Question:

A function \( f : \mathbb{R} \to A \) defined as \( f(x) = x^2 + 1 \) is onto, if \( A \) is: {5pt}

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For quadratic functions \( f(x) = ax^2 + bx + c \): - The range depends on the vertex of the parabola. - For \( a>0 \), the range starts from the minimum value.
Updated On: Jan 29, 2025
  • \( (-\infty, \infty) \)
  • \( (1, \infty) \)
  • \( [1, \infty) \)
  • \( [-1, \infty) \)
    {5pt}
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The Correct Option is C

Solution and Explanation


Step 1: Understanding the function. 
The function is defined as \( f(x) = x^2 + 1 \), where \( x \in \mathbb{R} \). The term \( x^2 \) is always non-negative, and adding 1 shifts the range of \( x^2 \) to start from 1. 
Step 2: Range of \( f(x) \). 
The minimum value of \( x^2 \) is 0, so the minimum value of \( f(x) = x^2 + 1 \) is: \[ f(x) = 0 + 1 = 1. \] Thus, the range of \( f(x) \) is \( [1, \infty) \). 
Step 3: Conditions for \( f(x) \) to be onto. 
For \( f(x) \) to be onto, the codomain \( A \) must include the entire range of \( f(x) \). Therefore, \( A = [1, \infty) \). 
Step 4: Conclusion. 
The function \( f(x) = x^2 + 1 \) is onto if \( A = [1, \infty) \). {10pt}

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