Porisma

Original question & working

A turning point fixes a rational curve · source reference

2011/PJC/Promo/9 · Question pp. 4 · Working pp. 13, 14

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Printed question

The curve CC has equation y=ax+bx2+cy=\frac{ax+b}{x^2+c}, where a,b,ca,b,c are constants. It is given that x=2x=2 is an asymptote of CC.

(i) State the value of cc. [1]

(ii) Given also that CC has a turning point at (6,14)(6,\frac14), find the values of a,ba,b. [4]

(iii) Sketch CC, stating the equations of the asymptotes and the coordinates of the turning points. [You need not indicate the coordinates of intersections with the axes.] [3]

(iv) Hence find the range of values of kk for which the equation kx2−3x+10−4k=0kx^2-3x+10-4k=0 has no real root. [2]

[2011/PJC/Promo/9]

Printed working

(i) c=−4c=-4.

(ii) Sub the point (6,14)(6,\frac14) and c=−4c=-4 into equation of curve,

y=ax+bx2−4y=\frac{ax+b}{x^2-4}.

14=6a+b62−4\frac14=\frac{6a+b}{6^2-4}.

6a+b=86a+b=8.

b=8−6ab=8-6a — (1).

dydx=(x2−4)a−(ax+b)(2x)(x2−4)2=ax2−4a−2ax2−2bx(x2−4)2=−ax2−2bx−4a(x2−4)2\frac{dy}{dx}=\frac{(x^2-4)a-(ax+b)(2x)}{(x^2-4)^2}=\frac{ax^2-4a-2ax^2-2bx}{(x^2-4)^2}=\frac{-ax^2-2bx-4a}{(x^2-4)^2}.

Since there is a turning point at (6,14)(6,\frac14),

⇒dydx=0\Rightarrow\frac{dy}{dx}=0 at x=6x=6.

⇒−a(6)2−2b(6)−4a=0\Rightarrow-a(6)^2-2b(6)-4a=0.

10a+3b=010a+3b=0 — (2).

Solving simultaneously, 10a+3(8−6a)=010a+3(8-6a)=0, 8a=248a=24.

⇒a=3,b=−10\Rightarrow a=3,b=-10.

(iii) Graph below: turning points (23,94),(6,14)(\frac23,\frac94),(6,\frac14); asymptotes x=−2,x=2,y=0x=-2,x=2,y=0.

(iv) kx2−3x+10−4k=0kx^2-3x+10-4k=0.

kx2−4k=3x−10kx^2-4k=3x-10.

k=3x−10x2−4k=\frac{3x-10}{x^2-4}.

Hence, for kx2−3x+10−4k=0kx^2-3x+10-4k=0 to have no real roots, the graph of y=3x−10x2−4y=\frac{3x-10}{x^2-4} and y=ky=k must not intersect. From the sketch, they will not intersect for 14<k<94\frac14<k<\frac94.

Native source reconstruction, question 9 diagram iii

Teacher notes

When x=2x=2, denominator =0⇒22+c=0=0\Rightarrow2^2+c=0.

CC has turning pt at (6,14)⇒(6,\frac14)\Rightarrow 1) (6,14)(6,\frac14) lies on CC, so satisfies eqn of curve AND 2) At x=6x=6, dydx=0\frac{dy}{dx}=0.

Re-express equation such that one side of it is ax+bx2+c\frac{ax+b}{x^2+c}, where a=3,b=−10,c=−4a=3,b=-10,c=-4.

Note: CANNOT use b2−4ac<0b^2-4ac<0 as question says “Hence” which means must use the curve sketched in the previous part.

Printed answer key

(i) c=−4c=-4. (ii) a=3,b=−10a=3,b=-10. (iii) Native graph above. (iv) 14<k<94\frac14<k<\frac94.

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.