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Difference between revisions of "2017 AMC 12A Problems/Problem 15"

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Between 3 and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either.
 
Between 3 and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either.
  
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Between <math>\pi</math> and <math>\pi+\frac{\pi}{4} < 4</math>, <math>f(x)</math> starts at -2 and increases to <math>-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0</math>. There is a zero here by the Intermediate Value Theorem!
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Between <math>\pi</math> and <math>\pi+\frac{\pi}{4} < 4</math>, <math>f(x)</math> starts at -2 and increases to <math>-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0</math>. There is a zero here by the Intermediate Value Theorem. Therefore, the answer is <math>\boxed{(D)}</math>.
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Therefore, the answer is <math>\boxed{(D)}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2017|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:11, 8 February 2017

Problem

Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$. In what interval does the smallest positive value of $x$ for which $f(x) = 0$ lie?

$\textbf{(A)}\ (0,1) \qquad \textbf{(B)}\ (1, 2) \qquad\textbf{(C)}\ (2, 3) \qquad\textbf{(D)}\ (3, 4) \qquad\textbf{(E)}\ (4,5)$

Solution

We must first get an idea of what $f(x)$ looks like:

Between 0 and 1, $f(x)$ starts at $2$ and increases; clearly there is no zero here.

Between 1 and $\frac{\pi}{2}$, $f(x)$ starts at a positive number and increases to $\infty$; there is no zero here either.

Between $\frac{\pi}{2}$ and 3, $f(x)$ starts at $-\infty$ and increases to some negative number; there is no zero here either.

Between 3 and $\pi$, $f(x)$ starts at some negative number and increases to -2; there is no zero here either.

Between $\pi$ and $\pi+\frac{\pi}{4} < 4$, $f(x)$ starts at -2 and increases to $-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0$. There is a zero here by the Intermediate Value Theorem. Therefore, the answer is $\boxed{(D)}$.

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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