Year 11 Methods — Unit 1 Paper 2

The range tells us the smallest and largest values \(f\) reaches. Solve \(f(x) = -5\) and \(f(x) = 4\) to find the corresponding x-values. Match the brackets: round bracket on \(-5\) → round bracket on the x-value too; square bracket on \(4\) → square bracket on the x-value.

Period: take the coefficient of \(x\) inside the sine — here \(\dfrac{2}{5}\) — and divide \(2\pi\) by it.
Range: the amplitude (number out front) is 3, so sine oscillates between \(-3\) and \(+3\). Shift down by 2 gives a range of \([-5, 1]\).

Axis of symmetry formula: \(x = -\dfrac{\text{coefficient of }x}{2 \cdot \text{coefficient of }x^{2}}\). The coefficient of \(x\) here is \(-2b\), NOT \(2b\) — careful with the sign.

The basic tan curve has period \(\pi\) (not \(2\pi\) like sin and cos). For \(\tan(kx)\), the period is \(\dfrac{\pi}{|k|}\). Here \(k = \dfrac{\pi}{4}\).

Two transformations: the \((x + 1)\) inside means shift LEFT by 1 (opposite sign of what's written). The \(\dfrac{1}{2}\) outside means halve the y-value. So \((x, y) \to (x - 1,\ y/2)\).

Swap \(x\) and \(y\), then solve for \(y\). Take the POSITIVE square root because the original domain is \([0, \infty)\). The domain of the inverse is the range of the original — the smallest value \(g\) reaches is \(g(0) = 3\).

\(\mathbb{R}\setminus\{5\}\) means everything except \(x = 5\). So plugging in \(x = 5\) must make the function undefined — usually a zero denominator.

The graph of \(f^{-1}\) is the reflection of \(f\) in the line \(y = x\). Swap the x and y axes. Wherever the original curve goes, the inverse goes with x and y swapped.

The centre of the clock sits 15 cm above the base (since the radius is 15). The tip's height above the centre is \(10\sin(\text{angle})\). A full rotation is 60 min, so the angular speed is \(\dfrac{2\pi}{60} = \dfrac{\pi}{30}\) rad/min.
At \(t = 0\) the hand points LEFT, so the tip is at clock-centre height: \(h(0) = 15\). The sine model that starts at 0 and goes UP first matches the clockwise motion (the tip rises towards 12 first).

Reflect in the y-axis: replace \(x\) with \(-x\).
Dilate from the y-axis by factor \(\dfrac{1}{2}\): this squashes the graph horizontally, which means inputs are scaled UP by 2. So replace \(x\) with \(2x\).

For independent events, \(\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)\). Let \(p = \Pr(A)\), so \(\Pr(B) = 2p\) and \(\Pr(A \cap B) = 2p^{2}\). Now sub into the addition rule:
\(\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)\). Solve the resulting quadratic in \(p\).

'Two real solutions' means discriminant \(\Delta > 0\). Here \(a = 1,\ b = 2,\ c = -k\).
So \(\Delta = b^{2} - 4ac = 4 - 4(-k) = 4 + 4k\).

The \(x/2\) inside scales x-inputs by 2 (graph gets twice as wide). The \(+5\) outside shifts everything up by 5. So \((x, y) \to (2x,\ y + 5)\).

Both angles are in the fourth quadrant (\(\dfrac{3\pi}{2}\) to \(2\pi\)). In Q4: cos is positive, sin is negative.
Use the Pythagorean identity \(\sin^{2}\theta + \cos^{2}\theta = 1\) to find the missing function, then attach the sign based on the quadrant.

Either both red, or both blue. Add the two probabilities.
\(\Pr(RR) = \dfrac{6}{10} \times \dfrac{5}{9}\) (after one red, only 5 reds and 9 total left).
\(\Pr(BB) = \dfrac{4}{10} \times \dfrac{3}{9}\).

(a)–(c) Just substitute into \(h\). Jump point comes out at \((50, 200)\).
(d)–(e) Helicopter goes from take-off \(x=0\) to jump \(x=50\), so domain \([0, 50]\).
(f) The parachute path is \(f(x) = a(x-h)^2 + k\) where \((h,k)=(50,200)\) is the jump point. Use the point \((100,100)\) to find \(a\).
(g) Parachute is active from jump \(x=50\) to chute-pull \(x=100\), so domain \([50,100]\).
(j) Inverse: swap \(x\) and \(y\), solve for \(y\). You get \(\pm\) — pick the sign that matches the descending path \((100,100)\to(200,50)\).
(l) Test \(p(200)\). If it equals \(50\), he lands on the HQ roof.
(m) Stitch three pieces together. Use \([0, 50)\), \([50, 100)\), \([100, 200]\) so domains don't overlap.

(b) Period = time for one rotation = \(\dfrac{60}{30} = 2\,\text{s}\).
(c) Period \(= \dfrac{2\pi}{a}\), so \(a = \dfrac{2\pi}{2} = \pi\).
(e)/(f) Blade 2 is opposite Blade 1 (they're 180° apart on the same fan). So shift the curve by half a period (1 second), giving \(B_{2}(t) = \cos(\pi(t - 1)) + \dfrac{11}{10} = -\cos(\pi t) + \dfrac{11}{10}\).
(g) 'Both blades above 0.5 m' means the minimum of the two curves is above 0.5. Solve to find the times when one of them dips to exactly 0.5, then measure the gap between those crossings.

\(\Pr(Y) = \dfrac{1}{10}\), \(\Pr(N) = \dfrac{9}{10}\) for each digit. Digits are independent (with replacement).
IMPORTANT for (c): 'Guess the code INCORRECTLY' means the safe doesn't open — i.e. the code is wrong, which happens when ANY digit is wrong. So it's the COMPLEMENT of all three correct, not all three wrong.
For 'at least two correct': add the probability of all three right and the probability of exactly two right. Exactly two right has three orderings (YYN, YNY, NYY).

Total cut length: \(x + 2y = 30\), so \(y = \dfrac{30 - x}{2}\).
The cut traces out a rectangle of width \(x\) and height \(y\): \(A = x \cdot y = \dfrac{x(30-x)}{2}\). This is a downward parabola in \(x\).
Domain: both \(x \geq 5\) and \(y \geq 5\). \(y \geq 5\) gives \(\dfrac{30-x}{2} \geq 5 \Rightarrow x \leq 20\). So \(x \in [5, 20]\).
Maximum at the vertex of the parabola: \(x = 15\). Then \(y = \dfrac{30 - 15}{2} = 7.5\).