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.