y = ax + b; x = forehand length; y = height

X	Y
35	56
45	65
50	69.5
55	74

For all the x-values, there is an image which is 'y
Moreover, the difference between two consecutive 'y' values is constant
In y = ax + b,
(i) The Relation 
R = { (35, 56), (45, 65), (50, 69.5), (55, 74) } is a function
(ii) In y = ax + b
when x = 35,y = 56
56 = 35a + b                 ..............(1)
when x = 45,y = 65
65 = 45a + b                ..............(2)
Solving (1) and (2), we get a = 0.90 and b = 24.5
(iii) Given, forehand length is 40 cm
i.e., when x = 40,y = ax + b
So, y = (0.90) (40) + 24.5 = 60.5
Height of person is 60.5 inches.
(iv) Given height is 53.3 inches
i.e. when y = 53.3, x = ?
53.3 = 0.9x + 24.5
53.3 - 24.5 = 0.9x
x = \(\frac{28.8}{0.9}\)
x = 32 cm
Length of fore hand is 32 cm.

f: A ⟶ B
f(x) =\(\frac { x }{ 2 } \)-1,
f(2) = \(\frac {2 }{ 2 } -1 = 0\)
f(2) = \(\frac {4 }{ 2 } -1 = 1\)
f(2) = \(\frac {6 }{ 2 } -1 = 2\)
f(2) = \(\frac {8 }{ 2 } -1 = 4\)
f(2) = \(\frac {12 }{ 2 } -1 = 5\)
A = {2,4,6,10,12} B = {0,1,2,4,5,9}
(i) Set of ordered pairs
= {(2,0),(4,1),(6,2),(10,4),(12,5)}
(ii) Table
x 2 4 6 10 12 f(x) 0 1 2 4 5 (iii) Arrow diagram

(iv) Graph
 

Given t (C) = F where \(F=\frac{9 C}{5}+32\)
C - Celsius, F - Fahrenheit
\(\therefore t(C)=\frac{9 C}{5}+32 \)
(i) \(t(0) =\frac{9(0)}{5}+32=0+32=32^{\circ} \mathrm{F} \)
(ii) \(t(28) =\frac{9(28)}{5}+32=\frac{252}{5}+32 \)
= 50.4 + 32 = 82.4^oF
(iii) \(t(-10)=\frac{9(-10)}{5}+32\) = -18 + 32 - 14^oF
(iv) Given t (C) = 212
\(\therefore \frac{9 C}{5}+32 =212 \Rightarrow \frac{9 C}{5}=212-32 \)
\(C =180 \times \frac{5}{9}=100^{\circ} C \)
(v) The temperature when the Celsius value is equal to the Fahrenheit value.
F = C
\(\frac{9 C}{5}+32=C \)
\(\frac{9 C}{5}-C=-32 \Rightarrow \frac{9 C-5 C}{5}=-32 \)
\(4 C=-32 \times 5 \Rightarrow C=-\frac{160}{4} \)
^oC = -40

gff(x) = g[f{f(x)}] (This means "g of f of f of x")
= g[f(3x + 1)] = g[3(x + 1) + 1] = g(9x + 4)
g(9x + 4) = [(9x + 4) + 3] = 9x + 7
fgg(x) = f[g{g(x)}] (This means " f of g of g of x")
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

f(x) = x⁵ , g(x) = x⁴
f(x) = x⁵
For any value of  'x', f (x) gives us a different value (image) in co domain.
f (x) is one - one function
g(1) = 1; g(- 1) = 1
Hence g(x) is not one-one function
f o g = f[g(x)] = f(x⁴) = (x⁴)⁵ = x²⁰
(fog) (1) = 1 and (f o g) (-1) = 1
f o g is not one - one function.

f(x) = \(\sqrt { 1+\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } } \)
\( f(x)=\sqrt{1-t} \)
\(where\ t=\sqrt{1-\sqrt{1-x^{2}}}\)
\(1-t \geq 0 \)
\(t \leq 1 \)
\(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
Squaring \(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
\(-\sqrt{1-x^{2}} \leq 0 \)
\(\sqrt{1-x^{2}} \geq 0 \)
\(1-x^{2} \geq 0 \)
\(x^{2} \leq 1 \)
= x [-1, 1] i.e., {- 1, 0, 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 x (B ∩ C) = (A x B) ∩ (A x C)
A = {0,1} , B = {2,3,4} , C = {3,5} , B∩C = {3}
\(A\cap (B\cap C)=\{ 0,1\} \times \{ 3\} \)
= {(0,3),(1, 3)}        ...(1)
A x B = {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
A x C = {(0,3),(0,5),(1,3),(1,5)}
\((A\times B)\cap (A\times C)\) = {(0,3),(1,3)}              ...(2)
From (1) and (2), it is clear that
A x (B ∩ C) = (A x B) ∩ (A x C)
Hence verified

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.

f(x) = 2x + 3, g(x) = 1 - 2x, h(x) = 3x
Now, (f o g)(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 5 - 4x
Then, (f o g) o h(x) = (f o g)(3x) = 5 - 4(3x) = 5 - 12x..(1)
(g o h)(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x
So, f o (g o h)(x) = f(1 - 6x) = 2(1 - 6x) + 3 = 5 - 12x...(2)
From (1) and (2), we get (f o g) oh = f o (g o h)