Use the periodic extension to identify the incorrect inequality.

01 / Standard#Your turn

f(x+2)=f(x),\quad f(x)=1+|x-5|\quad(4\le x\le6)

Official paper · jm01-2025 · I.15 · PDF 3

Skills and prerequisite lessons

Option A $f(\sin\frac\pi6)>f(\cos\frac\pi6)$
Option B $f(\sin\frac\pi3)>f(\cos\frac\pi3)$
Option C $f(\cos\pi)<f(\sin\pi)$
Option D $f(\sin\frac{2\pi}3)<f(\cos\frac{2\pi}3)$
Option E $f(\sin\frac\pi2)<f(\cos\frac\pi2)$

Working and explanation

BUILD THE REASONING

Hint 1

Reduce all arguments to [0,2] using periodicity.

Hint 2

On [0,1], the function equals 2−x.

Worked solution

The translated formula is a V-shape centred at 1.
$f(x)=1+|x-1|\ (0\le x\le2),\quad f(0)=2,\ f(1)=1$
Evaluate the relevant half-angle inputs.
$f(1/2)=f(-1/2)=\frac32,\quad f(\sqrt3/2)=2-\frac{\sqrt3}2<\frac32$
Thus B reverses the correct inequality; A,C,D,E hold.

B is the incorrect inequality.

Checks and common pitfalls: The function is periodic but not increasing on each whole period.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The translated formula is a V-shape centred at 1.
$f(x)=1+|x-1|\ (0\le x\le2),\quad f(0)=2,\ f(1)=1$
Evaluate the relevant half-angle inputs.
$f(1/2)=f(-1/2)=\frac32,\quad f(\sqrt3/2)=2-\frac{\sqrt3}2<\frac32$
Thus B reverses the correct inequality; A,C,D,E hold.

Think first. Reveal a hint when the class is ready.

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