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

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.

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
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

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