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Chapter 1: Problem 19

Let function \(f: R \rightarrow R\) be defined by \(f(x)=2 x+\sin x\) for \(x \in \mathrm{R} .\) Then \(f\) is: (a) One-one and onto (b) One-one but not onto (c) Onto but not one-one (d) Neither one-one nor onto

Short Answer

Expert verified
The function \(f(x)=2 x+\sin x\) is both one-one and onto. So the correct answer is (a) One-one and onto.

Step by step solution

01

Determine if the function is one-one

Analyze if the function is one-one. This can be done by checking if for every x, there is a unique y associated with it in the function. Check the first derivative of the function. If \(f'(x)\) > 0 for all x, then the function is increasing and hence one-one. So, first you derive the function \(f'(x) = 2 + \cos x\). Since the result will always be greater than zero, because the cosine function oscillates between -1 and 1, the function is one-one.
02

Determine if the function is onto

Analyze if the function is onto. This can be done by checking if for every y in the co-domain there exists an x in the domain such that y=f(x). The range of the sine function is -1 to 1. Since \(f(x)=2 x+\sin x\), \(f(x)\) should be able to take any value in R, making the function onto. It is not limited to a specific range.

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