What is (1-x)f(\sqrt{x})+xf(\sqrt{x}+1) equal to?
For the following two (02) items: Consider the function f(x)=\frac{x}{1-x} (x \gt 0, x \neq 1).
- A. -f(x)
- B. f(x)
- C. x
- D. 0 ✓
Correct Answer: D. 0
Explanation
Compute the two terms separately. First, f(\sqrt{x}) = \frac{\sqrt{x}}{1-\sqrt{x}}. Multiplying by (1-x) = (1-\sqrt{x})(1+\sqrt{x}) gives \sqrt{x}(1+\sqrt{x}) = \sqrt{x} + x. Second, f(\sqrt{x}+1) = \frac{\sqrt{x}+1}{1-(\sqrt{x}+1)} = \frac{\sqrt{x}+1}{-\sqrt{x}}. Multiplying by x gives -\sqrt{x}(\sqrt{x}+1) = -x - \sqrt{x}. Summing the two parts yields (\sqrt{x} + x) + (-x - \sqrt{x}) = 0.
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