2017 AMC 12A Problem 15

Question

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?

(0,1)

(1,2)

(2,3)

(3,4)

(4,5)

For

0\lt x\lt\dfrac{\pi}{2}

all three terms are positive, so there is no root there

Just past

\dfrac{\pi}{2}

the term

3\tan x

makes

f

negative; locate the first sign change after

\pi

For

0\lt x\lt\dfrac{\pi}{2}

all three terms are positive, so

f(x)\gt0.

For

\dfrac{\pi}{2}\lt x\lt\pi,

set

s=\sin x\gt0

and

t=-\cos x\gt0.

Since

0\lt t\le1,

\begin{aligned} f(x)&=s-2t-\dfrac{3s}{t} \\ &=-2t-s\left(\dfrac{3}{t}-1\right)\lt0. \end{aligned}

Thus no root occurs before

x=\pi.

At

x=\pi,

f(\pi)=0+2(-1)+0=-2\lt0.

At

x=\dfrac{5\pi}{4},

\tan x=1

so

f=-\dfrac{\sqrt2}{2}+3\gt0.

By the intermediate value theorem the smallest positive root lies in

\left(\pi,\dfrac{5\pi}{4}\right).

Since

\pi\gt3

and

\dfrac{5\pi}{4}\lt4,

this interval sits inside

(3,4).

Thus, the correct answer is D.

Problem 15 in other years