\(\mathrm{A}-\mathrm{B} =\{1,2 ; 4\}-\{2,4,5\} =\{1\} \)
\(\mathrm{B}-\mathrm{C} =\{2,4,5\}-\{2,5\} =\{4\} \)
\(\therefore \ (\mathrm{A}-\mathrm{B}) \times(\mathrm{B}-\mathrm{C}) =\{1\} \times\{4\} =\{(1,4)\} \)

\(A \times B=\{1,2,3\} \times\{3,4\}\)
\(=\{(1,3) ;(1,4),(2,3),(2,4),(3,3),(3,4)\}\)
\(\mathrm{A} \times \mathrm{C}=\{1,2,3\} \times\{4,5,6\}\)
\(=\{(1,4),(1,5),(1,6),(2,4),(2,5)(2,6),(3,4),(3,5),(3,6)\}\)
\(\therefore(A \times B) \cup(A \times C)=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\}\)

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.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

A = {5, 6, 8, to}, B = {19, 15,9, 11}
f(x) = 2x - 1
f(5) = 2(5) - 1 = 9
f(8) = 2(5)-1 = 15
\(\therefore\) a = 9; b = 15

f(2) - f(4)
f(2) = 2x - 1
= 2(2) - 1 = 3
f(4) = 3x² - 10
= 3(42) - 10 = 38
\(\therefore\) f(2) - f(4) = 3 - 38 = 35

f(3) =. 2x - 1
= 2(3) - 1 = 6 - 1 = 5

F(5) = 3x² - 10
= 3(5)²- 10 = 75 - 10 = 65

A = {1,2,3,4,5}
B = {1,2,3,4, ... }
f: A \(\rightarrow\)B, f(x) = x²
\(\therefore\)  f(1) = 1² = 1
f(2)  = 2²= 4
f(3) = 3² = 9
f(4)  = 4²= 16
f(5) = 5²= 25
\(\therefore\) Range of f = {1, 4, 9, 16, 25}
Elements in A have been connected with different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

f(x) = 1 + x
g(x) = (2x - 1)
fog(x) = f(g(x) =  f(2x - 1)
= 1 + 2x - 1 = 2x ...(1)
gof(x) = g(f(x) = g(1 + x) = 2(1 + x) - 1
= 2 + 2x - 1
= 2x + 1      ...(2)
(1) \(\neq \)(2)
\(\therefore\) fog(x) \(\neq \) gof(x)
It is verified

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
For f(1) & f(3)
Since 1 & 3 lies between -5 ≤ x ≤2, We take, f(x) = x +5
f(1) = 1 + 5 = 6
f(-3) = -3 +5 = 2
For f(4): Since 4 lies between 2 < x < 6 We take, f(x) = x -1
f(4) = 4 -1 = 3
For f(-6): Since -6 lies between -7≤ x < -5 ,We take, f(x) = x² + 2x +1
f(-6) = (-6)² + 2(-6) + 1 = 36 -12 + 1 = 24 +1 = 25
Now ,\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) =\(\frac{ 4(2) +2(3) }{ 25 - 3(6)}\)
= \(\frac{ 8 + 6 }{ 25 -18}\)
= \(\frac{14 }{ 7}\) = 2
Hence, the value of \(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) = 2

f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 x 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

The distance travelled by the particle in time t hours is given by S(t)=\(\frac { { t }^{ 2 }+t }{ 2 } \).
(i) t = 3.5 hours. Therefore, S(3.5)=\(\frac { (3.5)^{ 2 }+3.5 }{ 2 } =\frac { 15.75 }{ 2 } \)=7.875
The distance travelled in 3.5 hours is 7.875 kms.
(ii) t = 8.25 hours. Therefore, S(8.25)=\(\frac { (8.25)^{ 2 }+8.25 }{ 2 } =\frac { 76.3125 }{ 2 } \)=38.15625
The distance travelled in 8.25 hours is 38.16 kms, approximately.