Compare a rational branch to the upper hyperbola · source reference

Printed question

The curve $C$ has equation $y=ax+b+2a+\frac{2(b+2a)}{x-2}$, where $a,b$ are real constants such that $a>0,b\ne-2a,x\ne2$.

(i) By using differentiation, find the range of values of $b$ in terms of $a$ such that $C$ has no stationary points. [4]

(ii) Given that $b=-3a$, sketch $C$, stating the equations of any asymptotes and the coordinates of the points where $C$ crosses the axes. [3]

(iii) On the same diagram as $C$, sketch the graph of $(\frac ya)^2-(x-1)^2=4$, showing clearly the equations of the asymptotes and turning points. Hence solve the inequality $x-1-\frac2{x-2}\ge\sqrt{(x-1)^2+4}$. [4]

[2018/AJC/Prelim/II/3]

Printed working

(i) $y=ax+b+2a+\frac{2(b+2a)}{x-2}$.

$\frac{dy}{dx}=a-\frac{2(b+2a)}{(x-2)^2}$.

When $\frac{dy}{dx}=0$, $\frac{2(b+2a)}{(x-2)^2}=a$.

$a(x-2)^2=2(b+2a)$ — (*).

For $C$ to have no stationary points, $2(b+2a)<0$ (since $a>0$).

$b<-2a$.

Alternatively,

$\frac{dy}{dx}=0$, $\frac{2(b+2a)}{(x-2)^2}=a$.

$a(x-2)^2=2(b+2a)$.

$ax^2-4ax-2b=0$.

For $C$ to have no stationary points,

Discriminant $=16a^2-4a(-2b)=8a(2a+b)<0$.

Since $a>0$, $2a+b<0\Rightarrow b<-2a$.

(ii) For $b=-3a$, $C:y=ax-a+\frac{-2a}{x-2}$.

Equations of asymptotes: $y=ax-a$, $x=2$.

(iii) $(\frac ya)^2-(x-1)^2=4\Rightarrow(\frac y{2a})^2-(\frac{x-1}2)^2=1$ (Hyperbola).

Asymptotes: $\frac y{2a}=\pm\frac{x-1}2\Rightarrow y=ax-a$ or $y=-ax+a$.

Vertices: $(1,-2a),(1,2a)$.

$x-1-\frac2{x-2}\ge\sqrt{(x-1)^2+4}$.

From the graphs of $y=ax-a-\frac{2a}{x-2}$ and $y=a\sqrt{(x-1)^2+4}$ (i.e. $y>0$), $1\le x<2$.

Teacher notes

Since $(x-2)^2>0$ for all $x\in\mathbb R$, for $C$ to have no stationary pts, equation (*) has no real roots $\Rightarrow a(x-2)^2<0$ since $a>0$, i.e. $2(b+2a)<0$.

1. To find asymptotes: Let $(\frac y{2a})^2-(\frac{x-1}2)^2=0$.

2. To find vertices: Put $(\frac{x-1}2)^2=0\Rightarrow x=1$, $(\frac y{2a})^2=1\Rightarrow y=\pm2a$.

$y=ax-a-\frac{2a}{x-2}=a(x-1-\frac2{x-2})$, and $y=a\sqrt{(x-1)^2+4}\Leftrightarrow(\frac ya)^2-(x-1)^2=4$ where $y>0$.

So $x-1-\frac2{x-2}\ge\sqrt{(x-1)^2+4}\Rightarrow$ range of values of $x$ for which the part of curve $y=ax-a-\frac{2a}{x-2}$ is on or above curve $y=a\sqrt{(x-1)^2+4}$.

Printed answer key

(i) $b<-2a$. (ii), (iii) $1\le x<2$; native graph above.

Editorial qualification

The teacher note’s statement “(x−2)²>0 for all x∈ℝ” needs x≠2; the question already excludes 2. Its line a(x−2)²<0 is a shorthand impossibility argument, not a feasible inequality. The printed transformed reciprocal branch continues to respect x≠2.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.