# Algebra

### Exercise: 1 'A'

1.

Given functions are:

f(x) = 3x - 4, h(x) = x + 3, g(x) = -2x + 1

a.

Here, fh = f(h(x)) = f(x + 3)

= 3 (x + 3) - 4

= 3x + 9 - 4 = 3x + 5

b.

Here, hh = h(h(x)) = h(x + 3)

= (x + 3) + 3 = x + 6

c.

Here, gh(f)(x) = gh(f(x))

= gh(3x - 4)

= g [h (3x - 4)]

= g [(3x - 4) + 3]

= g(3x - 1)

= -2(3x - 1) + 1

= -6x + 3

d.

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

= f[g(x + 3)]

= f[-2(x + 3) + 1]

= f(-2x - 6 + 1) = f(-2x - 5)

= 3(-2x - 5) - 4

= -6x - 15 - 4 = -6x - 19

2.

Here, from question no. 1 ,

f(x) = 3x - 4 , h(x) = x + 3 and g(x) = -2x + 1

a.

From 1"A",

fh = fh(x) = 3x + 5 [fh(x) = 3x + 5]

Here, putting x = 2,

fh(2) = 3.2 + 5 = 6 + 5 = 11

b.

From 1"B",

hh = x + 6[

so, hh(x) = x + 6

Here, putting x = 2,

hh(2) = 2 + 6 = 8

c.

From 1"D",

f(gh) = fh(x) = -6x - 19

f(gh)(x) = -6x - 19

Here, putting x = 3,

f(gh)(3) = -6.3 - 19 = -18 - 19 = -37

d.

f(hh) = fh(h) [h(x)= x + 3]

= fh(x + 3)

= fh{(x + 3)+ 3}

= f(x + 6) = 3(x + 6) - 4 [f(x) = 3x - 4]

= 3x + 18 - 4 = 3x + 14

So, f(hh)(x) = 3x + 14

Here, putting x = -5,

f(hh)(-5) = 3 - 5 + 14= -15 + 14 = -1

3.

Here, given functions are:

f(x) = $\frac{1}{x}$ ; x θ R , x θ 0.

a.

Here, ff(x) = f[f(x)] = f ($\frac{1}{x}$< /FONT>)

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

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) = x3­­ 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

 θ°=x 0° 15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165° 180° 2θ 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° Sin2θ=y 0 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0

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,

 θ°=x 0° 15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165° 180° 2θ 0° 45° 90° 135° 180° 225° 270° 315° 360° 405° 450° 495° 540° Sin2θ=y 0 0.71 1 0.71 0 -0.71 -1 -0.71 0 0.71 1 0.71 0

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

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