If a real-valued function f is defined by f(x | vedclass

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(C) Given $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$. The domain is determined by $a^2 - x^2 \geq 0$,so $x \in [-|a|, |a|]$ (excluding $x=0$).For $x \i

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both one-one and onto

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(C) Given $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$. The domain is determined by $a^2 - x^2 \geq 0$,so $x \in [-|a|, |a|]$ (excluding $x=0$).For $x \in (0, |a|]$,$f(x) = \frac{a}{b} + \frac{1}{b}\sqrt{\frac{a^2}{x^2} - 1}$.Let $g(x) = \frac{a^2}{x^2} - 1$. As $x$ increases from $0$ to $|a|$,$g(x)$ decreases from $\infty$ to $0$.Thus,$f(x)$ is a strictly monotonic function on its domain intervals.Since the function is strictly monotonic on its domain,it is one-one.For the range,as $x 	o 0^+$,$f(x) 	o \infty$,and as $x 	o |a|$,$f(x) = \frac{a^2}{b|a|} = \frac{|a|}{b}$.Since the function covers all real values in its codomain (assuming $R$ as codomain),it is onto.Therefore,$f$ is both one-one and onto.

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

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