04-BS-16, December 2014, Q6 | Nivaar Exam Prep

Question 6

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Sine restricted to two different domains; the squaring function $f(x)=x^2$ on $R$; two specific real numbers $2.3,1.6$; two floor/ceiling identity claims.

Find. (a) Injectivity and range for each restricted sine. (b) The image $f(A)$ for each set $A$. (c) Two floor/ceiling values. (d) Truth value of each identity, with a counterexample if false.

Approach. Use monotonicity to settle injectivity, sweep the domain to determine the range/image, and apply the floor/ceiling definitions directly; test the two identities algebraically (they turn out to be genuinely true, unlike Q2c).

a1) $f:[-\pi/2,\pi/2]\to R,\ f(x)=\sin x$. $\sin$ is strictly increasing on this interval (its derivative $\cos x\ge0$ there, $=0$ only at the endpoints), so $f$ is $\boxed{\text{one-to-one}}$, with range $\boxed{[-1,1]}$ (endpoints attained at $x=\pm\pi/2$).
a2) $f:[0,\pi]\to R,\ f(x)=\sin x$. Here $\sin(\pi/6)=\sin(5\pi/6)=\tfrac12$ with $\pi/6\ne5\pi/6$ both in $[0,\pi]$, so $f$ is $\boxed{\text{NOT one-to-one}}$; the range is still $\boxed{[0,1]}$ (minimum 0 at the endpoints, maximum 1 at $x=\pi/2$).
b1) $f(A)$ for $A=[-7,2]$. $x^2$ attains its minimum $0$ at $x=0\in A$ and its maximum on $A$ at whichever endpoint has larger $|x|$: $|-7|=7>|2|=2$, so the max is $(-7)^2=49$. Since $x^2$ is continuous, every value between is attained. $\boxed{f(A)=[0,49]}$.
b2) $f(A)$ for $A=(-4,-3]\cup[5,6]$. On $(-4,-3]$: $x^2$ is decreasing as $x\to-3^-$ increases toward $-3$ (further from 0), so as $x$ ranges over $(-4,-3]$, $x^2$ ranges over $[9,16)$ (value $9$ attained at $x=-3$; value $16$ approached but never attained since $x=-4$ is excluded). On $[5,6]$: $x^2$ increasing, giving $[25,36]$. $\boxed{f(A)=[9,16)\cup[25,36]}$.
c) Floor/ceiling arithmetic. $\lfloor2.3\rfloor=2$, $\lfloor1.6\rfloor=1$, so (1) $2-1=\boxed{1}$. $\lceil2.3\rceil=3$, so (2) $3-1=\boxed{2}$.
d1) $\lfloor a\rfloor=\lceil a\rceil-1$ for all $a\in R-Z$. For any non-integer $a$, write $a=n+f$ with $n=\lfloor a\rfloor\in Z$ and fractional part $0
d2) $-\lceil a\rceil=\lfloor-a\rfloor$ for all $a\in R$. $\lceil a\rceil$ is the smallest integer $\ge a$, i.e. $\lceil a\rceil=-\lfloor-a\rfloor$ by the standard floor/ceiling reflection identity (negating flips "smallest integer above" into "largest integer below" of the negated value). Rearranging gives exactly $-\lceil a\rceil=\lfloor-a\rfloor$. $\boxed{\text{TRUE for all } a\in R}$ (true for integers and non-integers alike, e.g. $a=2.3$: $-\lceil2.3\rceil=-3$, $\lfloor-2.3\rfloor=\lfloor-2.3\rfloor=-3$ — matches; $a=2$: $-\lceil2\rceil=-2=\lfloor-2\rfloor$ — matches).

Question 6 results
Part	Result
a1	1-1, range $[-1,1]$
a2	not 1-1, range $[0,1]$
b1	$[0,49]$
b2	$[9,16)\cup[25,36]$
c1	1
c2	2
d1	True
d2	True

Topic: Functions — injectivity, range/image, floor and ceiling.

Key relations: Strict monotonicity $\Rightarrow$ injective; the image of an interval under a continuous function is an interval (IVT); floor/ceiling reflection identity $\lceil a\rceil=-\lfloor-a\rfloor$.

Why this works: Restricting sine's domain to a monotone piece is exactly how $\arcsin$ is constructed; the squaring map's image on an interval is governed entirely by which endpoint (or $x=0$, if included) is farthest from the origin.

Common pitfall: On $(-4,-3]$ it is easy to mis-orient which endpoint gives the open vs. closed bound of the image — the excluded domain endpoint ($x=-4$) produces the excluded image endpoint (16), not the included one.

Source: Rosen, Discrete Mathematics and Its Applications, 7th ed., Ch. 2.3 (Functions).

Question 6 of 12

Question 6