The relation R is defined by xRy if x + 2y = 1 for x, y \(\in \)  N.
Reflexivity : Let x, y \(\in \)  N
xRx \(\Rightarrow\)  x + 2x = 1 \(\Rightarrow\) 3x = 1 \(\Rightarrow\) x = \(\frac { 1 }{ 3 } \notin N\)
\(\therefore\)  R is reflexive.
Symmetricity: xRy \(\Rightarrow\) yRx for x, y \(\in \)  N
xRy \(\Rightarrow\)  x + 2y = 1 which is not possible for any values of x, Y \(\in \) N
\(\therefore\) R is not symmetric
Transitivity: xRy and yRz \(\Rightarrow\) xRz.
xRy and yRz are not possible for any values of x, y, z \(\in \) N
\(\therefore\)  R is not transitive.
\(\therefore\) R is neither reflexive, nor symmetric and not transitive.

Given relation is aRb if a is not a sister of b.
Let a, b, C \(\in \)  A.
Reflexivity : aRa \(\Rightarrow\)  a is not a sister of a
\(\therefore\) R is reflexive.
Symmetricity: aRb \(\Rightarrow\)  bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a.
\(\therefore\) R is symmetric.
Transitivity : aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b, b is not a sister of C [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is a sister of C.
\(\therefore\) R is not transitive.
\(\therefore\) R is reflexive, symmetric and but not transitive.

Given relation is "aRb if a is not a sister of b". and a, b, c \(\in \) A.
Reflexivity: aRa \(\Rightarrow\) a is not a sister of a
\(\therefore\)  R is reflexive.         
Symmetric : aRb \(\Rightarrow\) bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a
\(\therefore\) R is not symmetric.
Transitivity: aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b and b is not a sister of C. [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is not a sister of C.
\(\therefore\)  R not is transitive.

Let N = {x \(\in \) N: x is a rational number}
\(\Rightarrow N=\left\{ \frac { 1 }{ 1 } ,\frac { 2 }{ 1 } ,\frac { 3 }{ 1 } ,\frac { 4 }{ 1 } ,....\infty \right\} \)
\(\Rightarrow\)  N is an infinite set.  

Let D = {x \(\in \) R : x is a rational number}
\(\Rightarrow\) D = {set of all rational number}
\(\Rightarrow\) D is an infinite set.

Let C = {x \(\in \) Z : x is even and < 10}
\(\Rightarrow\) C = {2, 4, 6, 8} 
\(\Rightarrow\) C is a finite set.

Let B = {x\(\in \)N : x is an odd prime number}
\(\Rightarrow\) B = {3, 5, 7, 11, .......}
\(\Rightarrow\) B is an infinite set.

Let A = { x \(\in \) N: x is an even prime number}
\(\Rightarrow\) A = {2} 
\(\Rightarrow\) A is a finite set.

{x \(\in \)​ ​​​​​​N : 4x + 9 < 52}
Let C = {x\(\in \)N:4x + 9 < 52}
\(\Rightarrow\) C = {x\(\in \)N:4x < 52 - 9}
\(\Rightarrow\) C = {x\(\in \)N: 4x < 43}
\(\Rightarrow\) C = \(\left\{ x\in N:x<\frac { 43 }{ 4 } \right\} \) \(\Rightarrow\) C = {x \(\in \) N : x < 10.75}
\(\Rightarrow\) C = {1,2,3,4,5,6,7,8,9,10}.

LetA = (a, b),B = (a,b,c)
A \(\times\)B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}
B \(\times\)A = {(a, a),(a, b), (b, a), (b, b), (c, a),(c, b)}
(A \(\times\)B) ⋂ (B \(\times\)A) = {(a, a), (a,b), (b, a): (b, b)}
= A \(\times\)A

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation

\(\text { If } x \in(-\infty,-2), \text { Let } x=-3\)
\(\text { then } f(x)=|-5|+|1|=6=-2 x\)
\(\text { If } x \in(-2,-2), \text { Let } x=0 \text { then }\)
\(f(x)=|0-2|+|0+2|=4\)
\(\text { If } x \in(2, \infty), \text { Let } x=4\)
\(\text { then } f^{}(x)=|2|+|6|=8=2 x\)
\(\therefore \mathrm{f}(x)=\left\{\begin{array}{rl} -2 x & x \in(-\infty,-2] \\ 4 & x \in(-2,2] \\ 2 x & x \in(2, \infty) \end{array}\right.\)

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)

X = {a, b, c, d} 
R = {(a, a), (b, b), (a, c)} 
(i) To make R reflexive we need to include (c, c) and (d, d) 
(ii) To make R symmetric we need to include (c, a) 
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c) 
To make R symmetric we need to include (c, c) and (c, a) for transitive 
∴ The relation now becomes 
R = {(a, a), (b, b), (a, c), (c, c), (c, a)} 
∴ R is equivalence relation.

(i) \(y=-x^{ \left( \frac { 1 }{ 3 } \right) }\)

\(Let \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\)
\(Then \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\) is the reflection of the graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)  about the x-axis.
(ii) \(y=x^{1\over 3}+1\)

Let \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)
Then \(y=x^{ ^{ \frac { 1 }{ 3 } } }+1\) is the x graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the upward for one unit
(iii) \(y=x^{1\over 3}-1\)

Let \(y=x^{1\over 3}\)
Then \(y=x^{1\over 3}\)-1 is the graph of \(x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the downward for one unit.
(iv)  \(y=(x+1)^{1\over 3}\)

\(y=(x+1)^{1\over 3}\) causes the graph of \({x}^{\frac{1}{3}}\), shifts to the left for one unit.

Step 1 :
Draw the Graph of y = x²

Step 2 :
The graph of y =(x -1)², shifts to the right for one unit.
Step 3 :
The graph of y = 3 (x - 1)², compresses towards the Y-axis that is moves away from the X-axis since the multiplying factor is 3 which is greater than 1.
Step 4:
The graph of y = 3 (x - 1)²+ 5, causes the shift to the upward for 5 units.