2021 H2 Maths Paper 1 Q1 — Worked Solution

Solution

5 marks

Approach

four conditions on $f(x) = ax^{3} + bx^{2} + cx + d$ give four equations in $a,b,c,d$ . Adding and subtracting the two point conditions eliminates two unknowns at once, which is faster than substituting condition by condition.

Since the graph passes through $(1,5)$ , $f(1) = 5$ :

a + b + c + d = 5\quad\text{...(1)}

Since the graph passes through $( - 1, - 3)$ , $f( - 1) = - 3$ :

- a + b - c + d = - 3\quad\text{...(2)}

Since there is a turning point at $x = 1$ , $f'(1) = 0$ . Differentiating, $f'(x) = 3ax^{2} + 2bx + c$ :

3a + 2b + c = 0\quad\text{...(3)}

Since $\int_{0}^{1}f(x)\, dx = 6$ ,

\int_{0}^{1}\left( ax^{3} + bx^{2} + cx + d \right)\, dx = \left\lbrack \frac{ax^{4}}{4} + \frac{bx^{3}}{3} + \frac{cx^{2}}{2} + dx \right\rbrack_{0}^{1}

= \frac{a}{4} + \frac{b}{3} + \frac{c}{2} + d

\frac{a}{4} + \frac{b}{3} + \frac{c}{2} + d = 6\quad\text{...(4)}

Adding (1) and (2),

2b + 2d = 2

\Rightarrow b + d = 1

Subtracting (2) from (1),

2a + 2c = 8

\Rightarrow a + c = 4

So $d = 1 - b$ and $c = 4 - a$ . Substituting into (3),

3a + 2b + (4 - a) = 0

\Rightarrow 2a + 2b + 4 = 0

\Rightarrow a + b = - 2

\Rightarrow b = - 2 - a

Then $d = 1 - b = 1 - ( - 2 - a)$ :

d = 3 + a

Substituting $b = - 2 - a$ , $c = 4 - a$ , $d = 3 + a$ into (4),

\frac{a}{4} + \frac{- 2 - a}{3} + \frac{4 - a}{2} + (3 + a) = 6

Multiplying through by $12$ (the LCM of $4,3,2$ ),

3a + 4( - 2 - a) + 6(4 - a) + 12(3 + a) = 72

\Rightarrow 3a - 8 - 4a + 24 - 6a + 36 + 12a = 72

\Rightarrow 5a + 52 = 72

\Rightarrow 5a = 20

\Rightarrow a = 4

Substituting back,

b = - 2 - 4 = - 6,\quad c = 4 - 4 = 0,\quad d = 3 + 4 = 7

To check your answer: with

f(x) = 4x^{3} - 6x^{2} + 7

f(1) = 4 - 6 + 7 = 5

✓,

f( - 1) = - 4 - 6 + 7 = - 3

✓,

f'(x) = 12x^{2} - 12x

, so

f'(1) = 12 - 12 = 0

✓, and

\int_{0}^{1}\left( 4x^{3} - 6x^{2} + 7 \right)\, dx = \left\lbrack x^{4} - 2x^{3} + 7x \right\rbrack_{0}^{1} = 1 - 2 + 7 = 6

✓.

$\therefore$ $a = 4$ , $b = - 6$ , $c = 0$ , $d = 7$ .

2021 H2 Maths Paper 1, Question 1

Solution

This is how we mark, and how we teach.