Question

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

1. f :R→R defined by f(x) =3−4x
2. f : R→R defined by f(x) =1+x2

Answer 1

(i) f: R → R is defined as f(x)=3−4x.

Let x1, x2 ∈ R such that $f(x_1)=f(x_2)$.
$⇒ 3-4x_1=3-4x_2$
$⇒ -4x_1=-4x_2$
$⇒ x_1=x_2$
∴ f is one-one.

For any real number (y) in R , there exists $\frac{3-y}{4}$ in R such that
$f(\frac{3-y}{4})=3-4(\frac{3-y}{4})=y$.
∴f is onto.

Hence, f is bijective.

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(ii) f: R → R is defined as
$f(x)=1+x^2$.

Let $x_1, x_2 ∈ R$ such that $f(x_1)=f(x_2)$
$⇒ 1+x_1^2 = 1+x_2^2$
$⇒ x_1^2=x_2^2$
$⇒ x_1=±x_2$
∴$f(x_1)=f(x_2)$ does not imply that $x_1=x_2$.

For instance,
$f(1)=f(-1)=2$
∴ f is not one-one.

Consider an element −2 in co-domain R.
It is seen that $f(x)=1+x^2$ is positive for all $x ∈ R$.
Thus, there does not exist any $x$ in domain R such that $f(x) = −2$.
∴ f is not onto.

Hence, f is neither one-one nor onto.