Cancel whole periods before integrating · source reference

Printed question

It is given that

$$f(x)=\begin{cases}\tan(\pi x/(4a)),&-a<x\le a,\\\sin(\pi x/(2a)),&a<x\le3a,\end{cases}$$

and $f(x)=f(x-4a)$ for all real $x$, where $a$ is a positive real constant.

(a) Sketch $y=f(x)$ for $-4a\le x\le7a$. [3]

(b) Using the sketch, write down an equation relating $f(x)$ and $f(-x)$. [1]

(c) Show $\int_{-2a}^{4a}f(x)dx=(ka/\pi)(1+\ln m)$, determining constants $k,m$. [5]

[2023/HCI/Prelim/I/7]

Printed working

(a)

The native graph labels $(-4a,0),(7a,-1)$, each multiple $-3a,-2a,-a,0,a,2a,3a,4a,5a,6a,7a$ and levels $1,-1$.

(b)

From the sketch, $f(-x)=-f(x)$.

(c)

$$\begin{aligned}\int_{-2a}^{4a}f(x)dx&=\int_{-2a}^{-a}f(x)dx+\int_{-a}^af(x)dx+\int_a^{3a}f(x)dx+\int_{3a}^{4a}f(x)dx\\&=\int_{2a}^{3a}f(x)dx+0+0+\int_{-a}^0f(x)dx\\&=\int_{2a}^{3a}\sin(\pi x/(2a))dx+\int_{-a}^0\tan(\pi x/(4a))dx\\&=\left[\frac{-\cos(\pi x/(2a))}{\pi/(2a)}\right]_{2a}^{3a}+\left[\frac{\ln|\sec(\pi x/(4a))|}{\pi/(4a)}\right]_{-a}^0\\&=\frac{2a}\pi[-\cos(3\pi/2)+\cos\pi]+\frac{4a}\pi[\ln|\sec0|-\ln|\sec(-\pi/4)|]\\&=\frac{2a}\pi[-1]+\frac{4a}\pi[-\ln\sqrt2]=\frac{-2a}\pi(1+2\ln\sqrt2)\\&=\frac{-2a}\pi(1+\ln2).\end{aligned}$$

$k=-2,m=2$ (shown).

Teacher notes

The smaller source sketch labels a positive region A on $[0,a]$ and its negative counterpart on $[-a,0]$. $\int_0^af(x)dx=A$ is the area bounded by curve, $x$-axis and $x=0,a$. $\int_{-a}^0f(x)dx=-A$ since the curve lies below the $x$-axis. Thus $\int_{-a}^af=-A+A=0$. Similarly $\int_a^{3a}f=\int_a^{2a}f+\int_{2a}^{3a}f=B-B=0$.

Printed answer key

(a) labelled native graph below; (b) $f(-x)=-f(x)$; (c) $k=-2,m=2$.

Source diagrams in native form