Question:

The real valued function \( f: \mathbb{R} \to \left[ \frac{5}{2}, \infty \right) \) defined by \( f(x) = \left| 2x + 1 \right| + \left| x - 2 \right| \) is:

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For functions involving absolute values, consider the symmetry and intervals created by the absolute value expressions when analyzing one-to-one and onto properties. Additionally, visualize the behavior of the function on different intervals to check its injectivity and surjectivity.
Updated On: May 18, 2025
  • One-one function but not onto
  • Onto function but not one-one
  • Bijection
  • Neither one-one function nor onto
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The Correct Option is B

Approach Solution - 1

We are given the function \[ f(x) = \left| 2x + 1 \right| + \left| x - 2 \right|. \]

Step 1: Checking if the function is one-one A function is one-one (injective) if distinct values of \(x\) produce distinct values of \(f(x)\). 
The function involves absolute value terms, and for certain intervals of \(x\), the absolute values might produce the same value for different \(x\)'s. For example:
- If \(x = 0\), \( f(0) = \left| 2(0) + 1 \right| + \left| 0 - 2 \right| = 1 + 2 = 3 \)
- If \(x = 1\), \( f(1) = \left| 2(1) + 1 \right| + \left| 1 - 2 \right| = 3 + 1 = 4 \)
- However, values for different \(x\)'s could also repeat based on symmetry in the absolute terms, which shows that the function is not one-one. 

Step 2: Checking if the function is onto A function is onto (surjective) if every value in the target set (the range) is the output of some input in the domain.
Here, we are given that the range of \(f(x)\) is \( \left[ \frac{5}{2}, \infty \right) \).
By testing extreme values and considering the nature of the absolute values, we can see that \(f(x)\) can take any value greater than or equal to \( \frac{5}{2} \), so the function is onto. 
Thus, the function is onto but not one-one.

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Approach Solution -2

Problem: Consider the real-valued function \[ f: \mathbb{R} \to \left[ \frac{5}{2}, \infty \right) \] defined by \[ f(x) = |2x + 1| + |x - 2|. \] Determine whether the function is one-one, onto, or both.

Step 1: Analyze the function - The function \( f(x) \) is a sum of two absolute value expressions. - Since absolute values are always non-negative, \( f(x) \geq 0 \). - The given range is \(\left[ \frac{5}{2}, \infty \right)\), so the minimum value of \( f(x) \) is \( \frac{5}{2} \).

Step 2: Check if \( f \) is onto - The range starts from \( \frac{5}{2} \) and extends to infinity. - By definition, the function attains all values in this interval. - Hence, \( f \) is onto \( \left[ \frac{5}{2}, \infty \right) \).

Step 3: Check if \( f \) is one-one - Because \( f \) involves absolute values, it is generally not strictly monotonic. - For example, \( f(-1) = |2(-1) + 1| + |-1 - 2| = | -2 + 1| + | -3| = 1 + 3 = 4 \), - And \( f(0) = |1| + | -2| = 1 + 2 = 3 \), so \( f \) changes but not strictly in one direction. - Also, different values of \( x \) can give the same \( f(x) \), so \( f \) is not one-one.

Final conclusion: The function \( f \) is onto but not one-one.

Answer: Onto function but not one-one.
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