Exponential and trigonometric graphs

(a) $3.2$ hours

1. $$3+(300-270)/(405-270)$$
2. 300 grams lies between the points (3, 270) and (4, 405).
3. The increase of 30 grams is 30/(405 $-$ 270) = 2/9 of this one-hour segment.
4. The estimate is 3 + 2/9 = 3.222… hours, or 3.2 hours to 1 decimal place.

• P1 Establishing $3+(300-270)/(405-270)$ or an equivalent valid method.
• A1 Correct answer: $3.2$ hours

(a) $30, 150$

1. $$180-30$$
2. The exact value sin $30^\circ$ is 1/2.
3. Sine is positive in the first and second quadrants, with equal values at x and $180^\circ$ $-$ x.
4. The solutions are $30^\circ$ and $150^\circ.$

• P1 Establishing $180-30$ or an equivalent valid method.
• A1 Correct answer: $30, 150$

(a) $a = 5$, $b = 3$

1. At $x = 0$: $y = a \times b^0 = a$, so $a = 5$.
2. At $x = 2$:
   $$5b^2 = 45 \Rightarrow b^2 = 9 \Rightarrow b = 3 \ (b > 0)$$

• P1 Using $b^0 = 1$ to get $a = 5$.
• P1 Forming $5b^2 = 45$.
• A1 $a = 5$, $b = 3$.

(b) $\frac{5}{3}$

1. $y = 5 \times 3^{-1} = \frac{5}{3}$.

• B1 The correct answer, $\frac{5}{3}$.

(a) $3498$

1. $2000 \times 1.15^4 = 2000 \times 1.74900625 = 3498.0125$, so about 3498 fish.

• M1 Writing $2000 \times 1.15^4$.
• A1 The correct answer, $3498$.

(b) $12$ years

1. Need $1.15^t > 5$.
2. $1.15^{11} = 4.65\ldots$ and $1.15^{12} = 5.35\ldots$
3. So the number first passes 10 000 after 12 years.

• P1 Setting up $2000 \times 1.15^t > 10\,000$ or trialling values of $t$.
• P1 Showing $t = 11$ gives under 10 000 (9304) and $t = 12$ gives over (10 700).
• A1 The correct answer, 12 years.

(a) $90^\circ,210^\circ$

1. Divide by the amplitude to obtain sine equal to one half.
2. Use both 30 and 150 degrees before adding the phase shift.
   $$60+150$$
3. Therefore $90^\circ,210^\circ$.

• M1 Divide by the amplitude to obtain sine equal to one half.
• M1 Use both 30 and 150 degrees before adding the phase shift.
• A1 Correct answer: $90^\circ,210^\circ$

(b) $2$

1. Sine cannot exceed 1, and the amplitude is 2.
   $$2\times 1$$
2. Therefore $2$.

• M1 Sine cannot exceed 1, and the amplitude is 2.
• A1 Correct answer: $2$

(a) $(1,7)$

1. Cosine ranges from −1 to 1 before scaling and translating.
   $$4+3\times (-1)$$
2. Therefore $(1,7)$.

• M1 Cosine ranges from −1 to 1 before scaling and translating.
• A1 Correct answer: $(1,7)$

(b) $90^\circ,270^\circ$

1. The condition reduces to cos x = 0.
2. Therefore $90^\circ,270^\circ$.

• M1 The condition reduces to cos x = 0.
• A1 Correct answer: $90^\circ,270^\circ$

(a) $1.5$

1. Use 2⁻¹ = 1/2.
   $$3\times 2^{-1}$$
2. Therefore $1.5$.

• M1 Use 2⁻¹ = 1/2.
• A1 Correct answer: $1.5$

(b) $5$

1. Divide by 3 and write 32 as a power of 2.
   $$2^x=32$$
2. Therefore $5$.

• M1 Divide by 3 and write 32 as a power of 2.
• A1 Correct answer: $5$

(c) 2^x is positive for every x, so 3 × 2^x is always greater than 0.

1. Therefore 2^x is positive for every x, so 3 × 2^x is always greater than 0.

• C1 Correct conclusion with supporting reasoning: 2^x is positive for every x, so 3 × 2^x is always greater than 0.

(a) $140^{\circ}$

1. The sine graph is symmetrical about $90^{\circ}$: $180^{\circ} - 40^{\circ} = 140^{\circ}$.

• B1 The correct answer, $140^{\circ}$.

(b) $220^{\circ}$

1. The graph from 180 to 360 is the graph from 0 to 180 turned upside down: $\sin(180^{\circ} + 40^{\circ}) = -\sin 40^{\circ}$.
2. So $180^{\circ} + 40^{\circ} = 220^{\circ}$.

• M1 Using $180^{\circ} + 40^{\circ}$ or the rotational symmetry about $(180, 0)$.
• A1 The correct answer, $220^{\circ}$.

(c) $50^{\circ}$

1. $\cos x = \sin(90^{\circ} - x)$, so $\cos 50^{\circ} = \sin 40^{\circ} = 0.643$.

• P1 Using $\cos x = \sin(90^{\circ} - x)$ (the cosine graph is the sine graph shifted by $90^{\circ}$).
• A1 The correct answer, $50^{\circ}$.