Functions Question 510
Question: If $ f:R\to R $ is given by $ f(x)=\frac{x^{2}-4}{x^{2}+1}, $ then the function f is
Options:
A) many-one onto
B) many-one into
C) one-one into
D) one-one onto
Show Answer
Answer:
Correct Answer: B
Solution:
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First, let’s consider the domain and range of the function.
Domain: The function is defined for all real numbers except where the denominator is zero. But $ x^2 + 1 $ > 0 for all real x, so the domain is all real numbers.
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Now, let’s consider the range:
As x approaches ±∞, f(x) approaches 1. When x = 0, f(x) = -4 When x = ±2, f(x) = 0
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Let’s find the derivative to check for extrema:
f’(x) = $ \frac{(2x(x^2+1) - (x^2-4)(2x))}{(x^2+1)^2} $
= $ \frac{(2x^3 + 2x - 2x^3 + 8x)}{(x^2+1)^2} $
= $ \frac{10x}{(x^2+1)^2} $
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The derivative is zero when x = 0, which gives us the minimum value of -4.
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From this analysis, we can conclude that the range of f is [-4, 1).
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Now, is the function one-to-one?
For a function to be one-to-one, each element in the codomain should be paired with at most one element in the domain.
Consider f(x) = f(-x): $\frac{(x^2 - 4)}{(x^2 + 1)} = \frac{((-x)^2 - 4)}{((-x)^2 + 1)}$
This is true for all x, which means f(x) = f(-x) for all x ≠ 0. Therefore, the function is not one-to-one.
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Is the function onto?
The range is [-4, 1), which is not equal to the entire codomain (R). Therefore, the function is not onto.
Conclusion: The function is many-to-one (not one-to-one) and not onto (into).
BETA
