2016 H2 Maths Paper 1 Q8 — Worked Solution

Part (i)

8 marks

Since $f(x) = \tan(ax + b)$ , by the chain rule,

f'(x) = a\sec^{2}(ax + b)

Using the identity $\sec^{2}\theta = 1 + \tan^{2}\theta$ ,

f'(x) = a\left( 1 + \tan^{2}(ax + b) \right)

Since $y = f(x) = \tan(ax + b)$ ,

f'(x) = a + ay^{2}.\ \ \text{(shown)}

Approach

Rather than differentiating $\tan(ax + b)$ afresh, we differentiate the shown result $f'(x) = a + ay^{2}$ using the chain rule, treating $y$ as a function of $x$ and writing $y'= f'(x)$ in terms of $y$ again each time. This keeps every derivative expressed in $a$ and $y$ only, as the question asks.

Differentiating $f'(x) = a + ay^{2}$ with respect to $x$ ,

f''(x) = 2ay\, y'

Since $y'= f'(x) = a + ay^{2} = a\left( 1 + y^{2} \right)$ ,

f''(x) = 2ay \cdot a\left( 1 + y^{2} \right)

\therefore\ f''(x) = 2a^{2}y\left( 1 + y^{2} \right).

Differentiating again with respect to $x$ ,

f'''(x) = 2a^{2}\left( y'+ 3y^{2}y'\right) = 2a^{2}y'\left( 1 + 3y^{2} \right)

Substituting $y'= a\left( 1 + y^{2} \right)$ ,

f'''(x) = 2a^{2} \cdot a\left( 1 + y^{2} \right) \cdot \left( 1 + 3y^{2} \right)

\therefore\ f'''(x) = 2a^{3}\left( 1 + y^{2} \right)\left( 1 + 3y^{2} \right).

Part (ii)

Approach

The Maclaurin series needs $f(0),f'(0),f''(0),f'''(0)$ . Since $y = f(x)$ , evaluating $y$ at $x = 0$ first lets us substitute a single number into the part (i) results instead of re-differentiating.

When $b = \frac{1}{4}\pi$ , $y(0) = f(0) = \tan\left( \frac{\pi}{4} \right) = 1$ .

Substituting $y = 1$ into the part (i) results,

f'(0) = a + a(1)^{2} = 2a

f''(0) = 2a^{2}(1)\left( 1 + (1)^{2} \right) = 4a^{2}

f'''(0) = 2a^{3}\left( 1 + (1)^{2} \right)\left( 1 + 3(1)^{2} \right) = 2a^{3}(2)(4) = 16a^{3}

Using the Maclaurin series $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + \ldots$ ,

f(x) = 1 + 2ax + \frac{4a^{2}}{2!}x^{2} + \frac{16a^{3}}{3!}x^{3} + \ldots

\therefore\ f(x) = 1 + 2ax + 2a^{2}x^{2} + \frac{8}{3}a^{3}x^{3} + \ldots

Part (iii)

3 marks

Approach

$\tan 2x$ is $f(x)$ of part (i) with $a = 2$ and $b = 0$ , so instead of expanding from scratch we substitute these values into the part (i) results, this time with $y(0) = \tan(0) = 0$ .

Taking $a = 2$ , $b = 0$ in part (i), $y(0) = f(0) = \tan(0) = 0$ . Substituting $y = 0$ ,

f'(0) = a + a(0)^{2} = a = 2

f''(0) = 2a^{2}(0)\left( 1 + (0)^{2} \right) = 0

f'''(0) = 2a^{3}\left( 1 + (0)^{2} \right)\left( 1 + 3(0)^{2} \right) = 2a^{3} = 2(2)^{3} = 16

Note: $f''(0) = 0$ , so the $x^{2}$ term vanishes; the first two non-zero terms are the $x^{1}$ and $x^{3}$ terms.

Using the Maclaurin series with $f(0) = 0$ , $f'(0) = 2$ , $f''(0) = 0$ , $f'''(0) = 16$ ,

\tan 2x = 0 + 2x + \frac{0}{2!}x^{2} + \frac{16}{3!}x^{3} + \ldots = 2x + \frac{8}{3}x^{3} + \ldots

To check your answer: by GC, expanding

\tan(2x)

to this order gives

2x + \frac{8}{3}x^{3} + \frac{64}{15}x^{5} + \ldots

, matching the first two non-zero terms found here. ✓

$\therefore$ the first two non-zero terms are $2x + \frac{8}{3}x^{3}$ .

2016 H2 Maths Paper 1, Question 8

Part (i)

Part (ii)

Part (iii)

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