A range proof needs degenerate cases · source reference

Complete question

(a) Without a calculator solve $(x+5)/(x-1)>9/x$. [3] Hence solve $(e^x+5)/(e^x-1)>9e^{-x}$. [2] (b) For $y=(1+kx)/(x^2-1)$ with real x≠±1 and $-1\le k\le1$, find the range in terms of k. [4]

[2018/ACJC/Prelim/I/8]

Printed working

Part (a)

$$\begin{aligned} \frac{x+5}{x-1}-\frac9x&>0,\qquad x\ne0,1\\ \frac{x^2-4x+9}{(x-1)x}&>0\\ \frac{(x-2)^2+5}{(x-1)x}&>0. \end{aligned}$$

Since $(x-2)^2+5>0$ for all real values of $x$, $(x-1)x>0$, giving $x<0$ or $x>1$.

For $\dfrac{e^x+5}{e^x-1}>9e^{-x}$, replace $x$ by $e^x$ in the previous result. Then $e^x<0$ (rejected) or $e^x>1$, hence $x>0$.

Part (b): Method 1 — algebraic, without GC

$$\begin{aligned} y&=\frac{1+kx}{x^2-1}\\ yx^2-y&=1+kx\\ yx^2-kx-(y+1)&=0. \end{aligned}$$

For values that $y$ can take, i.e. for the equation to have real roots, discriminant $\ge0$:

$$\begin{aligned} (-k)^2-4y(-y-1)&\ge0\\ 4y^2+4y+k^2&\ge0.\qquad(*) \end{aligned}$$

Consider $4y^2+4y+k^2=0$:

$$y=\frac{-1\pm\sqrt{1-k^2}}2.$$

The printed conclusion for (*) is

$$y\le\frac{-1-\sqrt{1-k^2}}2\quad\text{or}\quad y\ge\frac{-1+\sqrt{1-k^2}}2.$$

This is the original printed conclusion; exceptional cases are qualified below.

Part (b): Method 2 — graphical

The source begins by setting $dy/dx=0$ and gives

$$kx^2+2x+k=0,\qquad x=\frac{-1\pm\sqrt{1-k^2}}k.$$

The stationary points are printed as

$$\left(\frac{-1-\sqrt{1-k^2}}k,\frac{-1+\sqrt{1-k^2}}2\right),\qquad \left(\frac{-1+\sqrt{1-k^2}}k,\frac{-1-\sqrt{1-k^2}}2\right).$$

The accompanying sketch labels the vertical lines $x=-1$, $x=1$, horizontal asymptote $y=0$, and stationary ordinate levels $(-1+\sqrt{1-k^2})/2$ and $(-1-\sqrt{1-k^2})/2$. It repeats the same printed range conclusion as Method 1.

Exceptional cases — reviewed supplement

For $0<|k|<1$, the printed range is correct. For $k=0$, the range is $(-\infty,-1]\cup(0,\infty)$. For $k=\pm1$, cancellation leaves a deleted input, so the range is $\mathbb R\setminus\{0,-\tfrac12\}$. The discriminant condition alone does not enforce the non-quadratic case or the excluded inputs.

Teacher margin notes

As $x$ is a variable, $x$ and $x-1$ can be either negative or positive. As such, DO NOT cross-multiply.

Express the inequality in the form $f(x)/g(x)>0$. Since numerator $f(x)$ is always positive, denominator $g(x)$ has to be positive for $f(x)/g(x)>0$.

$e^x$ is always $>0$, i.e. $e^x>0$ for all $x\in\mathbb R$, so there is no real solution for $e^x<0$.

NOTE: Let $y=p$, $p$ constant. The range of values that $y$ can take is the values of $y$ (i.e. $p$) for which the line $y=p$ intersects the curve: $(1+kx)/(x^2-1)=p$, i.e. $px^2-kx-(p+1)=0$ has real roots, hence discriminant $\ge0$.

Source diagrams

Review qualification

Source discrepancy to review: solution PDF pp.4–5 applies the discriminant formula to all −1≤k≤1 without checking k=0 or cancelled roots at k=±1. Porisma separates those cases; the printed source remains available.

The corrected or qualified authored derivation is retained in the question article.

Provenance

Question page references: 1. Solution page references: 4, 5. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.