Points and gradients identify coefficients · source reference

Printed question

A curve $C$ has equation $y=a(2x+1)^2+bx+ce^{2x}$, where $a,b,c$ are constants. It is given that $C$ passes through the point $(1,17-e^2)$ and the gradient of $C$ at the point $(0,1)$ is 5. Find the equation of $C$. [5]

[2021/TMJC/Prelim I/1]

Printed working

$y=a(2x+1)^2+bx+ce^{2x}$.

Differentiate wrt $x$:

$$\frac{dy}{dx}=2a(2x+1)2+b+2ce^{2x}\qquad(*)$$

Sub $(1,17-e^2)$ into $C$:

$$9a+b+ce^2=17-e^2\qquad(1).$$

Sub $(0,1)$ into $C$:

$$a+0b+c=1\qquad(2).$$

Sub $(0,1)$ and gradient 5 into $(*)$:

$$4a+b+2c=5\qquad(3).$$

Using GC, $a=2,b=-1,c=-1$.

$\therefore$ Equation of $C$ is $y=2(2x+1)^2-x-e^{2x}$.

Teacher notes

$C$ passes through $(1,17-e^2)$.

$C$ passes through $(0,1)$.

$C$ passes through $(0,1)$ and gradient of $C$ at $(0,1)$ is 5.

Printed answer key

Equation of $C$ is $y=2(2x+1)^2-x-e^{2x}$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.