✦ Exam-Style Practice
20 Core Concept
Essential Questions
Full syllabus coverage — Algebra, Statistics, Calculus, Probability, Modelling & more. IB exam-style difficulty, instant feedback, detailed solutions.
§1 — Number & Algebra
C
Topic 1 · Number & Algebra
Arithmetic & Geometric Sequences
🔑 Key Formulae to Memorise
Arithmetic: $u_n = u_1 + (n-1)d$ | $S_n = \dfrac{n}{2}(2u_1 + (n-1)d)$
Geometric: $u_n = u_1 \cdot r^{n-1}$ | $S_n = \dfrac{u_1(r^n - 1)}{r-1}$ | $S_\infty = \dfrac{u_1}{1-r},\ |r|<1$
Geometric: $u_n = u_1 \cdot r^{n-1}$ | $S_n = \dfrac{u_1(r^n - 1)}{r-1}$ | $S_\infty = \dfrac{u_1}{1-r},\ |r|<1$
Memory tip: "AP adds, GP multiplies." For sum, always check — finite or infinite? Infinite geometric only works when $|r| < 1$.
Quick Example
AP: $u_1=3,\; d=5$. Find $S_{10}$. $\Rightarrow S_{10} = \frac{10}{2}(6+45) = 5 \times 51 = \mathbf{255}$
✓ Answer: 255
Q01
Arithmetic Sequences
Medium
An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 4$.
Find the sum of the first 20 terms, $S_{20}$.
Find the sum of the first 20 terms, $S_{20}$.
Use $S_n = \dfrac{n}{2}(2u_1 + (n-1)d)$:
$S_{20} = \dfrac{20}{2}(2 \times 7 + 19 \times 4) = 10(14 + 76) = 10 \times 90 = \mathbf{900}$
$S_{20} = \dfrac{20}{2}(2 \times 7 + 19 \times 4) = 10(14 + 76) = 10 \times 90 = \mathbf{900}$
Q02
Geometric Sequences
Medium
A geometric sequence has first term $u_1 = 2$ and common ratio $r = 3$.
Find the 6th term, $u_6$.
Find the 6th term, $u_6$.
$u_n = u_1 \cdot r^{n-1}$, so $u_6 = 2 \times 3^{6-1} = 2 \times 3^5 = 2 \times 243 = \mathbf{486}$
Q03
Logarithms
Hard
Solve the equation $2\log x = \log(3x + 4)$, where $x > 0$.
$2\log x = \log x^2$, so $\log x^2 = \log(3x+4)$
$\Rightarrow x^2 = 3x + 4 \Rightarrow x^2 - 3x - 4 = 0$
$\Rightarrow (x-4)(x+1) = 0 \Rightarrow x = 4$ or $x = -1$
Since $x > 0$, we reject $x = -1$. Answer: $\mathbf{x = 4}$
$\Rightarrow x^2 = 3x + 4 \Rightarrow x^2 - 3x - 4 = 0$
$\Rightarrow (x-4)(x+1) = 0 \Rightarrow x = 4$ or $x = -1$
Since $x > 0$, we reject $x = -1$. Answer: $\mathbf{x = 4}$
§2 — Functions
C
Topic 2 · Functions
Composite & Inverse Functions
🔑 Key Concepts
Composite: $(f \circ g)(x) = f(g(x))$ — apply $g$ first, then $f$
Inverse: $f^{-1}(x)$: swap $x \leftrightarrow y$, solve for $y$
Domain of $f^{-1}$ = Range of $f$
Inverse: $f^{-1}(x)$: swap $x \leftrightarrow y$, solve for $y$
Domain of $f^{-1}$ = Range of $f$
Memory tip: "Composite reads right to left." $(f \circ g)(x)$: first do $g$, then $f$. Like putting on socks, then shoes.
Quick Example
$f(x) = x^2,\ g(x) = x+1$. $(f \circ g)(3) = f(4) = \mathbf{16}$
✓ NOT $g(f(3)) = g(9) = 10$
Q04
Composite Functions
Medium
Let $f(x) = 2x + 1$ and $g(x) = x^2 - 3$.
Find the value of $(f \circ g)(2)$.
Find the value of $(f \circ g)(2)$.
Step 1: $g(2) = 2^2 - 3 = 4 - 3 = 1$
Step 2: $f(g(2)) = f(1) = 2(1) + 1 = \mathbf{3}$
Step 2: $f(g(2)) = f(1) = 2(1) + 1 = \mathbf{3}$
§3 — Statistics & Probability
C
Topic 4 · Statistics & Probability
Normal Distribution & Binomial Distribution
🔑 Key Formulae
Normal: $X \sim N(\mu, \sigma^2)$, standardise $z = \dfrac{x-\mu}{\sigma}$
Binomial: $X \sim B(n,p)$, $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
Mean $= np$, Variance $= np(1-p)$
Binomial: $X \sim B(n,p)$, $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
Mean $= np$, Variance $= np(1-p)$
Memory tip: Normal = continuous, bell-shaped. Binomial = discrete, fixed $n$ trials, each success/fail. Always check: is this counting (binomial) or measuring (normal)?
Q05
Normal Distribution
Medium
The heights of students follow a normal distribution with mean $\mu = 50$ kg and standard deviation $\sigma = 8$ kg. Find $P(X > 58)$.
Standardise: $z = \dfrac{58-50}{8} = \dfrac{8}{8} = 1$
$P(X > 58) = P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = \mathbf{0.1587}$
(Using GDC or standard normal tables.)
$P(X > 58) = P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = \mathbf{0.1587}$
(Using GDC or standard normal tables.)
Q06
Binomial Distribution
Medium
A fair die is rolled 10 times. The probability of rolling a 6 on any roll is $p = \frac{1}{3}$. Let $X \sim B(10, 0.3)$.
Find $P(X = 3)$, correct to 4 decimal places.
Find $P(X = 3)$, correct to 4 decimal places.
$P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7$
$= 120 \times 0.027 \times 0.0824 = \mathbf{0.2668}$
GDC: binompdf(10, 0.3, 3) $= 0.2668$
$= 120 \times 0.027 \times 0.0824 = \mathbf{0.2668}$
GDC: binompdf(10, 0.3, 3) $= 0.2668$
Q07
Chi-Squared Test
Medium
A chi-squared test of independence is performed on a contingency table with 3 rows and 4 columns. What is the number of degrees of freedom?
Degrees of freedom $= (r-1)(c-1) = (3-1)(4-1) = 2 \times 3 = \mathbf{6}$
where $r$ = number of rows, $c$ = number of columns.
where $r$ = number of rows, $c$ = number of columns.
Q08
Conditional Probability
Hard
Events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B \mid A) = 0.6$, $P(B \mid A') = 0.3$.
Find $P(A \mid B)$, correct to 4 significant figures.
Find $P(A \mid B)$, correct to 4 significant figures.
Total law of probability:
$P(B) = P(B|A)P(A) + P(B|A')P(A') = 0.6(0.4) + 0.3(0.6) = 0.24 + 0.18 = 0.42$
Bayes' theorem:
$P(A|B) = \dfrac{P(B|A) \cdot P(A)}{P(B)} = \dfrac{0.6 \times 0.4}{0.42} = \dfrac{0.24}{0.42} \approx \mathbf{0.5714}$
$P(B) = P(B|A)P(A) + P(B|A')P(A') = 0.6(0.4) + 0.3(0.6) = 0.24 + 0.18 = 0.42$
Bayes' theorem:
$P(A|B) = \dfrac{P(B|A) \cdot P(A)}{P(B)} = \dfrac{0.6 \times 0.4}{0.42} = \dfrac{0.24}{0.42} \approx \mathbf{0.5714}$
Q09
Poisson Distribution
Hard
The number of goals scored per match follows a Poisson distribution with mean $\lambda = 3$.
Find $P(X \leq 2)$, correct to 4 decimal places.
Find $P(X \leq 2)$, correct to 4 decimal places.
$P(X \leq 2) = P(0)+P(1)+P(2)$
$= e^{-3}\!\left(\dfrac{3^0}{0!}+\dfrac{3^1}{1!}+\dfrac{3^2}{2!}\right) = e^{-3}(1+3+4.5) = 0.0498 \times 8.5 = \mathbf{0.4232}$
GDC: poisscdf(3, 2) $= 0.4232$
$= e^{-3}\!\left(\dfrac{3^0}{0!}+\dfrac{3^1}{1!}+\dfrac{3^2}{2!}\right) = e^{-3}(1+3+4.5) = 0.0498 \times 8.5 = \mathbf{0.4232}$
GDC: poisscdf(3, 2) $= 0.4232$
§4 — Calculus
C
Topic 5 · Calculus
Differentiation & Integration Essentials
🔑 Key Rules
Power rule: $\dfrac{d}{dx}[x^n] = nx^{n-1}$ | $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$
Exponential: $\dfrac{d}{dx}[e^{kx}] = ke^{kx}$ | Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$
Local max: $f'(x)=0$ and $f''(x)<0$ | Local min: $f'(x)=0$ and $f''(x)>0$
Exponential: $\dfrac{d}{dx}[e^{kx}] = ke^{kx}$ | Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$
Local max: $f'(x)=0$ and $f''(x)<0$ | Local min: $f'(x)=0$ and $f''(x)>0$
Memory tip: "Derivative → bring down, reduce power. Integral → increase power, divide by new power." Second derivative test: negative = max (frowning curve), positive = min (smiling curve).
Q10
Differentiation
Medium
Let $f(x) = 3e^{2x} - 5x$. Find $f'(1)$, correct to 3 significant figures.
$f'(x) = 6e^{2x} - 5$
$f'(1) = 6e^2 - 5 = 6(7.389) - 5 = 44.334 - 5 = \mathbf{39.3}$ (3 s.f.)
$f'(1) = 6e^2 - 5 = 6(7.389) - 5 = 44.334 - 5 = \mathbf{39.3}$ (3 s.f.)
Q11
Integration
Medium
Evaluate $\displaystyle\int_0^2 (3x^2 + 2x - 1)\,dx$.
$\displaystyle\int_0^2(3x^2+2x-1)\,dx = \Big[x^3 + x^2 - x\Big]_0^2$
$= (8 + 4 - 2) - (0) = \mathbf{10}$
$= (8 + 4 - 2) - (0) = \mathbf{10}$
Q12
Max & Min
Hard
Let $f(x) = x^3 - 6x^2 + 9x + 2$. The local maximum value of $f$ is:
$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) = 0$
Critical points: $x = 1$ and $x = 3$
$f''(x) = 6x - 12$: $f''(1) = -6 < 0$ → local maximum
$f(1) = 1 - 6 + 9 + 2 = \mathbf{6}$ ✓
Critical points: $x = 1$ and $x = 3$
$f''(x) = 6x - 12$: $f''(1) = -6 < 0$ → local maximum
$f(1) = 1 - 6 + 9 + 2 = \mathbf{6}$ ✓
Q13
Numerical Integration
Hard
Use the trapezoid rule with $n = 4$ equal strips to approximate $\displaystyle\int_0^4 (x^2 + 1)\,dx$.
$h = \frac{4-0}{4} = 1$. Values at $x = 0,1,2,3,4$:
$f(0)=1,\ f(1)=2,\ f(2)=5,\ f(3)=10,\ f(4)=17$
$T = \frac{h}{2}[f_0 + 2(f_1+f_2+f_3) + f_4]$
$= \frac{1}{2}[1 + 2(2+5+10) + 17] = \frac{1}{2}[1+34+17] = \frac{52}{2} = \mathbf{26.0}$
(Exact value: $\frac{76}{3} \approx 25.33$; trapezoid overestimates for convex functions.)
$f(0)=1,\ f(1)=2,\ f(2)=5,\ f(3)=10,\ f(4)=17$
$T = \frac{h}{2}[f_0 + 2(f_1+f_2+f_3) + f_4]$
$= \frac{1}{2}[1 + 2(2+5+10) + 17] = \frac{1}{2}[1+34+17] = \frac{52}{2} = \mathbf{26.0}$
(Exact value: $\frac{76}{3} \approx 25.33$; trapezoid overestimates for convex functions.)
Q14
Differential Equations
Hard
Given $\dfrac{dy}{dx} = x + y$ with $y(0) = 1$, use Euler's method with step $h = 0.1$ to find the approximate value of $y(0.2)$.
Euler: $y_{n+1} = y_n + h \cdot f(x_n, y_n)$ where $f(x,y)=x+y$
Step 1: $y_1 = 1 + 0.1(0+1) = 1 + 0.1 = 1.1$
Step 2: $y_2 = 1.1 + 0.1(0.1+1.1) = 1.1 + 0.1(1.2) = 1.1 + 0.12 = \mathbf{1.22}$
Step 1: $y_1 = 1 + 0.1(0+1) = 1 + 0.1 = 1.1$
Step 2: $y_2 = 1.1 + 0.1(0.1+1.1) = 1.1 + 0.1(1.2) = 1.1 + 0.12 = \mathbf{1.22}$
§5 — Geometry & Trigonometry
C
Topic 3 · Geometry & Trigonometry
Sine Rule, Cosine Rule & Area
🔑 Key Formulae
Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$
Area: $\text{Area} = \dfrac{1}{2}ab\sin C$
Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$
Area: $\text{Area} = \dfrac{1}{2}ab\sin C$
Memory tip: Use sine rule when you have angle-side pairs. Use cosine rule when you have SAS (2 sides + included angle) or SSS (all 3 sides). Area formula needs 2 sides and included angle.
Q15
Sine Rule
Medium
In triangle $ABC$, $a = 8$ cm, $\hat{A} = 40°$, $\hat{B} = 65°$. Find the length $b$, correct to 3 significant figures.
Sine rule: $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$
$b = \dfrac{8 \times \sin 65°}{\sin 40°} = \dfrac{8 \times 0.9063}{0.6428} = \dfrac{7.250}{0.6428} \approx \mathbf{11.3}$ cm
$b = \dfrac{8 \times \sin 65°}{\sin 40°} = \dfrac{8 \times 0.9063}{0.6428} = \dfrac{7.250}{0.6428} \approx \mathbf{11.3}$ cm
§6 — Matrices & Graphs
C
Topic 1 HL · Matrices
Determinants, Inverses & Systems
🔑 Key Formulae
$\det\begin{pmatrix}a&b\\c&d\end{pmatrix} = ad - bc$
$A^{-1} = \dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ (exists iff $\det A \neq 0$)
System $A\mathbf{x} = \mathbf{b}$: unique solution iff $\det A \neq 0$
$A^{-1} = \dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ (exists iff $\det A \neq 0$)
System $A\mathbf{x} = \mathbf{b}$: unique solution iff $\det A \neq 0$
Memory tip: $\det = ad - bc$: "main diagonal minus off diagonal." If determinant is zero, matrix is singular — no inverse exists, system may have no solution or infinitely many.
Q16
Matrix Determinant
Medium
Find the determinant of the matrix $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.
$\det(A) = (3)(4) - (1)(2) = 12 - 2 = \mathbf{10}$
Since $\det A \neq 0$, the matrix is invertible.
Since $\det A \neq 0$, the matrix is invertible.
Q17
Graph Theory · MST
Hard
A weighted graph has vertices $A$, $B$, $C$, $D$ with edges:
$AB=4,\ AC=6,\ BC=3,\ BD=5,\ CD=2$
Using Kruskal's algorithm, find the total weight of the minimum spanning tree (MST).
$AB=4,\ AC=6,\ BC=3,\ BD=5,\ CD=2$
Using Kruskal's algorithm, find the total weight of the minimum spanning tree (MST).
Kruskal's: sort edges by weight ascending:
$CD=2$ ✓ add (connects C–D)
$BC=3$ ✓ add (connects B–C–D)
$AB=4$ ✓ add (connects A–B–C–D — all 4 vertices, MST complete!)
Skip $BD=5$ (creates cycle B–C–D–B) and $AC=6$ (creates cycle).
Total MST weight $= 2+3+4 = \mathbf{9}$
$CD=2$ ✓ add (connects C–D)
$BC=3$ ✓ add (connects B–C–D)
$AB=4$ ✓ add (connects A–B–C–D — all 4 vertices, MST complete!)
Skip $BD=5$ (creates cycle B–C–D–B) and $AC=6$ (creates cycle).
Total MST weight $= 2+3+4 = \mathbf{9}$
§7 — Modelling & Regression
C
Topic 4 · Statistics
Correlation, Regression & Mathematical Models
🔑 Key Concepts
Pearson's $r$: $-1 \leq r \leq 1$
$|r| > 0.85$ → strong | $|r| < 0.5$ → weak | $r > 0$ → positive
Regression line: $\hat{y} = a + bx$ passes through $(\bar{x}, \bar{y})$
Logistic model: $P(t) = \dfrac{L}{1 + Ce^{-kt}}$, carrying capacity $= L$
$|r| > 0.85$ → strong | $|r| < 0.5$ → weak | $r > 0$ → positive
Regression line: $\hat{y} = a + bx$ passes through $(\bar{x}, \bar{y})$
Logistic model: $P(t) = \dfrac{L}{1 + Ce^{-kt}}$, carrying capacity $= L$
Memory tip: "Correlation is NOT causation." Regression line always passes through the mean point. Logistic models have an S-shape — growth slows as population nears carrying capacity $L$.
Q18
Correlation Coefficient
Medium
A data set yields a Pearson's correlation coefficient of $r = 0.87$. Which statement best describes this?
$r = 0.87 > 0$ → positive correlation
$|r| = 0.87 > 0.85$ → strong linear correlation
Answer: Strong positive linear correlation ✓
Note: $r$ does not imply causation.
$|r| = 0.87 > 0.85$ → strong linear correlation
Answer: Strong positive linear correlation ✓
Note: $r$ does not imply causation.
Q19
Logistic Model
Hard
A population is modelled by $N(t) = \dfrac{1000}{1 + 49e^{-0.4t}}$, where $t$ is in years.
What is the initial population $N(0)$ and the carrying capacity?
What is the initial population $N(0)$ and the carrying capacity?
$N(0) = \dfrac{1000}{1+49e^0} = \dfrac{1000}{1+49} = \dfrac{1000}{50} = \mathbf{20}$
As $t \to \infty$: $e^{-0.4t} \to 0$, so $N(\infty) = \dfrac{1000}{1+0} = \mathbf{1000}$
The carrying capacity $L = 1000$.
As $t \to \infty$: $e^{-0.4t} \to 0$, so $N(\infty) = \dfrac{1000}{1+0} = \mathbf{1000}$
The carrying capacity $L = 1000$.
§8 — Optimisation & Linear Programming
C
Topic 2 · Functions
Linear Programming & Optimisation
🔑 Key Method
1. Identify objective function $z = ax + by$
2. Graph all constraint inequalities → find feasible region
3. Find vertices (corner points) of feasible region
4. Evaluate $z$ at each vertex → max/min is at a vertex
2. Graph all constraint inequalities → find feasible region
3. Find vertices (corner points) of feasible region
4. Evaluate $z$ at each vertex → max/min is at a vertex
Memory tip: "The optimal solution always occurs at a corner point of the feasible region." Find all intersections of boundary lines (solve simultaneous equations), then test them all in the objective function.
Q20
Linear Programming
Hard
Maximise $z = 5x + 4y$ subject to:
$x + y \leq 6$, $3x + 2y \leq 15$, $x \geq 0$, $y \geq 0$
What is the maximum value of $z$?
$x + y \leq 6$, $3x + 2y \leq 15$, $x \geq 0$, $y \geq 0$
What is the maximum value of $z$?
Vertices of feasible region:
$(0,0): z=0$ | $(5,0): z=25$ | $(0,6): z=24$
Intersection of $x+y=6$ and $3x+2y=15$:
$y=6-x \Rightarrow 3x+2(6-x)=15 \Rightarrow x+12=15 \Rightarrow x=3, y=3$
$(3,3): z=5(3)+4(3)=15+12=\mathbf{27}$ ← Maximum!
Check: $3+3=6\leq6$ ✓, $9+6=15\leq15$ ✓
$(0,0): z=0$ | $(5,0): z=25$ | $(0,6): z=24$
Intersection of $x+y=6$ and $3x+2y=15$:
$y=6-x \Rightarrow 3x+2(6-x)=15 \Rightarrow x+12=15 \Rightarrow x=3, y=3$
$(3,3): z=5(3)+4(3)=15+12=\mathbf{27}$ ← Maximum!
Check: $3+3=6\leq6$ ✓, $9+6=15\leq15$ ✓
🏆
Quiz Complete!
0
Correct
0
Incorrect
0%
Score
–
Time Used
IB GRADE SCALE: 1 — 2 — 3 — 4 — 5 — 6 — 7
Answer Key & Solutions
Q01 · Arithmetic Sequences
Answer: B — S₂₀ = 900
S₂₀ = (20/2)(2×7 + 19×4) = 10(14+76) = 10×90 = 900
S₂₀ = (20/2)(2×7 + 19×4) = 10(14+76) = 10×90 = 900
Q02 · Geometric Sequences
Answer: C — 486
u₆ = 2 × 3⁵ = 2 × 243 = 486
u₆ = 2 × 3⁵ = 2 × 243 = 486
Q03 · Logarithms
Answer: B — x = 4
log x² = log(3x+4) → x²=3x+4 → (x−4)(x+1)=0; reject x=−1
log x² = log(3x+4) → x²=3x+4 → (x−4)(x+1)=0; reject x=−1
Q04 · Composite Functions
Answer: B — 3
g(2)=4−3=1; f(1)=2(1)+1=3
g(2)=4−3=1; f(1)=2(1)+1=3
Q05 · Normal Distribution
Answer: A — 0.1587
z=(58−50)/8=1; P(Z>1)=1−0.8413=0.1587
z=(58−50)/8=1; P(Z>1)=1−0.8413=0.1587
Q06 · Binomial Distribution
Answer: C — 0.2668
P(X=3)=C(10,3)(0.3)³(0.7)⁷=120×0.027×0.0824=0.2668
P(X=3)=C(10,3)(0.3)³(0.7)⁷=120×0.027×0.0824=0.2668
Q07 · Chi-Squared Test
Answer: C — 6
df=(3−1)(4−1)=2×3=6
df=(3−1)(4−1)=2×3=6
Q08 · Conditional Probability
Answer: C — 0.5714
P(B)=0.6(0.4)+0.3(0.6)=0.42; P(A|B)=0.24/0.42≈0.5714
P(B)=0.6(0.4)+0.3(0.6)=0.42; P(A|B)=0.24/0.42≈0.5714
Q09 · Poisson Distribution
Answer: B — 0.4232
P(X≤2)=e⁻³(1+3+4.5)=0.0498×8.5=0.4232
P(X≤2)=e⁻³(1+3+4.5)=0.0498×8.5=0.4232
Q10 · Differentiation
Answer: C — 39.3
f'(x)=6e²ˣ−5; f'(1)=6e²−5≈39.3
f'(x)=6e²ˣ−5; f'(1)=6e²−5≈39.3
Q11 · Integration
Answer: C — 10
[x³+x²−x]₀²=(8+4−2)−0=10
[x³+x²−x]₀²=(8+4−2)−0=10
Q12 · Max & Min
Answer: B — f=6 at x=1
f'(x)=3(x−1)(x−3)=0; x=1: f''=−6<0 → max; f(1)=6
f'(x)=3(x−1)(x−3)=0; x=1: f''=−6<0 → max; f(1)=6
Q13 · Numerical Integration
Answer: B — 26.0
T=(1/2)[1+2(2+5+10)+17]=(1/2)(52)=26.0
T=(1/2)[1+2(2+5+10)+17]=(1/2)(52)=26.0
Q14 · Differential Equations
Answer: C — 1.22
y₁=1+0.1(1)=1.1; y₂=1.1+0.1(1.2)=1.22
y₁=1+0.1(1)=1.1; y₂=1.1+0.1(1.2)=1.22
Q15 · Sine Rule
Answer: B — 11.3 cm
b=8×sin65°/sin40°=7.250/0.6428≈11.3 cm
b=8×sin65°/sin40°=7.250/0.6428≈11.3 cm
Q16 · Matrix Determinant
Answer: C — 10
det=3×4−1×2=12−2=10
det=3×4−1×2=12−2=10
Q17 · Graph Theory MST
Answer: B — 9
Kruskal: CD=2+BC=3+AB=4=9; skip BD and AC (cycles)
Kruskal: CD=2+BC=3+AB=4=9; skip BD and AC (cycles)
Q18 · Correlation
Answer: B — Strong positive linear correlation
r=0.87>0 (positive); |r|=0.87>0.85 (strong)
r=0.87>0 (positive); |r|=0.87>0.85 (strong)
Q19 · Logistic Model
Answer: B — N(0)=20, capacity=1000
N(0)=1000/(1+49)=20; as t→∞, N→1000
N(0)=1000/(1+49)=20; as t→∞, N→1000
Q20 · Linear Programming
Answer: C — z=27 at (3,3)
Corner (3,3): z=5(3)+4(3)=27; verify 3+3=6≤6✓, 9+6=15≤15✓
Corner (3,3): z=5(3)+4(3)=27; verify 3+3=6≤6✓, 9+6=15≤15✓