Question

If a real valued function \( f \) is defined by \( f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx} \), then \( f \) is:

• A. only one-one
• B. only onto
• C. both one-one and onto
• D. neither one-one nor onto

Hint:

Always examine domain restrictions carefully when roots and denominators are involved. Use boundary analysis and symmetry to test injectivity and surjectivity.

Answer 1

The Correct Option is B

We are given the function: \[ f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx} \] where \( a \) and \( b \) are constants. Step 1: Determine the domain of \( f(x) \) - The square root expression \( \sqrt{a^2 - x^2} \) implies: \( a^2 - x^2 \geq 0 \Rightarrow x \in [-a, a] \) - The denominator \( bx \neq 0 \Rightarrow x \neq 0 \) So, the domain of \( f(x) \) is: \[ x \in [-a, a] \setminus \{0\} \] Step 2: Test for injectivity (One-One) Suppose \( f(x_1) = f(x_2) \). Then: \[ \frac{ax_1 + \sqrt{a^2 - x_1^2}}{bx_1} = \frac{ax_2 + \sqrt{a^2 - x_2^2}}{bx_2} \] This equation does not imply \( x_1 = x_2 \), and is not generally solvable to prove injectivity. Hence, \( f \) is not one-one. Example: Take \( a = 1, b = 1 \), then: \[ f\left(\frac{1}{2}\right) = \frac{\frac{1}{2} + \sqrt{1 - \left(\frac{1}{2}\right)^2}}{\frac{1}{2}} = \frac{\frac{1}{2} + \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 1 + \sqrt{3} \] \[ f\left(-\frac{1}{2}\right) = \frac{-\frac{1}{2} + \sqrt{1 - \left(\frac{1}{2}\right)^2}}{-\frac{1}{2}} = \frac{\frac{\sqrt{3} - 1}{2}}{-\frac{1}{2}} = -( \sqrt{3} - 1 ) = 1 - \sqrt{3} \] These are not equal, but no general injectivity is evident. Try plotting or checking symmetry — ultimately, \( f \) fails the horizontal line test across its domain. Conclusion: Not one-one. Step 3: Test for surjectivity (Onto) Let us observe the function behavior: \[ f(x) = \frac{a}{b} + \frac{\sqrt{a^2 - x^2}}{bx} \] As \( x \to 0^+ \), \( f(x) \to +\infty \); As \( x \to 0^- \), \( f(x) \to -\infty \); As \( x \to \pm a \), the square root \( \to 0 \), and the value tends to \( \frac{a}{b} \) Hence, \( f(x) \) sweeps all values in \( \mathbb{R} \), depending on \( x \)'s sign and proximity to 0, making it onto.