Count roots on either side of a pole · source reference

Printed question

The curve $C$ has equation $y=\frac{ax^2+x-4}{x+2}$, where $a$ is a constant.

(i) Find $\frac{dy}{dx}$ in terms of $a$, and deduce the range of values of $a$ if $C$ has two distinct stationary points. [4]

(ii) Given that the asymptotes of $C$ intersect at the point $(-2,-7)$, show that $a=2$. [2]

(iii) Sketch $C$, indicating clearly the asymptotes and stationary points. [2]

(iv) For $r\ge0$, find the range of values of $r$, correct to 3 decimal places, such that the equation $\ln(x-r)=\frac{x+2}{2x^2+x-4}$ has exactly one real root. [2]

[2011/NJC/Prelim/I/12b]

Printed working

(i) The curve $C$ has equation $y=\frac{ax^2+x-4}{x+2}$, where $a$ is a constant.

$\frac{dy}{dx}=\frac{(2ax+1)(x+2)-(1)(ax^2+x-4)}{(x+2)^2}=\frac{ax^2+4ax+6}{(x+2)^2}$.

If $C$ has two distinct stationary points, then $\frac{dy}{dx}=0$ has 2 distinct real roots.

$\Rightarrow ax^2+4ax+6=0$ has two distinct roots.

Discriminant $>0$.

$(4a)^2-24a>0$.

$8a(2a-3)>0$.

$\therefore a>\frac32$ or $a<0$.

(ii) $y=\frac{ax^2+x-4}{x+2}=ax+(1-2a)+\frac{4a-6}{x+2}$ (by long division).

Vertical Asymptote: $x=-2$.

Oblique Asymptote: $y=ax+(1-2a)$.

At $(-2,-7)$, $-7=a(-2)+(1-2a)\Rightarrow a=2$.

(iii) $y=2x-3+\frac2{x+2}$, stationary points $(-1,-3),(-3,-11)$.

(iv) $\ln(x-r)=\frac{x+2}{2x^2+x-4}\Rightarrow\frac{2x^2+x-4}{x+2}=\frac1{\ln(x-r)}$ — (*).

Note: It would be advisable to get graph of $y=\frac1{\ln(x-r)}$ from the graph of $y=\ln(x-r)$.

From GC, the curve cuts the $x$-axis at $A(-1.186,0)$ (to 3 d.p.).

For exactly 1 real root, i.e., exactly 1 point of intersection, $r\ge1.186$ (3 d.p.).

Teacher notes

Curve $y=\frac1{\ln(x-r)}$ excludes the pt $(r,0)$ and has a vertical asymptote at $x=r+1$.

For eqn (*) to have 1 real root, the pt $(r,0)$ must be the pt $A$ ($r=1.186$) or to the right of pt $A$ ($r>1.186$).

Printed answer key

(i) $\frac{ax^2+4ax+6}{(x+2)^2}$; $a>\frac32$ or $a<0$. (iii) Native graph above. (iv) $r\ge1.186$ (3 d.p.). The short key omits (ii).

Editorial qualification

The source states A(-1.186,0), but the relevant positive zero is A(1.186,0): (−1+√33)/4≈1.186. The original negative sign is preserved as a printed error; the final r bound uses the positive value. At the excluded original pole x=-2 an algebraic derivative root must not be retained.

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.