Question

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f ( x ) = 3 − 4 x (ii) f : R → R defined by f ( x ) = 1 + x 2

Answer 1

(i)

The given function f:R→R is defined by f(x)=3−4x.

Assume x 1 ∈R and x 2 ∈R such that f( x 1 )=f( x 2 ).

⇒3−4 x 1 =3−4 x 2 ⇒−4 x 1 =−4 x 2 ⇒ x 1 = x 2

Therefore, f is one-one.

For y∈R, there exists 3−y 4 in R such that,

f( 3−y 4 )=3−4( 3−y 4 ) =y

Therefore, f is onto.

Function f is both one-one and onto. So, fis bijective.

(ii)

The given function f:R→R is defined by f(x)=1+ x 2 .

Assume x 1 and x 2 ∈R such that f( x 1 )=f( x 2 ).

⇒1+ x 1 2 =1+ x 2 2 ⇒ x 1 2 = x 2 2 ⇒ x 1 =± x 2

Therefore, f( x 1 )=f( x 2 ) does not imply x 1 = x 2 . So, f is not one-one.

Assume any element −2 in co-domain of R. The function f(x)=1+ x 2 is positive for all x∈R.

Therefore, there does not exist any x in R such that f( x )=−2. So f is not onto.

Hence, f is neither one-one nor onto.