IBDP Maths SL 5.1 Concept of a limit AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
Using the quotient rule: Let \( u = (2x + a)^3 \) and \( v = (x + 5)^2 \). Then \( u’ = 6(2x + a)^2 \) and \( v’ = 2(x + 5) \).

Quotient rule: \( f'(x) = \frac{u’v – uv’}{v^2} \). \( f'(x) = \frac{6(2x + a)^2 \cdot (x + 5)^2 – (2x + a)^3 \cdot 2(x + 5)}{(x + 5)^4} \) Factor \( 2(2x + a)^2 (x + 5) \) from numerator: \( f'(x) = \frac{2(2x + a)^2 (x + 5)\big[3(x + 5) – (2x + a)\big]}{(x + 5)^4} \)

Simplify: \( f'(x) = \frac{2(2x + a)^2 (3x + 15 – 2x – a)}{(x + 5)^3} = \frac{2(2x + a)^2 (x + 15 – a)}{(x + 5)^3} \) \( \boxed{f'(x) = \dfrac{2(2x + a)^2 (x + 15 – a)}{(x + 5)^3}} \)

(b)
The gradient of the tangent at \( x = 1 \) is \( f'(1) \). This gradient also equals \( \tan(70^\circ) \) because the angle with the horizontal is \( 70^\circ \). \( \tan(70^\circ) \approx 2.747477 \).

Substitute \( x = 1 \) into \( f'(x) \): \( f'(1) = \frac{2(2 \cdot 1 + a)^2 (1 + 15 – a)}{(1 + 5)^3} = \frac{2(2 + a)^2 (16 – a)}{216} \) Set equal to \( \tan(70^\circ) \): \( \frac{2(2 + a)^2 (16 – a)}{216} = 2.747477 \) \( \frac{(2 + a)^2 (16 – a)}{108} = 2.747477 \) \( (2 + a)^2 (16 – a) \approx 296.727 \)

Expand or solve numerically: \( (4 + 4a + a^2)(16 – a) \approx 296.727 \) Solving gives: \( a \approx 2.72844 \) or \( a \approx 14.96968 \).

To three significant figures: \( \boxed{a \approx 2.73 \text{ and } a \approx 15.0} \)