2024 H2 Maths Paper 1 Q6 — A Tangent Through the Origin, and the Area It Bounds

The curve and its tangentaiming the tangent

The constant

a

4.00

Fill the region up to

x

0.40

The curve

C

The tangent at

T

The region

Part (a)

5 marks

A curve $C$ has equation $y=2e^{x^{3}}+a$ , with $a$ a positive constant, and $T$ is the point on $C$ with $x$ -coordinate 1. Use calculus to find the equation of the tangent at $T$ , in the form $y=e(px+q)+ra$ .

Approach

A tangent needs one point and one gradient. The gradient is the chain rule applied to $e^{x^{3}}$ ; the point is the curve evaluated at $x=1$ . The prescribed form $y=e(px+q)+ra$ is a hint about the answer's shape — every $e$ should end up inside the bracket's coefficient and the constant $a$ should stand alone.

Differentiating, with the inner function $x^{3}$ :

\frac{dy}{dx}=2e^{x^{3}}\cdot3x^{2}=6x^{2}e^{x^{3}}

At $x=1$ the point and the gradient are

y(1)=2e^{1}+a=2e+a,\qquad \left.\frac{dy}{dx}\right|_{x=1}=6(1)^{2}e^{1}=6e

so $T=(1,\,2e+a)$ and the tangent through it with gradient $6e$ is

\begin{aligned}y-(2e+a)&=6e(x-1)\\y&=6ex-6e+2e+a\\y&=6ex-4e+a\\y&=e(6x-4)+a\end{aligned}

To check your answer:

At $x=1$ the tangent gives $y=e(6-4)+a=2e+a$ , which is exactly the $y$ -coordinate of $T$ ✓ — a tangent must at least pass through its own point of contact.

$\therefore\ y=e(6x-4)+a$ , so $p=6$ , $q=-4$ , $r=1$ .

Where the marks go

The constant $a$ differentiates to zero, so it does not appear in the gradient — but it very much appears in the point. Dropping it from $y(1)$ gives a parallel line through the wrong place, and part (b) then has no answer at all. The other slip is the chain rule: $\frac{d}{dx}e^{x^{3}}=3x^{2}e^{x^{3}}$ , not $e^{3x^{2}}$ and not $3x^{2}e^{3x^{2}}$ .

Part (b)(i)

1 mark

The tangent to $C$ at $T$ passes through the origin. Find the exact value of $a$ .

"Passes through the origin" means $x=0$ , $y=0$ satisfies the tangent's equation:

0=e\left(6(0)-4\right)+a=-4e+a

a=4e

To check your answer:

$a=4e\approx10.87$ , which is positive as the question requires ✓. The tangent then becomes $y=6ex-4e+4e=6ex$ — a line through the origin, as intended.

$\therefore\ a=4e$ .

Note — one mark, and it is the y-intercept

Written as $y=e(6x-4)+a$ , the tangent's $y$ -intercept is $-4e+a$ : raising $a$ lifts the tangent bodily, without changing its gradient. So there is exactly one value of $a$ that puts the intercept at zero, and no need to solve anything harder than a linear equation. Slide $a$ in the model and watch the intercept slide with it.

Part (b)(ii)

2 marks

Find the area of the region bounded by $C$ , the line $x=0$ and the tangent at $T$ . Give your answer correct to 1 decimal place.

Approach

Area between two curves is $\int(\text{upper}-\text{lower})\,dx$ , so the first job is to decide which is which and where the limits are. $\int e^{x^{3}}dx$ has no elementary antiderivative — the "1 decimal place" in the question is the examiner telling you a calculator is expected for that piece, and only that piece.

With $a=4e$ the two boundaries are

C:\ y=2e^{x^{3}}+4e,\qquad \text{tangent}:\ y=6ex

They touch at $x=1$ and are separated by the line $x=0$ on the other side, so the limits are 0 and 1. At $x=0$ the curve gives $y=2+4e\approx12.87$ while the tangent gives $y=0$ , so the curve is the upper boundary; and since a tangent to this curve meets it only at the point of contact, it stays upper throughout $0\leq x\leq1$ . Hence

\text{Area}=\int_{0}^{1}\left[\left(2e^{x^{3}}+4e\right)-6ex\right]dx

Split it into the piece that needs a calculator and the pieces that do not:

\text{Area}=\int_{0}^{1}2e^{x^{3}}\,dx+\int_{0}^{1}4e\,dx-\int_{0}^{1}6ex\,dx

\int_{0}^{1}4e\,dx=4e,\qquad \int_{0}^{1}6ex\,dx=3e\left[x^{2}\right]_{0}^{1}=3e

and the remaining integral, evaluated numerically:

\int_{0}^{1}2e^{x^{3}}\,dx\approx2.6838

\text{Area}\approx2.6838+4e-3e=2.6838+e\approx2.6838+2.7183=5.4021

To check your answer:

The region sits inside the triangle-ish box from $(0,0)$ to $(1,\,2e+4e)$ . Its height at $x=0$ is $2+4e\approx12.87$ and it narrows to nothing at $x=1$ , so an area of roughly half of $12.87\times1$ , a little over 5, is the right size ✓. Fill the region in the model and the running total lands on 5.4021.

$\therefore$ the area is $2.6838+e\approx\mathbf{5.4}$ (1 d.p.) units $^{2}$ .

Where the marks go

Reaching for a substitution or integration by parts on $\int e^{x^{3}}dx$ burns the time this two-mark part is worth: it has no answer in elementary functions, and "correct to 1 decimal place" is the signal. Equally, putting the whole integrand into the calculator in one go is fine and gets 5.4 — but separating out the exact pieces is what lets you write $2.6838+e$ and check the arithmetic afterwards.

What this question was really testing

The chain rule on an awkward exponent. $\frac{d}{dx}e^{x^{3}}=3x^{2}e^{x^{3}}$ . Everything in part (a) rests on getting the inner derivative in the right place.
A parameter that survives to the answer. $a$ vanishes from the gradient but not from the point, and part (b) exists only because it survived.
Reading the prescribed form. $y=e(px+q)+ra$ tells you how the answer should look before you have it — a free check on your algebra.
Knowing when to stop integrating. Some integrands have no elementary antiderivative. "1 decimal place" is the examiner saying so; recognising it is worth more than any technique.

A tangent aimed at the origin, and the sliver it leaves behind

Part (a)

Part (b)(i)

Part (b)(ii)

What this question was really testing

This is how we mark, and how we teach.