Overview
The IB offers two SL mathematics courses: Analysis and Approaches (AA SL) emphasises mathematical rigour, proof, and algebraic/calculus thinking. Applications and Interpretation (AI SL) emphasises modelling, technology (GDC required), statistics, and real-world problem-solving.
Exam: Paper 1 (no GDC, 80 marks, 90 min) + Paper 2 (GDC, 80 marks, 90 min). External 80% + Internal Assessment 20%.Topic AA SL content AI SL content ~Marks 1 — Number & Algebra Sequences, logs, binomial theorem Financial maths, TVM 16 2 — Functions Domain/range, transformations, inverse Modelling, regression 21 3 — Geometry & Trig Circle, triangles, vectors 2D Voronoi, 3D trig 25 4 — Statistics & Probability Binomial, Normal, prob Chi-squared, Spearman 27 5 — Calculus Differentiation, integration Slope fields, Euler method 28
Exam format (both courses): Paper 1 (no GDC, 90 min, 80 marks) + Paper 2 (GDC required, 90 min, 80 marks). Both papers include short-answer and extended-response questions.
Number & Algebra
Topic 1 covers sequences, series, logarithms, and (for AI SL) financial mathematics. The emphasis is on recognising patterns, applying formulae, and choosing between arithmetic and geometric models.
\[u_n=u_1+(n-1)d,\quad S_n=\frac{n}{2}(u_1+u_n)=\frac{n}{2}[2u_1+(n-1)d].\]
\[u_n=u_1\,r^{n-1},\quad S_n=\frac{u_1(r^n-1)}{r-1}\;(r\neq1),\quad S_\infty=\frac{u_1}{1-r}\;(|r|<1).\]
\[a^m\cdot a^n=a^{m+n},\quad\frac{a^m}{a^n}=a^{m-n},\quad(a^m)^n=a^{mn},\quad a^0=1.\]\[\log_a(xy)=\log_a x+\log_a y,\quad\log_a(x/y)=\log_a x-\log_a y,\quad\log_a(x^n)=n\log_a x.\]\[\log_b a=\frac{\ln a}{\ln b}\quad(\text{change of base}).\]
\[(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r,\quad\binom{n}{r}=\frac{n!}{r!(n-r)!}.\]
\[\text{Compound interest: }FV=PV\!\left(1+\frac{r}{100k}\right)^{kn}.\] Use GDC TVM solver: \(N\) = total periods, \(I\%\), \(PV\), \(PMT\), \(FV\), \(P/Y=C/Y\).
Problem. The 5th term of a geometric sequence is 48 and the 8th term is 384. Find \(u_1\) and \(r.\)
\(u_5=u_1r^4=48,\quad u_8=u_1r^7=384.\)
Divide: \(r^3=384/48=8\Rightarrow r=2.\quad u_1=48/2^4=48/16=3.\)
Problem. Find the coefficient of \(x^4\) in \((2x-3)^6.\)
Term: \(\binom{6}{2}(2x)^4(-3)^2=15\cdot16x^4\cdot9=2160\,x^4.\) Coefficient: 2160.
Problem. Solve \(\log_3(x+2)+\log_3(x-4)=3.\)
\(\log_3[(x+2)(x-4)]=3\Rightarrow(x+2)(x-4)=27.\)
\(x^2-2x-8=27\Rightarrow x^2-2x-35=0\Rightarrow(x-7)(x+5)=0.\)
Reject \(x=-5\) (gives \(\log_3(-3)\) undefined). Answer: \(x=7.\)
An arithmetic sequence has first term \(a\) and common difference \(d\). The 6th term is 17 and the sum of the first 8 terms is 88. Find \(a\) and \(d\). \(u_6=a+5d=17\) (1) \(S_8=\frac{8}{2}(2a+7d)=88\Rightarrow 2a+7d=22\) (2) From (1): \(a=17-5d.\) Sub into (2): \(2(17-5d)+7d=22\Rightarrow 34-3d=22\Rightarrow d=4.\) \(a=17-20=-3.\) Answer: \(a=-3,\;d=4.\)Show solution
Worked Example · Geometric sequence
Find the 8th term and sum of the first 8 terms of the sequence 3, 6, 12, …
Solution: \(u_1=3,\; r=2\)
\(u_8 = 3 \cdot 2^{7} = 384\)
\(S_8 = \dfrac{3(2^8-1)}{2-1} = 3\times 255 = 765\)
Worked Example · Change of base
Evaluate \(\log_4 128\).
Solution: \(\log_4 128 = \dfrac{\ln 128}{\ln 4} = \dfrac{7\ln 2}{2\ln 2} = \dfrac{7}{2} = 3.5\)
Functions
Functions describe mathematical relationships. IB SL requires understanding domain, range, inverse functions, composite functions, transformations, and key function families.
Domain: set of permitted inputs. Range: set of possible outputs. Composite: \((f\circ g)(x)=f(g(x)).\) Inverse: \(f^{-1}\) exists iff \(f\) is one-to-one; \(f(f^{-1}(x))=x.\) Graphically: \(f^{-1}\) is the reflection of \(f\) over \(y=x.\)
Transformation Equation Effect Vertical shift \(+k\) \(y=f(x)+k\) Up by \(k\) Horizontal shift \(+h\) \(y=f(x-h)\) Right by \(h\) Vertical stretch \(a\) \(y=af(x)\) Scale by \(a\) vertically Horizontal stretch \(\frac{1}{b}\) \(y=f(bx)\) Scale \(x\) by \(1/b\) Reflect \(x\)-axis \(y=-f(x)\) Flip vertically Reflect \(y\)-axis \(y=f(-x)\) Flip horizontally
Family Form Key feature Linear \(f(x)=mx+c\) Constant rate of change Quadratic \(f(x)=ax^2+bx+c\) Vertex, axis of symmetry Exponential \(f(x)=ka^x+c\) HA at \(y=c\); multiplicative growth Logarithmic \(f(x)=a\log_b x+c\) VA at \(x=0\); inverse of exponential Rational \(f(x)=\frac{ax+b}{cx+d}\) VA and HA; one-to-one Sinusoidal \(f(x)=a\sin(bx+c)+d\) Amplitude \(|a|\), period \(2\pi/b\)
| x | x² | g(x) |
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Grey is the base \(f(x)=x^2\); blue is \(y=a\,f(x-h)+k\) — exactly the transformation table above, live.
Problem. \(f(x)=2x+3,\;g(x)=x^2-1.\) Find \((f\circ g)(x)\) and \((g\circ f)(x).\)
\((f\circ g)(x)=f(x^2-1)=2(x^2-1)+3=2x^2+1.\)
\((g\circ f)(x)=g(2x+3)=(2x+3)^2-1=4x^2+12x+8.\)
Problem. Find \(f^{-1}(x)\) for \(f(x)=\dfrac{2x-1}{x+3}.\)
Let \(y=\dfrac{2x-1}{x+3}.\) Rearrange: \(y(x+3)=2x-1\Rightarrow xy-2x=-1-3y\Rightarrow x(y-2)=-3y-1.\)
\(f^{-1}(x)=\dfrac{-3x-1}{x-2}=\dfrac{3x+1}{2-x}.\)
Let \(f(x)=3x-1\) and \(g(x)=x^2+2.\) Find \(f^{-1}(x)\) and solve \((g\circ f)(x)=14.\) \(f^{-1}(x)=\dfrac{x+1}{3}.\) \((g\circ f)(x)=(3x-1)^2+2=14\Rightarrow(3x-1)^2=12\Rightarrow 3x-1=\pm2\sqrt{3}.\) \(x=\dfrac{1\pm2\sqrt{3}}{3}.\)Show solution
Worked Example · Completing the square
Write \(f(x)=2x^2-8x+5\) in vertex form and state the vertex.
Solution: \(f(x) = 2(x^2-4x)+5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3\)
Vertex: \((2,-3)\)
Worked Example · Inverse function
Find \(f^{-1}(x)\) for \(f(x) = \dfrac{3x+1}{x-2},\; x\neq2\).
Solution: Let \(y=\dfrac{3x+1}{x-2}\). Swap and solve: \(x(y-3)=1+2y\implies x=\dfrac{2y+1}{y-3}\).
So \(f^{-1}(x)=\dfrac{2x+1}{x-3},\; x\neq3\).
Geometry & Trigonometry
Topic 3 covers circle mensuration, triangle trigonometry, sinusoidal functions, and (AA SL) vectors in 2D. Real-world applications include navigation, surveying, and wave modelling.
\[\text{Circumference }=2\pi r,\quad\text{Area }=\pi r^2,\quad\text{Arc length }=r\theta,\quad\text{Sector area }=\tfrac{1}{2}r^2\theta.\]
\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\quad(\text{Sine rule}).\]\[c^2=a^2+b^2-2ab\cos C\quad(\text{Cosine rule}).\]\[\text{Area}=\tfrac{1}{2}ab\sin C.\]
\[\sin^2\theta+\cos^2\theta=1,\quad\tan\theta=\frac{\sin\theta}{\cos\theta},\]\[\sin2\theta=2\sin\theta\cos\theta,\quad\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta.\]
\[\mathbf{v}=\begin{pmatrix}v_1\\v_2\end{pmatrix},\quad|\mathbf{v}|=\sqrt{v_1^2+v_2^2},\quad\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=|\mathbf{u}||\mathbf{v}|\cos\theta.\] Vector equation of a line: \(\mathbf{r}=\mathbf{a}+t\mathbf{b}.\)
Problem. The tide height (m) is modelled by \(h(t)=3\sin\!\left(\dfrac{\pi t}{6}\right)+5.\) Find: amplitude, period, max and min heights.
Amplitude \(=3\) m. Period \(=\dfrac{2\pi}{\pi/6}=12\) hours. Max \(=5+3=8\) m. Min \(=5-3=2\) m.
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Defaults reproduce Worked Example 3.1 exactly. max/min \(=k\pm A\) — drag any slider and the readouts update to match.
Problem. A vertical pole AB stands on horizontal ground. From point C, 20 m from B, the angle of elevation of A is 35°. Find the height of the pole.
\(\tan35°=\dfrac{AB}{BC}=\dfrac{AB}{20}.\quad AB=20\tan35°\approx20(0.700)=14.0\text{ m}.\)
In triangle ABC, \(AB=9\) cm, \(BC=7\) cm, \(\angle ABC=110^\circ.\) Find AC and the area of triangle ABC. Cosine rule: \(AC^2=81+49-2(9)(7)\cos110^\circ=130+43.1=173.1.\) \(AC\approx13.2\) cm. Area\(=\frac{1}{2}(9)(7)\sin110^\circ\approx29.6\) cm\(^2.\)Show solution
Worked Example · Sine rule — non-right triangle
In triangle \(ABC\), \(a=7\), \(b=9\), \(\hat{A}=40°\). Find \(\hat{B}\).
Solution: \(\dfrac{\sin B}{9} = \dfrac{\sin 40°}{7}\)
\(\sin B = \dfrac{9\sin40°}{7} \approx 0.8264\implies B\approx55.8°\)
Worked Example · Area of a triangle
Find the area of triangle \(PQR\) where \(p=8\;\text{cm}\), \(q=11\;\text{cm}\), \(\hat{R}=62°\).
Solution: \(A = \tfrac{1}{2}pq\sin R = \tfrac{1}{2}(8)(11)\sin62° \approx 38.8\;\text{cm}^2\)
Statistics & Probability
IB SL statistics covers descriptive statistics, probability laws, distributions, and correlation — with extensive use of the GDC for computations and hypothesis tests (AI SL).
\[\bar{x}=\frac{\sum f_ix_i}{\sum f_i}\quad(\text{mean from frequency table}),\quad IQR=Q_3-Q_1.\] A value is an outlier if \(x<Q_1-1.5\,IQR\) or \(x>Q_3+1.5\,IQR.\)
\[P(A\cup B)=P(A)+P(B)-P(A\cap B),\quad P(A\mid B)=\frac{P(A\cap B)}{P(B)}.\] Independent: \(P(A\cap B)=P(A)\cdot P(B).\) Mutually exclusive: \(P(A\cap B)=0.\)
Binomial \(X\sim B(n,p)\): \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\;\mu=np,\;\sigma^2=np(1-p).\) Normal \(X\sim N(\mu,\sigma^2)\): \(z=\dfrac{x-\mu}{\sigma}.\) Use GDC: Bivariate (AI SL): Pearson's correlation \(r\); LSRL \(\hat{y}=ax+b\). Use GDC for linear regression.normalcdf, invNorm.
\[\chi^2_{\text{calc}}=\sum\frac{(f_o-f_e)^2}{f_e},\quad f_e=\frac{\text{row total}\times\text{col total}}{\text{grand total}}.\] Reject \(H_0\) (independence) if \(\chi^2_{\text{calc}}>\chi^2_{\text{crit}}\) or \(p<\) significance level.
Problem. \(X\sim B(12,0.35).\) Find \(P(X\le4)\) using GDC, and find \(\mu\) and \(\sigma.\)
GDC: binomcdf(12, 0.35, 4) \(\approx0.583.\)
\(\mu=12(0.35)=4.2.\quad\sigma=\sqrt{12(0.35)(0.65)}\approx1.65.\)
Problem. Weights are \(N(68,\,5^2)\). Find \(P(60\le X\le75).\)
GDC: normalcdf(60, 75, 68, 5) \(\approx0.868.\)
Alternatively: \(z_1=(60-68)/5=-1.6,\;z_2=(75-68)/5=1.4.\)
\(P=\Phi(1.4)-\Phi(-1.6)=0.9192-0.0548=0.8644.\)
normalcdf, invNorm
A bag contains 4 red and 6 blue balls. Two are drawn without replacement. Find P(both same colour). \(P(RR)=\frac{4}{10}\cdot\frac{3}{9}=\frac{12}{90},\quadP(BB)=\frac{6}{10}\cdot\frac{5}{9}=\frac{30}{90}.\) \(P(\text{same})=\frac{42}{90}=\frac{7}{15}\approx0.467.\)Show solution
Worked Example · Normal distribution probability
\(X\sim N(50,\,9^2)\). Find \(P(44 < X < 62)\).
Solution: Standardise: \(z_1=\dfrac{44-50}{9}\approx-0.667\), \(z_2=\dfrac{62-50}{9}\approx1.333\)
\(P = \Phi(1.333)-\Phi(-0.667)\approx0.9088-0.2525=0.656\)
Worked Example · Expected value and variance
A discrete RV \(X\) takes values 1,2,3 with probabilities 0.2, 0.5, 0.3. Find \(E(X)\) and \(\text{Var}(X)\).
Solution: \(E(X)=1(0.2)+2(0.5)+3(0.3)=2.1\)
\(E(X^2)=1(0.2)+4(0.5)+9(0.3)=4.9\)
\(\text{Var}(X)=4.9-2.1^2=0.49\)
Calculus
IB SL Calculus covers limits, differentiation, integration, and applications including kinematics. AA SL goes deeper into integration techniques; AI SL focuses on graphical and numerical methods.
\[\frac{d}{dx}(x^n)=nx^{n-1},\quad\frac{d}{dx}(e^x)=e^x,\quad\frac{d}{dx}(\ln x)=\frac{1}{x},\]\[\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\cos x)=-\sin x,\quad\frac{d}{dx}(\tan x)=\sec^2 x.\]
\[\text{Product: }(uv)'=u'v+uv'.\quad\text{Quotient: }\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}.\quad\text{Chain: }\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x).\] Applications: tangent line at \((a,f(a))\): \(y-f(a)=f'(a)(x-a).\) Increasing/decreasing: \(f'>0\) increasing; \(f'<0\) decreasing. Local extrema: at critical points where \(f'=0.\)
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Curve is \(f(x)=x^3-3x\). The slope of the tangent line \(y-f(a)=f'(a)(x-a)\) is \(f'(a)\), read straight off the curve.
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\;(n\neq-1),\quad\int e^x\,dx=e^x+C,\quad\int\frac{1}{x}\,dx=\ln|x|+C.\]\[\int\sin x\,dx=-\cos x+C,\quad\int\cos x\,dx=\sin x+C.\]\[\int_a^b f(x)\,dx=F(b)-F(a)\quad(\text{Fundamental Theorem of Calculus}).\] Area between curves: \(\displaystyle\int_a^b[f(x)-g(x)]\,dx\) where \(f\ge g\) on \([a,b].\)
\[v(t)=s'(t),\quad a(t)=v'(t)=s''(t).\quad s(t)=\int v(t)\,dt.\] Displacement \(=\displaystyle\int_a^b v(t)\,dt.\) Total distance \(=\displaystyle\int_a^b|v(t)|\,dt.\)
Problem. Differentiate \(f(x)=(3x^2-1)^5.\)
Let \(u=3x^2-1\). \(f'(x)=5u^4\cdot u'=5(3x^2-1)^4\cdot6x=30x(3x^2-1)^4.\)
Problem. Find \(\displaystyle\int_0^{\pi/2}(2\sin x+3\cos x)\,dx.\)
\(\left[-2\cos x+3\sin x\right]_0^{\pi/2}=(-2\cos\tfrac{\pi}{2}+3\sin\tfrac{\pi}{2})-(-2\cos0+3\sin0)\)
\(=(0+3)-(-2+0)=3+2=5.\)
Problem. A particle has velocity \(v(t)=3t^2-12t+9\) m/s. Find: (a) when at rest; (b) displacement for \(0\le t\le4\).
(a) \(v=0:\;3(t-1)(t-3)=0.\quad t=1\) s and \(t=3\) s.
(b) \(s=\int_0^4(3t^2-12t+9)\,dt=\left[t^3-6t^2+9t\right]_0^4=(64-96+36)-0=4\) m.
Find the area enclosed between \(y=x^2\) and \(y=x+2.\) Intersect: \(x^2-x-2=0\Rightarrow x=-1,\;x=2.\) Area\(=\displaystyle\int_{-1}^{2}[(x+2)-x^2]\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}=\frac{10}{3}+\frac{7}{6}=\frac{9}{2}.\) Area \(=\frac{9}{2}\) sq units.Show solution
Worked Example · Differentiation — chain rule
Differentiate \(y=(3x^2-5)^4\).
Solution: \(\dfrac{dy}{dx} = 4(3x^2-5)^3\cdot6x = 24x(3x^2-5)^3\)
Worked Example · Definite integral
Evaluate \(\displaystyle\int_1^3 (2x^2-x+4)\,dx\).
Solution: \(\left[\dfrac{2x^3}{3}-\dfrac{x^2}{2}+4x\right]_1^3 = (18-4.5+12)-(\tfrac{2}{3}-0.5+4) = 25.5-4.167=21.3\)
Worked Example · Optimisation
A farmer has 80 m of fencing to enclose a rectangular plot. Find the maximum area.
Solution: Let width \(= x\). Then length \(= 40-x\). \(A = x(40-x) = 40x-x^2\).
\(\dfrac{dA}{dx}=40-2x=0\implies x=20\). Max area \(=400\;\text{m}^2\).
AI SL — Specific Topics
The following topics appear in Applications and Interpretation SL but not in Analysis and Approaches SL.
Selecting appropriate models from data: linear, quadratic, exponential, power, sinusoidal, or logistic. Use GDC regression to find best-fit equations; interpret \(r^2\) as the coefficient of determination.
\[r_s=1-\frac{6\sum d_i^2}{n(n^2-1)}\] where \(d_i\) is the difference in ranks. \(r_s\approx\pm1\) = strong; \(r_s\approx0\) = weak. Non-parametric alternative to Pearson's \(r.\)
Partition a plane into regions, each consisting of all points closest to a given site. Applications: nearest service location, cell coverage. The boundary between two sites is the perpendicular bisector of the segment joining them.
Slope fields: visual representation of \(\dfrac{dy}{dx}=f(x,y)\). Euler's method: \(y_{n+1}=y_n+h\cdot f(x_n,y_n),\;x_{n+1}=x_n+h.\)
Problem. Use Euler's method with \(h=0.2\) to estimate \(y(0.4)\) given \(\dfrac{dy}{dx}=x+y,\;y(0)=1.\)
Step 1: \(x_0=0,\;y_0=1.\) \(f(0,1)=0+1=1.\) \(y_1=1+0.2(1)=1.2.\)
Step 2: \(x_1=0.2,\;y_1=1.2.\) \(f(0.2,1.2)=1.4.\) \(y_2=1.2+0.2(1.4)=1.48.\)
Estimate: \(y(0.4)\approx1.48.\) (Exact: \(y=2e^x-x-1\Rightarrow y(0.4)\approx1.584.\))
Use Euler's method with \(h=0.1\) to estimate \(y(0.2)\) for \(\frac{dy}{dx}=y-x,\;y(0)=2.\) Step 1: \(f(0,2)=2.\;y_1=2+0.1(2)=2.2.\) Step 2: \(f(0.1,2.2)=2.1.\;y_2=2.2+0.1(2.1)=2.41.\) \(y(0.2)\approx2.41.\)Show solution
Worked Example · Voronoi diagram — nearest site
Points \(A(0,0)\) and \(B(6,0)\) are sites. The perpendicular bisector of \(AB\) is the Voronoi edge. Find it.
Solution: Midpoint \(= (3,0)\). \(AB\) is horizontal so the bisector is the vertical line \(x=3\).
Worked Example · Transition matrix
A Markov chain has transition matrix \(T=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\) and initial state \(s_0=\begin{pmatrix}1\\0\end{pmatrix}\). Find \(s_1\).
Solution: \(s_1 = T s_0 = \begin{pmatrix}0.7\\0.3\end{pmatrix}\).
Practice Set
The first term of a geometric sequence is 4 and the common ratio is 3. Find the sum of the first 5 terms.
Solve \(\log_2(x+3) + \log_2(x-1) = 5.\)
Find the range of \(f(x) = 3 - 2\sin(x).\)
The sum of the first \(n\) terms of an arithmetic series is \(S_n = 3n^2 + 2n.\) Find \(u_{10}\).
Solve \(2\cos^2\theta - \cos\theta - 1 = 0\) for \(0 \leq \theta \leq 2\pi.\)
A random variable \(X \sim B(12, 0.4).\) Find \(P(X \leq 3).\)
Find \(\displaystyle\int_1^4 \frac{3}{\sqrt{x}}\,dx.\)
The mean of a data set is 15 and the standard deviation is 3. A value of 21 is observed. Calculate its standardised score (z-score) and comment on whether it is an outlier.
Find the exact value of \(\displaystyle\int_0^{\pi/3} x\sin x\,dx.\)
\(\displaystyle\int x\sin x\,dx = -x\cos x + \int\cos x\,dx = -x\cos x + \sin x + C.\)
Evaluating: \(\Big[-x\cos x + \sin x\Big]_0^{\pi/3} = \left(-\dfrac{\pi}{3}\cdot\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\right) - 0 = \dfrac{\sqrt{3}}{2} - \dfrac{\pi}{6}.\)
A function is defined by \(f(x) = x^3 - 6x^2 + 9x + 1.\) Find all local maxima and minima and determine the intervals on which \(f\) is increasing.
\(f''(x) = 6x-12.\) At \(x=1\): \(f''(1)=-6<0\) → local max, \(f(1)=5.\) At \(x=3\): \(f''(3)=6>0\) → local min, \(f(3)=1.\)
\(f\) is increasing on \((-\infty,1)\cup(3,\infty).\)
Two events \(A\) and \(B\) are such that \(P(A)=0.5,\;P(B)=0.4,\;P(A\cup B)=0.7.\) Determine whether \(A\) and \(B\) are independent.
For independence: \(P(A)\cdot P(B) = 0.5\times0.4 = 0.2.\)
Since \(P(A\cap B) = P(A)\cdot P(B)\), events \(A\) and \(B\) are independent.
The graph of \(y=f(x)\) has a horizontal asymptote at \(y=2\) and passes through \((0,0).\) Given that \(f(x)=\dfrac{2x^2}{x^2+k}\) for some constant \(k>0\), find \(k\) if \(f'(1)=\tfrac{1}{2}.\)
At \(x=1\): \(\dfrac{4k}{(1+k)^2} = \dfrac{1}{2} \Rightarrow 8k = (1+k)^2 = 1+2k+k^2 \Rightarrow k^2-6k+1=0 \Rightarrow k = 3\pm 2\sqrt{2}.\)
Both values are positive; checking \(f(0)=0\) is satisfied for any \(k.\) So \(k = 3-2\sqrt{2}\approx0.17\) or \(k=3+2\sqrt{2}\approx5.83.\)
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