................. \\$

Answer:

$(\text { fofof })(x) =f[f(x))] \\ =f\left[\begin{array}{l} f\left(\frac{x}{\sqrt{1+x^{2}}}\right) \end{array}\right] \\ $

$=f\left(\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}\right) \\$

$ =f\left(\frac{x}{\sqrt{1+2 x^{2}}}\right)$
$\\\text { fofof }(x) =\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2 x^{2}}}} \\ =\frac{x}{\sqrt{1+3 x^{2}}}$

Question:52

If, $f(x) = [4 - (x - 7)^{3}]$, then $f^{-1}$(x) = .............

Answer:

$\\Here, f(x) = [4 - (x - 7)^{3}]\\$

$ Say, y = [4 - (x - 7)^{3}]\\ $

$Or, (x - 7)^{3} = (4 - y)\\ $

$Or, (x-7) = (4-y)^{1/3}\\$

$ Or, x = 7 + (4-y)^{1/3}\\$

$ Therefore, f^{-1}(x) = 7 + (4 - x)^{1/3}\\$

Question:53

Let R = {(3,1), (1,3), (3,3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive and not reflexive.

Answer:

False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}

As (1 ,1) R, R is not a reflexive one.

As (3, 1) R and (1, 3) belongs to R, then R is symmetric.

Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.

Question:54

Let f: R→R be the function defined by $f(x) = sin(3x+2)$, x R. Then f is invertible.

Answer:

False
Here, $f(x) = sin(3x+2)$, x R. This function is not a one-one onto function for all $x \in R.\\$
Again, $f(x) = sin(3x+2) = 0\\$
$or, 3x + 2 = n\pi, n \in \ Z$.
Hence, the function is not invertible.

Question:55

Every relation which is symmetric and transitive is also reflexive.

Answer:

False

Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}

Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.

Question:56

An integer m is said to be related to another integer n, if m is an integral multiple of n. This relation Z is reflexive, symmetric, and transitive.

Answer:

False.

Here, the given relation in the question is reflexive and transitive. However, it is not symmetric

Question:57

Let $A = \{ 0,1 \}$ and N be the set of natural numbers. Then the mapping $f: N \rightarrow A$ defined by $f(2n - 1) = 0, f(2n) = 1, n \in N, \\$is onto.

Answer:

True
Here, $A = \{ 0,1 \} , and\: \: f(2n - 1) = 0, f(2n) = 1, \forall A n \in N\\$
Therefore, the range of f is $\{ 0, 1 \} \\$
Hence, $f: N \rightarrow A$ mapping is onto.

Question:58

The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}

Answer:

False

As, $R = \{ (1,1), (1,2), (2,1), (3,3) \} \\$

Therefore, $(2,2) \notin R\\$

Hence, R is not reflexive.

Question:59

The composition function is cumulative.
Answer:

Question:60

The composition of function is associative.

Answer:

True.

fo(goh)(x) = (fog)oh

Question:61

Every function is invertible.

Answer:

False.

Only bijective functions are invertible.

Question:62

A binary operation on a set always has an identity element.

Answer:

False

‘+’ is a binary operation on the set N but it has no identity element.