Difference between revisions of "2010 AMC 12A Problems/Problem 4"

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<math>\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt[3]{x}</math>
 
<math>\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt[3]{x}</math>
  
[[2010 AMC 12A Problems/Problem 4|Solution]]
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== Solution ==
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<math>x</math> is negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use <math>-1</math>.
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<math>\textbf{(A)} \Rightarrow \frac{-1}{|-1|} = -1</math>
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<math>\textbf{(B)} \Rightarrow -(-1)^2 = -1</math>
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<math>\textbf{(C)} \Rightarrow -2^{(-1)} = -\frac{1}{2}</math>
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<math>\textbf{(D)} \Rightarrow -(-1)^{(-1)} = 1</math>
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<math>\textbf{(E)} \Rightarrow \sqrt[3]{-1} = -1</math>
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Obviously only <math>\boxed{\textbf{(D)}}</math> is positive.

Revision as of 16:30, 10 February 2010

Problem 4

If $x<0$, then which of the following must be positive?

$\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt[3]{x}$

Solution

$x$ is negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use $-1$.

$\textbf{(A)} \Rightarrow \frac{-1}{|-1|} = -1$

$\textbf{(B)} \Rightarrow -(-1)^2 = -1$

$\textbf{(C)} \Rightarrow -2^{(-1)} = -\frac{1}{2}$

$\textbf{(D)} \Rightarrow -(-1)^{(-1)} = 1$

$\textbf{(E)} \Rightarrow \sqrt[3]{-1} = -1$


Obviously only $\boxed{\textbf{(D)}}$ is positive.