Show that the function fx 3x 5 is onetoone Use trigonometr

Show that the function f(x) = 3x - 5 is one-to-one Use trigonometric identities to transform one of the equation into the other (0 lessthanorequalto theta

Solution

10)

All linear equations are one-one/injections.

Method I -
f: R-->R
In this case, you can substitue first few elements of R (Real numbers, domain) in the equation & you will observe that the image of f(1), f(2) etc is a real number. Hence, it exists in the range.

Since you haven\'t been given the domain and the range of the function, you can prove it by method II.

Method II -
x1 =/ x2 (x1 & x2 € R)
3.x1 =/ 3.x2
3.x1-5 =/ 3.x2 -5
Therefore, f(x1) =/ f(x2)

where
x1 is x subscript 1.
=/ stands for \'not equal to\'.

It needs to be a bijective function, for its inverse to exist.
For it to be a bijective function, it needs to be both onto & one-one.

Onto function :
For every y€R (co-domain of f), there exists an x €R(domain of f) such that
f(x) =3x-5
Let f(x) = y
Therefore, y= 3x-5

y=3x-5
(y+5)/3 = x

Substitute x value in f(x)

f(x) = 3x-5
= 3[(y+5)/3] - 5
= (y+5) -5
= y

f(x) = y
Therefore, f is one to one

11)sin(120)=cos(90-120)

=cos(-30)

=cos(30)

=sqrt(3)/2

b)cos300=sin(90-300)

=sin(-210)

=-sin(210)

=-sin(180+30)

=-(sin180*cos30+cos180*sin30) (sin(x+y)=sinxcosy+cosxsiny)

=-(0*sqrt(3/2)+(-1*1/2))

=-(-1/2)

cos300=1/2