Question

Let \( f: R_{+} \rightarrow[-5, \infty) \) be defined as \( f(x)=9 x^{2}+6 x-5 \), where \( R_{+} \)is the set of all non-negative real numbers. Then, \( f \) is : (A) one-one (B) onto (C) bijective (D) neither one-one nor onto

Answer 1

Correct option is (B) onto

f : R⁺ → [−5, ∞] given by f(x) = 9x² + 6x − 5

To show: f is one-one & onto

Let us assume that f is not one-one.

\(\therefore\) there exists 2 as more numbers for which images are same

For x₁, x₂ ∈ R + & x₁ ± x₂

Let f(x₁) = f(x₂)

Since x₁ & x₂ are positive

9(x₁ − x₂) + 6 > 0

\(\therefore\) x₁ − x₂ = 0 ⇒ x₁ = x₂

Therefore, it contradicts are assumption.

Hence the function is one-one

Now, let as prove that f is onto

A function f : X → Y is onto it for every y ∈ Y, these exists a pie image in X.

f(x) = 9x² + 6x − 5

= 9x² + 6x + 1  −6

= (3x + 1)² − 6

Now, for all x ∈ R + as [0, ∞), f(x) ∈ [−5 ,∞)

\(\therefore\) Range = co-domain

Hence f is onto

We have proved, that function is one-one & onto. Hence in turn it is proved that f is invertible.