f(x) = 2x + 1,g(x) = x² - 2
f o g(x) =  f(g(x) = f(x² - 2) = 2(x²- 2) + 1 = 2x² - 3
g o f(x) = g(f(x) = g(2x + 1) = (2x + 1)²-2 = 4 = x² + 4x - 1
Thus f o g = 2x² - 3,g o f = 4x² + 4x-1 .From the above, we see that f o g ≠ g o f.

Given 
A = {1,2,3,4,5,6,7}
B = {2,3,5,7}
C = {2}
A x ( B - C) = (A x B) - (A x C)
B - C = {2,3,5,7} - {2}
=  {3,5,7}
A x (B-C) = {1,2,3,4,5,6,7} x {3.5.7}
={(1,3),(1,5),(1,7),(2,3),(2,5),(2,7) (3,3),(3,5),(3,7),(4,3),(4,5),(4,7) (5,7),(5,3),(5,5) (6,3),(6,5),(6,7),(7,3),(7,5),(7,7)}    ....(1)
A x B = {1,2,3,4,5,6,7} x {2,3,5,7}
= {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7) (3,2),(3,3),(3,5),(3,7),(4,2),(4,3),(4,5),(4,7) (5,2),(5,3),(5,5),(5,7),(6,2),(6,3),(6,5),(6,7) (7,2),(7,3),(7,5),(7,7)}
 A x C = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2)(6,2),(7,2)}
(A x B) - (A x C) = {(1,3),(1,5),(1,7),(2,3) (2,5),(2,7),(3,3),(3,5) (3,7),(4,3),(4,5),(4,7), (5,3),(5,5),(5,7),(6,3) (6,5),(6,7),(7,3),(7,5),(7,7)}                     ..(2)
From (1) and (2), it is clear that
A x ( B - C) = (A x B) - (A x C)
Hence verified

(A U B) x  C = (A x C) U (B x C)
A = {0,1} , B = {2,3,4} , C = {3,5}
\(A\cup B\) = {0,1,2,3,4}
\((A\cup B)\times C\) = {0,1,2,3,4} x {3,5}
= {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}             ...(1)
A x C = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
B x C = {2,3,4} x {3,5}
= {(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
\((A\times C)\cup (B\times C)\) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}                .....(2)
From (1) and (2), it is clear that
(A U B) x  C = (A x C) U (B x C)
Hence verified.

f(x) = 6x + 8 , \(g(x)=\frac { x-2 }{ 3 } \)
(i) \(g\left(\frac{1}{2}\right)=\frac{\frac{1}{2}-2}{3}=\frac{1-4}{2 \times 3}=-\frac{3}{6}=-\frac{1}{2} \)
\(\therefore \ g g\left(\frac{1}{2}\right)=g\left[g\left(\frac{1}{2}\right)\right] \)
\(=g\left(-\frac{1}{2}\right)=\frac{-\frac{1}{2}-2}{3}=-\frac{1-4}{2 \times 3}=-\frac{5}{6} \)
(ii) gf (x) = g [f (x)] = g [6x + 8]
\(=\frac{6 x+8-2}{3}=\frac{6 x+6}{3} \)
\(=\frac{6(x+1)}{3}=2(x+1) \)

\(f(x)=\frac { x-1 }{ x+1 } ,x\neq 0\)
\(f(f(x))=f\left( \frac { x-1 }{ x+1 } \right) =\frac { \left( \frac { x-1 }{ x+1 } \right) -1 }{ \left( \frac { x-1 }{ x+1 } \right) +1 } \)
\(=\frac{\frac{\not x-1-x-1}{(\not x+1)}}{\frac{\not x-1+x+1}{(\not x+1)}}=\frac{-2}{2 x}=\frac{-1}{x}\)
Hence it is proved.

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

The function f is defined by three values in intervals I, II, III as shown by the side.
For a given value of x = a, find out the interval at which the point a is located, there after find
f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x - 2; f(4) = 3(4) = 10
(ii) x = -2 lies in the second interval
Therefore, f(x) = x² - 2; f(-2) = (-2)² - 2 = 2
(iii) From (i), f(4) =10.
To find f(1) first we see that x = 1 lies in the second interval.
Therefore, f(x) = x²-2 ⇒ f(1) = 1² - 2 = -1
So, f(4) + 2f(1) = 10 + 2(-1) = 8
(iv) We know that f(1) = -1 and f(4) = 10
For finding f(-3), we see that x = −3, lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, \(\frac { f(1)-3f(4) }{ f(-3) } =\frac { -2-3(10) }{ 1 } \) = - 31

Give the function f: x ⟶ x² - 5x + 6.
i) f(-1) = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12
ii) f(2a) = (2a)² - 5(2a) + 6 = 4a² - 10a + 6
iii) f(2) = 2² - 5(2) + 6 = 4 - 10 + 6 = 0
iv) f (x - 1)² - 5(x - 1) + 6
= x²- 2x + 1 - 5x + 5 + 6
= x²-7x + 12

Given A = {x \(\in \) W| x < 2} A = {0,1}
B = {x \(\in \) N| 1 < x ≤ 4} B = {2,3,4}
C = {3,5}
A x (B U C) = (A x B) U (A x C)
\(B\cup C\) = {2,3,4,5}
A x (B U C) = {0,1} x {2,3,4,5}
= {{0,2},(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}          ...(1)
A x B = {0,1} x {2,3,4}
= {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
A x C = {0,1} x {3,5}
= {{0,3},(0,5),(1,3),(1,5)}
\((A\times B)\cup (A\cup C)\) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ...(2)
From (1) x (2),it is clear that
\(A\times (B\cup C)=(A\times B)\cup (A\times C)\)
Hence verified

Given: A = {1,2,3} , B = {2,3,5} , C = {3,4} ,D = {1,3,5}
\(A\cap C\) = {3}
\(B\cap D\) = {3,5}
\((A\cap C)\times(B\cap D)=\{ 3\} \times \{ 3,5\} \)
= {(3,3),(3,5)}          ..(1)
A x B = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}
C x D = {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
\((A\times B)\cap (C\times D)\)  = {(3,3),(3,5)}                ....(2)
From (1) and (2),it is clear that
(A ∩ C) x (B ∩ D) = (A x B) ∩ (C x D)
Hence it is true.