= $\frac{1}{{\frac{1}{x}}} = x$

So, f(x) = $\frac{1}{x}$

So, ff(2) = 2

b.

Here, ff(x) = x [ from "A"]

so, ff(3) = 3

c.

Here, From "A".

ff(x) = x

so, ff(-3) = -3

d.

Here, from "A",

ff(x) = x

4.

Here, given;

f(x) = x^3­­ and g(x) = x²

Here, fg(x) = f[g(x)]

= f(x²) [g(x) = x²]

= (x²)³ [f(x) = x³]

=x⁶

Again, gf(x) = g[f(x)]

= g(x³) [f(x) = x³]

= (x³)² [g(x) = x²]

=x⁶

5.

a.

Here, given function is, f = {(1,1), (2,2), (3,3)}

Now, interchanging the position of domain and range of the function to get a inverse function f^{-1< /SUP>. Now, since f is a one-one union.}

(f^-1) = {(1,1), (2,2), (3,3)}

b.

Here, given function is, f = {(x,y):y = x, x, y θ R}

Now, interchanging the position of domain and range of the function to get a inverse function f^{-1< /SUP>. Now, since f is a one-one union.}

(f^-1) = {(y,x): x = y, x, y θ R}

c.

Here, given function is, f = {(x,2x + 1): x θ R}

Here, let us suppose, y = 2x + 1

Now, interchanging the position of x and y.

x = 2y + 1

or, x - 1 =2y [ y =$\: \frac{{x - 1}}{2}]$

now, inverse of f is;

(f^-1) = {(x,$\: \frac{{x - 1}}{2}$): x θ R}

d.

Here, given function is, f = {(x,y): y =$\: \frac{1}{x}\: ,\: $x θ R, xθ0}

Here, y =$\: \frac{1}{x}$

Now, interchanging the position of x and y.

So, x = $\frac{1}{y}$ and y = $\frac{1}{x}$

Now, the inverse of function f is:

f^-1 = {(y,x): x =$\: \frac{1}{y}\: ,\: $x θ R, yθ0}

6.

a.Here the given function is:

f:x θ 4x, or f(x) = 4x

Let us suppose, y = 4x

Now, interchanging the positions of x and y.

x= 4y and y = $\frac{x}{4}$

Now, inverse of the function f^-1(x) = $\frac{x}{4}$

b.

Here, the given function is:

g:x 3x + 4, or g(x) = 3x + 4

Let us suppose, y = 3x + 4

Now, interchanging the positions of x and y.

x= 3y + 4

or, x - 4 = 3y

so, y = $\frac{{x - 4}}{3}$

Now, inverse of the function g^-1: x = $\frac{{x - 4}}{3}$

So, g^-1(x) = $\frac{{x - 4}}{3}$

c.

Here, the given function is:

h:x-> $\frac{{x\: + \: 3}}{2}$, or h(x) = $\frac{{x\: + \: 3}}{2}$

Let us suppose, y = $\frac{{x\: + \: 3}}{2}$

Now, interchanging the positions of x and y.

x = $\frac{{y\: + \: 3}}{2}$

or, 2x = y + 3

so, y = 2x - 3

Now, inverse of the function h^-1: x = 2x - 3

So, h^-1(x) = 2x - 3

d.

Here, the given function is:

k:x-> $\frac{{2x\: + \: 5}}{{x + 2}}$, or k(x) = $\frac{{2x\: + \: 5}}{{x + 2}}$

Let us suppose, y =$\: \frac{{2x\: + \: 5}}{{x + 2}}$

Now, interchanging the positions of x and y.

x = $\frac{{2y\: + \: 5}}{{y + 2}}$

or, xy + 2x = 2y + 5

or, y(x - 2) = 5 - 2x

so, y = $\frac{{5 - 2x\: }}{{x - 2}}$

Now, inverse of the function h^-1: x = 2x - 3

So, k^-1(x) = $\frac{{5 - 2x\: }}{{x - 2}}$; xθ 2

7.

a.

Here, according to question 6(A) ,

f(x) = 4x and f^-1(x) = $\frac{x}{4}$

so, ff^-1(x) = f[f^-1(x)] = f ($\frac{x}{4}$) = 4$\left({\frac{x}{4}} \right)$

b.

Here, according to question 6(A) ,

g(x) = 3x + 4 and h^-1(x) = $2x - 3$

so,gh^-1(x) = g[h^-1(x)] = g(2x - 3)

= 3(2x - 3) + 4

= 6x - 9 + 4

= 6 x- 5

8.

Here, given;

The inverse of f(x) = x + 2 is f^-1 (x) = x - 2

a.

Here,f(-2)

Now, let us put x = -2 in f(x) = x + 2

So, f(-2) = -2 + 2 = 0

b.

Here, f^-1(-2)

Now, let us put x = -2 in f(x) = x + 2

So, f^-1(-2) = -2 - 2 = -4

c.

Here,

Now, let us put x = 4 in f(x) = x + 2

So, f (4) = 4 + 2 = 6

d.

Here,

Now, let us put x = -4 in f^-1(x) = x - 2

So, f (4) = -4 - 2 = -6

e.

Here, f(x -2)

Now, let us put x = -4 in f (x + 2) = (x - 2) + 2 = x - 2 + 2 = x

So, f (x-2) = x

f.

Here, f^-1(x +2)

Now, let us put x = -4 in f^-1 (x - 2) = (x + 2) - 2 = x

So, f^-1 (x-2) = x

9.

Here,

1. y = x + 1 , x θ N

2. y = x + 1, x θ Z and

3. y = x + 1, x θ R

Here, every function is in f(x) = mx + c.

so,

a. The given functions are linear functions.

b. Here given; (i) y - 5, x θ N, (ii) y = 5, x θ Z (iii) y = 5, x θ R.

Since y = fx = c, the functions are constant functions.

c.

Here given,

(i) y = x, x θ N

(ii) y = x, x θ Z

(iii) y = x, x θ R

Here, every function is in the form of y = f(x) = x

So, all the functions are 1 to 1 functions.

10.

To draw the graph of y = sinθ and y = sin3θ, θ is divided from 0° to 180° in the graph. And their respective trigonometric values are also added

a.

In the equation y = sin2θ

Now, plotting y = sin2θ in graph. We get,

b.

y = sin3θ

To draw the graph of y = sinθ and y = sin3θ, θ is divided from 0° to 180° in the graph. And their respective trigonometric values are also added

In the equation y = sin3θ,

Now, plotting y = sin2θ in graph. We get,