· H2 Mathematics — Ten-Year Series, worked ← All 11 questions

2023 A-Level · Paper 1 · Question 1

2023 H2 Maths Paper 1, Question 1

Differentiation 5 marks Applications of Differentiation
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Solution

5 marks

Differentiating implicitly with respect to xx, using the chain rule on the right-hand side (with u=11−5xu = 11 - 5x, so dudx=−5\frac{du}{dx} = - 5):

1ydydx=2(11−5x)⋅(−5)\frac{1}{y}\frac{dy}{dx} = 2(11 - 5x) \cdot ( - 5)
=−10(11−5x)= - 10(11 - 5x)

Multiplying through by yy gives a new equation for dydx\frac{dy}{dx} itself:

dydx=−10y(11−5x)\frac{dy}{dx} = - 10y(11 - 5x)

At x=2x = 2: 11−5(2)=111 - 5(2) = 1, so ln⁡y=12=1\ln y = 1^{2} = 1, giving y=ey = e.

Substituting x=2x = 2 and y=ey = e into the derivative:

dydx=−10e(1)\frac{dy}{dx} = - 10e(1)
=−10e= - 10e
To check your answer: at x=2x = 2, ln⁡y=1⇒y=e\ln y = 1 \Rightarrow y = e, and dydx=−10y(11−5x)\frac{dy}{dx} = - 10y(11 - 5x) evaluated at x=2x = 2 gives −10e(1)=−10e- 10e(1) = - 10e, consistent ✓

The tangent at (2,e)(2,e) has gradient −10e- 10e:

y−e=−10e(x−2)y - e = - 10e(x - 2)
=−10ex+20e= - 10ex + 20e

Adding ee to both sides:

y=−10ex+21ey = - 10ex + 21e

∴\therefore the equation of the tangent is y=21e−10exy = 21e - 10ex.

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