A discriminant reveals a forbidden output interval · source reference

Printed question

The curve $C$ has equation $y=\frac{2x^2+4}{3(x+\lambda+1)}$. It is given that $C$ has a vertical asymptote $x=0$.

(i) Determine the value of $\lambda$. [1]

(ii) Find the equation of the other asymptote of $C$. [2]

(iii) Prove, using an algebraic method, that $C$ cannot lie between $-\frac{4\sqrt a}{3}<y<\frac{4\sqrt a}{3}$, $a<0$, where $a$ is to be determined. [3]

(iv) When $\lambda=0$, the graph of $y=\frac{2x^2+4}{3(x+\lambda+1)}+k$ cuts the $x$-axis at 2 distinct points where $k$ is a positive integer.

(a) Sketch the graph of $y=\frac{2x^2+4}{3(x+\lambda+1)}+k$ when $\lambda=0$, indicating clearly all asymptotes and the coordinates of the axial intercepts. [4]

(b) Deduce the least possible integer value of $k$. [1]

[2009/NYJC/Prelim/I/7]

Printed working

(i) Since $C$ has a vertical asymptote at $x=0$, thus $\lambda+1=0\Rightarrow\lambda=-1$.

(ii) $y=\frac{2x^2+4}{3x}=\frac23x+\frac4{3x}$.

Oblique asymptote is $y=\frac23x$.

(iii) $y=\frac{2x^2+4}{3x}$.

$3xy=2x^2+4$.

$2x^2-3xy+4=0$.

The equation above has no real roots when discriminant $<0$.

$(-3y)^2-4(2)(4)<0$.

$9y^2-32<0$.

$(y+\frac{\sqrt{32}}3)(y-\frac{\sqrt{32}}3)<0$.

$-\frac{4\sqrt2}3<y<\frac{4\sqrt2}3$.

$\therefore a=2$.

(iv)(a) $y=\frac{2x^2+4}{3(x+1)}+k=\frac23x-\frac23+k+\frac6{3(x+1)}$.

$x$-intercepts are $-\frac{3k}4+\sqrt{\frac{9k^2}{16}-\frac{3k}2-2}$ and $-\frac{3k}4-\sqrt{\frac{9k^2}{16}-\frac{3k}2-2}$.

(b) For the graph of $y=\frac{2x^2+4}{3(x+1)}+k$ to cut the $x$-axis at 2 distinct points,

$\frac{9k^2}{16}-\frac{3k}2-2>0\Rightarrow$ least integer value of $k$ is 4.

OR: For 2 distinct pts, $\frac{2x^2+4}{3(x+1)}+k=0\Rightarrow2x^2+3kx+(4+3k)=0$ has 2 distinct real roots $\Rightarrow$ Discriminant $>0$, i.e. $9k^2-4(2)(4+3k)>0$.

Teacher notes

NOTE: Let $y=k$, $k$ constant. Range of values of $y$ that $C$ cannot lie in (i.e. range of values that $y$ cannot take) → the values of $y$ (i.e. $k$) for which the line $y=k$ does not intersect the curve $C$ at all.

$\Rightarrow\frac{2x^2+4}{3x}=k$, i.e. $2x^2-3kx+4=0$ has no real roots $\Rightarrow$ Discriminant $<0$.

Printed answer key

(i) $\lambda=-1$. (ii) $y=\frac23x$. (iv)(a) Native graph above, both x-intercepts as transcribed in working. (iv)(b) 4. The key omits (iii).

Editorial qualification

Part (iii) prints a<0 under √a, which is inconsistent over the reals. The worked solution gives a=2; retain the printed premise but correct the sign assumption to a>0 when using the answer. The abbreviated answer-key page omits part (iii).

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.