\(\cos \frac{\pi}{3}=\frac{1}{2} \text {, also } \cos \frac{5 \pi}{3}=\frac{1}{2} \text {, not one-one }\)

Given) f : R →→ R defined by f(x) = [x]
For one-one: We know by definition that for
a ≤ x i.e. for x₁,x²∈ (a, a+1) ,
x₁ ≠ x₂ ⇒ f(x₁) = f(x₂) = a
Hence, not one-one.
For onto: For y(non integer) in co-domain there does not exist x ∈ R in domain such that f(x) = y. Hence, not onto.

Given \(f(x)=x^3+1\)
For \(x_1 \neq x_2\)
\(\Rightarrow x_1^3 \neq x_2^3 \Rightarrow x_1^3+1 \neq x_2^3+1\)
\(\Rightarrow f\left(x_1\right) \neq f\left(x_2\right)\), hence, one-one

(i)  Let \(x_{ 1 },x_{ 2 }\in R\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow \frac { 4x_{ 1 }-3 }{ 5 } =\frac { 4x_{ 2 }-3 }{ 5 } \)
\(\Rightarrow \) \(4x_{ 1 }-3=4x_{ 2 }-3\Rightarrow 4x_{ 1 }=4x_{ 2 }\)
\(\Rightarrow \)  \(x_{ 1 }=x_{ 2 }\Rightarrow \quad 'f'\) one-one
Let \(y\in R.\) Let \(y=f(x_{ 0 })\).
Then \(\frac { 4x_{ 0 }-3 }{ 5 } =y\Rightarrow x_{ 0 }=\frac { 5y+3 }{ 4 } \Rightarrow x_{ 0 }\in R\)
\(f(x_{ 0 })=\frac { 4x_{ 0 }-3 }{ 5 } =\frac { 4 }{ 5 } \left( \frac { 5y+3 }{ 4 } \right) -\frac { 3 }{ 5 } \)
\(=y+\frac { 3 }{ 5 } -\frac { 3 }{ 5 } =y\)
\(\therefore \) For each \(y\in R\), there exists \(x_{ 0 }y\in R\) such that ​​​​​​
\(f(x_{ 0 })=y\)
\(\therefore \) 'f' is one-one and onto function. 

The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one, say (2, 3) to R₁, then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1).
Thus, the only equivalence relation bigger than R₁ is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two.

Partition the set N into pairwise disjoint subsets. The equivalent classes are as given by:
\(A_{ 0 }=\{ 5,10,15,20,............\} \)
\(A_{ 1 }=\{ 1,16,11,16,21,.......\} \)
\(A_{ 2 }=\{ 2,7,12,17,22,.........\} \)
\(A_{ 3 }=\{ 3,8,13,18,23,.........\} \)
\(A_{ 4 }=\{ 4,9,14,19,24,.........\} \)
Clearly the above five sets are pairwise disjoint and \(A_{ 0 }\cup A_{ 1 }\cup A_{ 2 }\cup A_{ 3 }\cup A_{ 4 }=\overset { 4 }{ \underset { i=0 }{ \cup } } A_{ i }=N\)

\([0]={ \pm 2,\pm 4,\pm 6,\quad .......\} . }\)

Reflexive, true as x s x ⇒ x - x + \(\sqrt3\)
= \(\sqrt3\) is an irrational number
Symmetric, fase e.g x =  \(\sqrt3\), y= 2
x S y ⇒ \(\sqrt3\) - 2 + \(\sqrt3\) = 2 \(\sqrt3\) -2 is an irrational number
but ySx ⇒ 2 - \(\sqrt3\)  + \(\sqrt3\) = 2 \(\sqrt3\) -2 is not irrational number
transitive, false e.g = x = 1 + \(\sqrt3\), y = 5 z = 2 \(\sqrt3\)
x S y ⇒ 1 + \(\sqrt3\) - 5 + \(\sqrt3\) = 2 \(\sqrt3\) - 4  is an irrational number
y S z ⇒ 5 -2 \(\sqrt3\)  + \(\sqrt3\) = 5 - \(\sqrt3\)  is an irrational number
But x S z ⇒ 1 + \(\sqrt3\)  - 2 \(\sqrt3\) + \(\sqrt3\) = 1  is an irrational number

Here (1,2) e R, (2,1) € R, if transitive (1,1) should belong to R.

T₁ and T₃ are similar as their sides are proportional.

The given relation is R = {(x, y):x, y ∈ Z and x-y is divisible by 3}.
To prove R is an equivalence relation, we have to prove R is reflexive, symmetric and transitive.
Reflexive : As, for any x ∈ Z, we have x - x = 0, which is divisible by 3.
⇒ (x - x) is divisible by 3
⇒ (x,x)∈ R, ∀ x ∈ Z
Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R, where x, y ∈ Z.
⇒ (x - y) is divisible by 3. [by definition of R]
⇒ x - y=3 A for some A ∈ Z.
⇒ y - x = 3(-A)
⇒ (y - x) is also divisible by 3. 
⇒ (y, x)∈R
Therefore, R is symmetric.
Transitive Let (x, y) ∈ R, where x, y ∈ Z.
⇒ (x - y) is divisible by 3.
⇒ x - y = 3A for some A∈Z
Again, let (y, z) ∈ R, where y, z∈ Z.
⇒ (y - z) is divisible by 3. 
⇒ y - z = 3B for some B ∈ Z. 
Now, (x - y)+(y - z) = 3A + 3B 
⇒ x - 2 = 3(A + B) 
⇒ (x - z) is divisible by 3 for some
(A + B)∈ Z ⇒ (x,z) ∈ R
Therefore, R is transitive.
Thus, R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

Given, function \(f: \mid-4,4] \rightarrow[0,4]\)
Defined by \(f(x)=\sqrt{16-x^2}\)
For one-one Let \(x_1, x_2 \in[-4,4]\)
Such that \(f\left(x_1\right)=f\left(x_2\right)\)
\(\Rightarrow  \sqrt{16-x_1^2}=\sqrt{16-x_2^2} \)
\(\Rightarrow  16-x_1^2=16-x_2^2\)
\(\Rightarrow  x_1^2=x_2^2 \Rightarrow\left(x_1^2-x_2^2\right)=0\)
\(\Rightarrow f(x)\) is not one-one function.
For onto
Let \(y \in[0,4]\) (codomain) be any arbitrary element.
Then, y=f(x)
\(\Rightarrow  y=\sqrt{16-x^2} \Rightarrow y^2=16-x^2 \)
\(\Rightarrow  x^2=16-y^2 \Rightarrow x=\sqrt{16-y^2}\)
Thus, for each \(y \in[0,4]\), there exists
\(x=\sqrt{16-y^2} \in[-4,4]\)
Such that f(x)=y
So, f(x) is onto.
Also, given \(f(a)=\sqrt{7} \Rightarrow \sqrt{16-a^2}=\sqrt{7}\)
\(\Rightarrow \quad 16-a^2=7 \quad \Rightarrow a^2=9 \Rightarrow a= \pm 3\)
\( \Rightarrow \quad\left(x_1-x_2\right)\left(x_1+x_2\right)=0 \Rightarrow x_1-x_2=0 \)
\( \text { or } \quad x_1+x_2=0 \Rightarrow x_1=x_2 \text { or } x_1=-x_2\)

Given, \(f(x)=\frac{4 x}{3 x+4}\) and \(f: R-\left\{-\frac{4}{3}\right\} \rightarrow R\)
For one-one Let \(f\left(x_1\right)=f\left(x_2\right)\), for some \(x_1, x_2 \in R-\left\{-\frac{4}{3}\right\}\)
\(\Rightarrow \frac{4 x_1}{3 x_1+4}  =\frac{4 x_2}{3 x_2+4}\)
\(\Rightarrow  \left(x_1\right)\left(3 x_2+4\right)  =\left(x_2\right)\left(3 x_1+4\right)\)
\(\Rightarrow  3 x_1 x_2+4 x_1  =3 x_1 x_2+4 x_2\)
\(\Rightarrow  4 x_1 =4 x_2 \Rightarrow x_1=x_2\)
\(\Rightarrow f(x) \text { is one-one function. }\)
For onto Let \(y=\frac{4 x}{3 x+4}\)
\(\Rightarrow  3 x y+4 y=4 x \Rightarrow 4 x-3 x y=4 y\)
\(\Rightarrow  x(4-3 y)=4 y \Rightarrow x=\frac{4 y}{4-3 y}\)
Clearly, x will not define, if 4-3 y=0
\(\Rightarrow \quad y=\frac{4}{3}\)
Therefore Range of \(f(x)=R-\left\{\frac{4}{3}\right\}\)
Therefore Range of \(f(x) \neq\) Codomain of f(x) So, f(x) is not an onto function.

Relation R is defined by (a, b) R (c, d) \(\Leftrightarrow\) ad = be for all (a, b), (c, d) \(\in\) N x N.
For reflexive: (a, b) R (a, b) \(\Leftrightarrow\) ab = ba, which is true in N. Hence, reflexive.
For symmetric: (a, b) R (c, d) \(\Leftrightarrow\) ad = be
\(\Leftrightarrow\) cb = da \(\Leftrightarrow\) (c, d) R (a, b).
Hence, symmetric.
For transitive: Consider (a, b) R (c, d) and
(c, d) R (e,f) \(\Leftrightarrow\) ad = be and
cf = de \(\Leftrightarrow\) ad-cf = be-de \(\Leftrightarrow\) af = be
\(\Leftrightarrow\) (a, b) R (e,f). Hence, transitive.
Since relation R is reflexive, symmetric and transitive.
Hence, relation R is an equivalence relation.

It is clear from the graph that it is a parabolic curve. So, let the equation of the function is
\(y=a x^{2}+b x+c \)
Points on the curve are A(1, 2), B(2, 3) and C(-1, 6) On putting the points A(1, 2), B(2, 3) and C(-1, 6) in the
Equation (i) one-by-one, we get
At \(A(1,2), 2=a(1)^{2}+b(1)+c \Rightarrow 2=a+b+c \)
At \(B(2,3), 3=a(2)^{2}+2 b+c \Rightarrow 3=4 a+2 b+c \)
and at \(C(-1,6), 6=a(-1)^{2}+b(-1)+c \)
⇒  6 = a - b + c
On solving Equation (ii), (iii) and (iv), we get
\(a=1, b=-2 \text { and } c=3\)
On putting the values of a, b and c in Equation (i), we get
\(y=x^{2}-2 x+3\)
\(\therefore\) Required equation of function is \(f(x)=y=x^{2}-2 x+3\)
Now,
\(y=(x-1)^{2}+2 \)
\(\Rightarrow (x-1)^{2}=y-2 \)
\(\Rightarrow (x-1) =\pm \sqrt{y-2}\)
\(\Rightarrow x =1 \pm \sqrt{y-2}\)
Let y = 3, then \(x=1 \pm \sqrt{3-2} \)
\(\Rightarrow x=1 \pm 1 \Rightarrow x=2,0 \)
Here, we see that for two different values of domain (i.e. x = 0, 2 ). We get same image (i.e. y = 3). Hence, f(x) is many-one function.

We have, \( y=f(x)=\sqrt{4-x^{2}, 0 \leq x \leq 2,0 \leq y \leq 2} \)
For one-one Let \( x_{1}, x_{2}\) be any two elements of the interval 0 \(\leq \)x \(\leq \)2,
such that \(f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow \sqrt{4-x_{1}^{2}}=\sqrt{4-x_{2}^{2}} \Rightarrow 4-x_{1}^{2}=4-x_{2}^{2} \Rightarrow \)
\(x_{1}^{2}=x_{2}^{2} \Rightarrow x_{1}=\pm x_{2} \Rightarrow x_{1}=x_{2}\)
\(\left[x_{1} \neq-x_{2},\right. since x_{1} and x_{2} are non-negative ] \)
So, f(x) is one-one. For onto Let k \(\in\)[0, 2] be any arbitrary element and let
\(f(x)=k \Rightarrow \sqrt{4-x^{2}}=k \)
On squaring both sides, we get \(4-x^{2}=k^{2}\) 
[\(\because\) x is non-negative, so we take positive]
 Also, for \(0 \leq k \leq 2\), we have \(0 \leq \sqrt{4-k^{3}} \leq 2 \Rightarrow 0 \leq x \leq 2\), which is true.
Thus, for each k \( \in\)[0, 2], there exists \(x=\sqrt{4-k^{2}} \in[0,2] \) such that f(x) = k .
So, f(x) is onto. Hence, f is bijective function.

We have a function\(f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1<x<1\}\) , defined as
\(f(x)=\frac{x}{1+|x|}=\left\{\begin{array} \frac{x}{1+x}, \text { if } x \geq 0 \\ \frac{x}{1-x}, \text { if } x<0 \end{array}\right. \)
For one-one Let \(x_{1}, x_{2} e R\) . Then, the following cases arise
Case T When both are lesg than 0
Let \(x_{1}, x_{2} e \ R \ such \ that \ x_{1}<0, x_{2}<0\) and\( f\left(x_{1}\right)=f\left(x_{2}\right) \)
\(\Rightarrow \quad \frac{x_{1}}{1-x_{1}}=\frac{x_{2}}{1-x_{2}} \Rightarrow x_{1}-x_{1} x_{2}=x_{2}-x_{1} x_{2} \)
\(\Rightarrow \quad x_{1}=x_{3} \)
Case II
When both are greater than or equal to 0 Let \(x_{1}, x_{2} \in R\)
such that \( x_{1} \geq 0, x_{2} \geq 0 and f\left(x_{1}\right)=f\left(x_{2}\right) \)
\(\frac{x_{1}}{1+x_{1}}=\frac{x_{2}}{1+x_{2}} \Rightarrow x_{1}+x_{1} x_{2}=x_{2}+x_{1} x_{2}\) 
Case III When one is non-negative and other is negative Let \( x_{1} \geq 0\ and\ x_{2}<0\)
Now, if \(f\left(x_{1}\right)=f\left(x_{2}\right)\) , then\( \frac{x_{1}}{1+x_{1}}=\frac{x_{2}}{1-x_{3}} \)
\(\Rightarrow x_{1}-x_{\mathrm{t}} x_{2}=x_{2}+x_{1} x_{2} \)
\(\Rightarrow \quad x_{1}-x_{2}=2 x_{1} x_{2} \), which is not possible as \(\mathrm{LHS}>0\)
and RHS \(\leq 0 . Thus, x_{1} \neq x_{2} \Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right) \)
From cases L , II and \(\Pi\) , we get f is one-one.
For onto Let \(y \in(-1,1)\) be any arbitrary element. Then. the following cases arise
Case I When y \(\geq\) 0 , t.e. 0 \(\leq\) y<1
\(\left[\because f(x)=\frac{x}{1+x} \geq 0\right.\)]
Cónsider, \(y=f(x)=\frac{x}{1+x}\)
\(\Rightarrow y+y x=x \Rightarrow y=x-y x \)
(\(\Rightarrow \ x=\frac{y}{1-y} \geq 0\), for 0
Thus, for each y \(\in\)[0, 1) , there exists \(x=\frac{y}{1-y} \in R\) such that f(x) = y .
Case II When y < 0, he. -1
\(\left[\because f(x)=\frac{x}{1-x}<0\right] \)
Consider, \(y=f(x)=\frac{x}{1-x} \)
\(\Rightarrow y=y x=x \Rightarrow y=y x+x=x(y+1) \)
 \( (x=\frac{y}{y+1}<0, \text { for }-1\) 
Thus, for each y \(\in\)(-1, 0) , there exist \(x=\frac{y}{y+1} \in R\) such that f(x) = y
From cases I and II, we get f is onto.

\(\text { Given function } f(x)=\left\{\begin{array}{ll} x+1, & \text { if } x \text { is odd } \\ x-1, & \text { if } x \text { is ever } \end{array}\right.\)
For one-one:
\(\text {(i) Let } x_{1}, x_{2} \in N \text { and } x_{1}, x_{2}, \text { are both even. }\)
\(x_{1} \neq x_{2} \Rightarrow x_{1}-1 \neq x_{2}-1 \Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)\)
\(\text {(ii) Let } x_{1}, x_{2} \in N \text { and } x_{1}, x_{2}, \text { are both odd. }\)
\(x_{1} \neq x_{2} \Rightarrow x_{1}+1 \neq x_{2}+1 \Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)\)
\(\text {(iii) Let } x_{1}, x_{2} \in N \text { and } x_{1} \text { is even and } x_{2} \text { is odd. Then } x_{1} \neq x_{2}\)
\(\text {Also, } f\left(x_{1}\right)=x_{1}-1(\text { odd }) \text { and } f\left(x_{2}\right)=x_{2}+1(\text { even })\)
\(\Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)\)
In all the three cases \(x_{1} \neq x_{2}\)
\(\Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)\)
Hence, function is one-one.
For onto: Let y \(\in\)N (co-domain)
If y is even
\(\Rightarrow y=x+1 \Rightarrow x=y-1 \in N \text { (domain) }\)
f (y - 1) = y -1 + 1 = y
If y is odd
\(\Rightarrow y=x-1 \Rightarrow x=y+1 \in N \text { (domain) }\)
f (y + 1) = y +1 - 1 = y
\(\therefore\) For every y \(\in\)N (co-domain), there exists x \(\in\) N (domain) such that y = f(x). Hence, function is onto.
\(\therefore\) f is both one-one and onto.

We have a function f :N ➝ N, defmed as
[(1) = [(2) = 1 and [(x) = x - 1, for every x > 2
For one-one Since, [(1) = [(2) = 1, therefore 1 and 2
have same image, namely 1. So, f is not one-one.
For onto Note that y = 1 has two pre-images, namely i and 2. Now, let y ∈ N, y ≠ 1 be any arbitrary element.
Then, y = [(x) ⇒ y = x -1
⇒ x = y + 1 > 2 for every y ∈ N, y ≠ 1
Thus, for every y ∈ N, y ≠ 1, there exists x = y +1 such that
[(x) = [(y +1) = Y+1-1 = y
Hence, f is onto.

We .have a relation R on N x N defined by (a, b)R(c, d), if
ad(b + c) = bc(a + d) .
Reflexive Let (a, b) ∈ N x N be any arbitrary element.
We have to show that (a, b) R (a, b), i.e.
ab(b + a) = ba(a + b), which is always true, as natural numbers are commutative under usual multiplication and addition .•
Since, (a, b) ∈ N x N was arbitrary, so R is reflexive.
Symmetric Let (a, b), (c, d) ∈ N x N
such that (a, b) R (c, d), i.e. ad(b + c) = bc(a + d)
We have to show that (c, d) R (a, b),
i.e. cb(d + a) = da(c + b)
From Equation (i), we have ad( b + c) = bc( a + d)
⇒ da(c + b) = eb(d + a)
[natural numbers are commutative under
usual addition and multiplication] [1]
⇒ cb(d+a) = da(c+b)
⇒ (c, d)R(a, b)
So, R is symmetric. 
Transitive Let (a, b), (c, d) and (e, f) ∈N x N such that (a, b) R (c, d) and (c, d) R (e, f).
⇒ (a, b)R(c, d) ⇒ ad(b+c) = be(a+d)
\( \Rightarrow \frac{b+c}{b c}=\frac{a+d}{a d} \)
\(\Rightarrow \frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\frac{1}{a}\) 
and \((c, d) R(e, f) \Rightarrow c f(d+e)=d e(c+f)\) 
\(\Rightarrow \frac{d+e}{d e}=\frac{c+f}{c f} \)
\( \Rightarrow \frac{1}{e}+\frac{1}{d}=\frac{1}{f}+\frac{1}{c}\) 
Now, adding Equation (ii) and (iii), we get
\(\left(\frac{1}{c}+\frac{1}{b}\right)+\left(\frac{1}{e}+\frac{1}{d}\right)=\left(\frac{1}{d}+\frac{1}{a}\right)+\left(\frac{1}{f}+\frac{1}{c}\right) \) 
\(\Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\) 
\(\Rightarrow \frac{e+b}{b e}=\frac{f+a}{a f}\) 
\(\Rightarrow a f(e+b)=b e(f+a)\) 
\(\Rightarrow a f(b+e)=b e(a+f) \)
\(\Rightarrow (a, b) R(e, f)\) 
So, R is transitive.
Hence, R is an equivalence relation.