Question

If \(f\) is a relation from the set of positive real numbers to the set of positive real numbers defined by \[ f(x)=3x^2-2, \] then \(f\) is

• A. one-one but not onto
• B. onto but not one-one
• C. a bijection
• D. not a function

Hint:

When checking whether a rule defines a function from a set \(A\) to a set \(B\), always verify that the image of every element of \(A\) lies inside \(B\). If even one image lies outside the codomain, the given rule is not a function from \(A\) to \(B\).

Answer 1

The Correct Option is D

Step 1: Recall the definition of a function.
A function from a set \(A\) to a set \(B\) must assign every element of \(A\) to exactly one element of \(B\). Further, for a function \[ f:A\to B, \] the image of every element of \(A\) must belong to \(B\).
Here, \[ A=B=\{\text{positive real numbers}\}. \]

Step 2: Check whether the image always belongs to the codomain.
Given, \[ f(x)=3x^2-2. \] Take \[ x=\frac{1}{2}, \] which is a positive real number. Then \[ f\left(\frac{1}{2}\right) =3\left(\frac{1}{2}\right)^2-2 =\frac{3}{4}-2 =-\frac{5}{4}. \]

Step 3: Verify the codomain condition.
Since \[ -\frac{5}{4}\lt 0, \] the value \[ f\left(\frac{1}{2}\right) \] is not a positive real number. Thus an element of the domain is mapped outside the codomain.

Step 4: Conclude about the relation.
Because the codomain is the set of positive real numbers, every image must be positive. However, \[ f\left(\frac{1}{2}\right)=-\frac{5}{4} \] is not positive. Therefore, the given relation does not define a function from positive real numbers to positive real numbers.

Step 5: Final conclusion.
Hence, \[ \boxed{\text{\(f\) is not a function}} \] and the correct option is \[ \boxed{(4)}. \]