Question

If f is a relation from 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
  % Telugu: ఏకైకము, కానీ సంగ్రస్తము కాదు
• B. onto but not one-one
  % Telugu: సంగ్రస్తము, కానీ ఏకైకము కాదు
• C. a bijection
  % Telugu: ద్విగుణ ప్రమేయము
• D. not a function % Telugu: ప్రమేయం కాదు

Hint:

For a mapping \(f: A \rightarrow B\) to be a function, two conditions must be met: 1. Every element in the domain A must have an image. 2. The image of every element in A must be an element of the codomain B.
Check if the range of \(f(x)\) for \(x\) in the domain is a subset of the specified codomain.

Answer 1

The Correct Option is D

The function is defined as \(f: R^+ \rightarrow R^+\), where \(R^+\) is the set of positive real numbers (\(x>0\)). The rule is \(f(x) = 3x^2-2\). For \(f\) to be a function from \(R^+\) to \(R^+\), for every \(x \in R^+\), \(f(x)\) must be in \(R^+\) (i.e., \(f(x)>0\)). Let's check if \(f(x)>0\) for all \(x \in R^+\). Consider small positive values of \(x\). If \(x=0.1\) (which is in \(R^+\)), then \(f(0.1) = 3(0.1)^2 - 2 = 3(0.01) - 2 = 0.03 - 2 = -1.97\). Since \(f(0.1) = -1.97\), which is not a positive real number (i.e., \(-1.97 \notin R^+\)), the output is not in the codomain for all inputs in the domain. This means that \(f(x) = 3x^2-2\) does not define a function from the set of positive real numbers to the set of positive real numbers because not all image values are positive. For a mapping to be a function from set A to set B, every element in A must map to exactly one element in B, and that element must be in B. Here, some \(f(x)\) values are negative, violating the codomain \(R^+\). Therefore, \(f\) as defined is not a function from \(R^+\) to \(R^+\). \[ \boxed{\text{not a function}} \]