Question

please solve this for me
1st puc mathematics pcmb
silly, wrong, unnecessary answers will be reported
NO SPAM​

Answer 1

Answer:

1.f:R→R is defined as f(x)=x

4

Let x,y∈R such that f(x)=f(y)

⇒x

4

=y

4

⇒x=±y

f(x

1

)=f(x

2

)



=x

1

=x

2

f(−1)=f(1)=1

F is not me -me

Consider an element 2 in codomain R

Then it is clear A any x∈R

Such that f(x)=2

So f is not onto

So f is neither one-one nor onto.

2.f

−1

(x)=

5

x−2

Given, f(x)=5x+2

Let f(x) be y

⇒f(x)=y=5x+2

⇒f

−1

(y)=x=

5

y−2

⇒f

−1

(x)=

5

x−2

I have solve only two sums