Core Concept
Master Quiz

General term: \(T_{r+1} = \binom{5}{r}(2x)^{5-r}\!\left(-\tfrac{1}{x}\right)^{\!r} = \binom{5}{r}2^{5-r}(-1)^r\,x^{5-2r}\)
For \(x^3\): set \(5-2r=3 \Rightarrow r=1\)
Coefficient \(= \binom{5}{1}\cdot 2^4 \cdot(-1)^1 = 5 \times 16 \times (-1) = \mathbf{-80}\)

Let \(y = \dfrac{3x+2}{x-1}\). Swap \(x\) and \(y\):
\(x(y-1) = 3y+2 \Rightarrow xy - x = 3y+2 \Rightarrow y(x-3) = x+2\)
\(\therefore\; f^{-1}(x) = \dfrac{x+2}{x-3}\), \(x \ne 3\)
Verify: \(f\!\left(\tfrac{7}{2}\right) = \dfrac{3(3.5)+2}{3.5-1} = \dfrac{12.5}{2.5} = 5 = f(f^{-1}(5))\) ✓

Let \(u = \sin x\): \(2u^2 - u - 1 = 0 \Rightarrow (2u+1)(u-1) = 0\)
• \(\sin x = 1 \Rightarrow x = \dfrac{\pi}{2}\) (1 solution)
• \(\sin x = -\dfrac{1}{2} \Rightarrow x = \dfrac{7\pi}{6},\;\dfrac{11\pi}{6}\) (2 solutions)
Total: \(\mathbf{3}\) solutions: \(\dfrac{\pi}{2},\;\dfrac{7\pi}{6},\;\dfrac{11\pi}{6}\)

Product rule with \(u = x^2\), \(v = \ln x\):
\(\dfrac{dy}{dx} = 2x \cdot \ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x = \mathbf{x(2\ln x + 1)}\)
Check at \(x=e\): \(e(2 \cdot 1 + 1) = 3e \approx 8.15\) ✓

Integration by parts: \(u = x\), \(dv = e^x\,dx \Rightarrow du = dx\), \(v = e^x\)
\(\displaystyle\int_0^1 xe^x\,dx = \left[xe^x\right]_0^1 - \int_0^1 e^x\,dx = e - \left[e^x\right]_0^1\)
\(= e - (e - 1) = \mathbf{1}\)

\(z = 2\!\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right) = 2\!\left(\tfrac{1}{2} + i\cdot\tfrac{\sqrt{3}}{2}\right) = \mathbf{1 + \sqrt{3}\,i}\)
Verify: \(|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\) ✓, \(\arg(z) = \arctan(\sqrt{3}/1) = \pi/3\) ✓

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2)+(-1)(1)+(2)(-1) = 2-1-2 = -1\)
\(|\mathbf{d}_1| = \sqrt{1+1+4} = \sqrt{6}\), \(|\mathbf{d}_2| = \sqrt{4+1+1} = \sqrt{6}\)
\(\cos\theta = \dfrac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \dfrac{1}{\sqrt{6}\cdot\sqrt{6}} = \dfrac{1}{6}\)
Acute angle \(= \mathbf{\arccos\!\left(\dfrac{1}{6}\right) \approx 80.4°}\)

\(\sigma^2 = 16 \Rightarrow \sigma = 4\). Standardize: \(Z = \dfrac{54-50}{4} = 1\)
\(P(X>54) = P(Z>1) = 1 - \Phi(1) = 1 - 0.8413 = \mathbf{0.1587}\)
Note: \(N(50, 16)\) means variance = 16, standard deviation = 4 (IB notation).

Since \(|r| = \dfrac{1}{2} < 1\), the infinite sum exists:
\(S_\infty = \dfrac{a}{1-r} = \dfrac{3}{1-\tfrac{1}{2}} = \dfrac{3}{\tfrac{1}{2}} = \mathbf{6}\)
Check: \(3 + \tfrac{3}{2} + \tfrac{3}{4} + \cdots\) clearly converges to 6. ✓

\(\log_2[x(x-2)] = 3 \Rightarrow x(x-2) = 8 \Rightarrow x^2-2x-8=0\)
\((x-4)(x+2)=0 \Rightarrow x=4\) or \(x=-2\)
Domain check: both \(x>0\) and \(x>2\) required, so \(x>2\).
\(x=-2\) is rejected. Answer: \(\mathbf{x=4}\)
Verify: \(\log_2 4 + \log_2 2 = 2+1 = 3\) ✓

Volume of cylinder: \(V = \pi r^2 h\). Given \(h = 2r\):
\(V = \pi r^2(2r) = 2\pi r^3 = 16\pi\)
\(r^3 = 8 \Rightarrow r = \mathbf{2}\text{ cm}\)
Note: \(\sqrt[3]{8} = 2\), so option D is equivalent to option C. The answer is \(\mathbf{2}\) cm.

\(P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.2}{0.5} = \mathbf{0.4}\)
Independence test: \(P(A)\cdot P(B) = 0.4 \times 0.5 = 0.2 = P(A\cap B)\) ✓
Since \(P(A|B) = P(A)\) (both equal 0.4), \(A\) and \(B\) are independent.

\(\det\begin{pmatrix}a&2\\3&1\end{pmatrix} = a(1) - 2(3) = a - 6\)
Set equal to 5: \(a - 6 = 5 \Rightarrow \mathbf{a = 11}\)
Verify: \(\det\begin{pmatrix}11&2\\3&1\end{pmatrix} = 11 - 6 = 5\) ✓

Separate variables: \(\dfrac{dy}{y} = 2x\,dx\)
Integrate: \(\ln|y| = x^2 + C \Rightarrow y = Ae^{x^2}\)
Apply \(y(0)=3\): \(3 = Ae^0 = A\), so \(y = 3e^{x^2}\)
At \(x=1\): \(y(1) = 3e^{1^2} = \mathbf{3e}\)

Start with: \(e^u = 1 + u + \dfrac{u^2}{2!} + \dfrac{u^3}{3!} + \cdots\)
Substitute \(u = -x^2\):
\(e^{-x^2} = 1 + (-x^2) + \dfrac{(-x^2)^2}{2!} + \dfrac{(-x^2)^3}{3!} + \cdots\)
\(= \mathbf{1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots}\)

The sum for \(n = p+1\) is: \(\displaystyle\sum_{k=1}^{p+1} k = \sum_{k=1}^{p} k + (p+1) = \dfrac{p(p+1)}{2} + (p+1)\)
Now verify this equals \(\dfrac{(p+1)(p+2)}{2}\):
\(\dfrac{p(p+1)}{2} + (p+1) = (p+1)\!\left(\dfrac{p}{2}+1\right) = \dfrac{(p+1)(p+2)}{2}\) ✓
Answer: \(\mathbf{\dfrac{p(p+1)}{2} + (p+1)}\)

Write \(1+i = \sqrt{2}\,e^{i\pi/4}\) (since \(|1+i|=\sqrt{2}\), \(\arg=\pi/4\))
By De Moivre: \((1+i)^8 = (\sqrt{2})^8\cdot e^{i\cdot 8\cdot\pi/4} = 16\cdot e^{i2\pi}\)
\(e^{i2\pi} = \cos 2\pi + i\sin 2\pi = 1\)
Therefore \((1+i)^8 = 16 \times 1 = \mathbf{16}\)

\(P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7\)
\(= 120 \times 0.027 \times 0.0823543\)
\(= 120 \times 0.002223566 = \mathbf{0.2668}\)
(to 4 s.f.)

Intersections: \(x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0\)
Limits: \(x=-1\) to \(x=2\)
\(\text{Area} = \displaystyle\int_{-1}^{2}(x+2-x^2)\,dx = \left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}\)
At \(x=2\): \(2+4-\tfrac{8}{3} = \tfrac{10}{3}\). At \(x=-1\): \(\tfrac{1}{2}-2+\tfrac{1}{3} = -\tfrac{7}{6}\)
Area \(= \dfrac{10}{3} - \left(-\dfrac{7}{6}\right) = \dfrac{20}{6}+\dfrac{7}{6} = \dfrac{27}{6} = \mathbf{\dfrac{9}{2}}\)

\(\overrightarrow{AB} = (-1,1,0)\), \(\overrightarrow{AC} = (-1,0,1)\)
Normal vector: \(\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&1&0\\-1&0&1\end{vmatrix}\)
\(= \mathbf{i}(1\cdot1-0\cdot0) - \mathbf{j}((-1)(1)-0(-1)) + \mathbf{k}(0-(-1)) = (1,1,1)\)
Plane: \(1(x-1)+1(y-0)+1(z-0)=0 \Rightarrow \mathbf{x+y+z=1}\)
Verify: \(A\!:\!1{+}0{+}0=1\)✓, \(B\!:\!0{+}1{+}0=1\)✓, \(C\!:\!0{+}0{+}1=1\)✓