S6 Mathematics II 2025 Past Paper Full Worked Solutions

The 2025 NESA Advanced Level Mathematics II examination  was sat on 9 July 2025 and covers all S6 mathematics combinations: MCB, MCE, MEG, MPC, MPG, and PCM. This article provides complete, step-by-step worked solutions to all 20 questions across Section A and Section B, designed to help Rwanda A-level students master the concepts tested and prepare effectively for future national examinations.

Whether you are a private tutor in Kigali, a student studying alone, or looking for S6 mathematics revision resources, these worked solutions explain not just the answer but the reasoning behind every step. Also you can browse all learning resources in on Mathrone Academy Like  Physics II  2025 past paper solutions. 

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SECTION A: Attempt All Questions (55 Marks)

Question 1: True or False: Matrices and Linear Transformations (3 marks)

a) FALSE. A matrix with 3 rows and 2 columns has dimensions $3\times 2$, not $2\times 3$. The convention is rows × columns.

b) FALSE. The identity matrix has 1s on the main diagonal and 0s everywhere else. A matrix of all 1s is not an identity matrix.

c) FALSE. When solving $AX=B$ using inverse matrices, we multiply both sides on the left by $A^{-1}$, giving $X=A^{-1}B$, not $X=B{A}^{-1}$ (matrix multiplication is not commutative).

d) TRUE. "Not invertible" and "singular" are synonymous — both describe a matrix whose determinant equals zero.

e) FALSE. If $det(A)\ne 0$, the matrix IS invertible. A non-zero determinant guarantees the existence of $A^{-1}$.

f) FALSE. The function $f(x)=\cos x$ is not a linear transformation because it does not satisfy $f(x+y)=f(x)+f(y)$ in general.

Question 2:True or False: Differential Equations (3 marks)

The given differential equation is:

a) TRUE. The degree is determined by the highest power of the highest-order derivative. Here $\frac{dy}{dx}$ appears to the first power, so the degree is 1.

b) FALSE. The order is the order of the highest derivative present. Here only $\frac{dy}{dx}$ appears (a first derivative), so the order is 1, not 2.

c) TRUE. Rearranging: $(3x^2-1)\frac{dy}{dx}=-y$. The equation is linear in $y$ and $\frac{dy}{dx}$ — no products of $y$ and its derivative, no powers of $y$ greater than 1.

Question 3: Logarithmic Inequality (4 marks)

The solution set based on the boundary values ${\log }_2(2)=1$ and ${\log }_2(32)=5$ is:

Question 4: Particular Solution of a Differential Equation (4 marks)

Solve ${\displaystyle \frac{d\theta }{dt}}=2e^{3t-2}$, given $\theta =0$ when $t=0$.

Step 1: Separate and integrate:

Step 2 : Apply initial condition $\theta (0)=0$:

Step 3: Write the particular solution:

Answer: (d)

Question 5: Vectors in ${\mathbb{R}}^3$ (5 marks)

Given: $\mathbf{u}=-\mathbf{i}+3\mathbf{j}+4\mathbf{k}$, $\mathbf{v}=4\mathbf{i}-3\mathbf{j}+2\mathbf{k}$, $\mathbf{w}=m\mathbf{i}+\mathbf{j}+n\mathbf{k}$.

Part (a) — Find $m$ and $n$ such that $\mathbf{u}=\mathbf{v}\times \mathbf{w}$ (3 marks)

We compute $\mathbf{v}\times \mathbf{w}$ using the determinant formula:

Matching components with $\mathbf{u}=(-1,3,4)$:

From the $\mathbf{k}$ component: $4+3m=4⟹m=0$

From the $\mathbf{j}$ component: $-(4n-0)=3⟹n=-\frac{3}{4}$

From the multiple-choice options provided, the best-fit answer is:

Part (b): Area of triangle with vertices $O,\mathbf{v},\mathbf{w}$ (2 marks)

Using $\mathbf{v}=(4,-3,2)$ and $\mathbf{w}=(1,1,0)$:

Question 6: Complex Numbers: Polar and Cartesian Forms (3 marks)

Given $z=\sqrt{3}+i$ and $w=1+i$. Find $\cos {\displaystyle \frac{5\pi }{12}}$ and $\sin {\displaystyle \frac{5\pi }{12}}$.

Step 1: Multiply in Cartesian form:

Step 2: Polar forms of each:

Step 3: Extract exact trig values from $zw=2\sqrt{2}(\cos {\displaystyle \frac{5\pi }{12}}+i\sin {\displaystyle \frac{5\pi }{12}})$:

Question 7: Probability (4 marks)

A bag contains 6 red, 4 white, and 8 blue balls. Three balls are drawn at random. Find the probability that one is red, one is white, and one is blue.

Total balls: $6+4+8=18$.

Total ways to choose 3:

Favourable outcomes (1 red AND 1 white AND 1 blue):

Question 8: System of Exponential Equations (3 marks)

System: $e^x+e^y=8$ and $2e^x+e^{2y}=16$.

Let $u=e^x$ and $v=e^y$:

From (1): $u=8-v$. Substitute into (2):

So $e^y=2⟹y=\ln 2$, and $e^x=8-2=6⟹x=\ln 6$.

Question 9: Mean and Variance (3 marks)

The mean of 5 observations is 4.4 and the variance is 8.24. Three observations are 1, 2, and 6. Find the other two.

Step 1 : Find the total sum:

Known sum: $1+2+6=9$. So:

Step 2 : Use the variance formula:

Known sum of squares: $1^2+2^2+6^2=41$. So:

Step 3 : Solve. Using $(x_4+x_5{)}^2=x_4^2+2x_4{x}_5+x_5^2$:

Step 4: Form the quadratic $t^2-13t+36=0$:

Question 10 :  Finding $\overrightarrow{AC}$ (3 marks)

Given $\overrightarrow{BA}=\left(\begin{array}{c}2\\ -3\end{array}\right)$ and $\overrightarrow{BC}=\left(\begin{array}{c}-3\\ 1\end{array}\right)$.

Question 11: Line in Symmetrical Form (3 marks)

A line in symmetrical form is written as ${\displaystyle \frac{x-x_0}{a}}={\displaystyle \frac{y-y_0}{b}}={\displaystyle \frac{z-z_0}{c}}$.

For the line through the origin with direction vector $(1,1,0)$:

Question 12:  Height of a Building (Angle of Elevation) (4 marks)

Let $h$ = height of the building and $d$ = horizontal distance after moving 120 m closer.

Subtracting the first from the second:

Question 13: Summation Using Standard Results (4 marks)

Find ${\displaystyle \sum _{r=36}^{48}(r-1)(3r-2)}$.

First expand: $(r-1)(3r-2)=3r^2-5r+2$.

Using the identity ${\displaystyle \sum _{r=a}^{b}f(r)=\sum _{r=1}^{b}f(r)-\sum _{r=1}^{a-1}f(r)}$, and standard results:

For $n=48$: $\sum r^2=\frac{48\cdot 49\cdot 97}{6}=38024$, $\sum r=\frac{48\cdot 49}{2}=1176$.

For $n=35$: $\sum r^2=\frac{35\cdot 36\cdot 71}{6}=14910$, $\sum r=\frac{35\cdot 36}{2}=630$.

Question 14: Volume of Revolution (4 marks)

The area between $y=x^2-1$ and the $x$-axis is rotated 360° about the $x$-axis.

The curve crosses the $x$-axis where $x^2-1=0$, so $x=\pm 1$.

Since the integrand is even:

Question 15: Exact Values of $\sin x$ and $\cos x$ (5 marks)

Given $\tan x=-{\displaystyle \frac{12}{5}}$ and ${\displaystyle \frac{3\pi }{2}}\le x\le 2\pi$ (fourth quadrant, where $\cos x>0$, $\sin x<0$).

Step 1 : Find the hypotenuse:

$$\text{hyp}=\sqrt{5^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13$$

Step 2:  Read off exact values (adjacent = 5, opposite = −12 in Q4):

$$\boxed{{\displaystyle \cos x=\frac{5}{13},\sin x=-\frac{12}{13}}}$$

Verification: $\tan x={\displaystyle \frac{-12/13}{5/13}}=-{\displaystyle \frac{12}{5}}$ 

SECTION B : Attempt Any THREE Questions (45 Marks)

Question 16: Correlation and Regression (15 marks)

Data:

Part (a): Coefficient of Correlation

Step 1 : Compute means:

$$\overline{x}=\frac{-5-1+3+10+13}{5}=\frac{20}{5}=4,\overline{y}=\frac{33+25+17+3-3}{5}=\frac{75}{5}=15$$

Step 2 : Compute required sums:

$$\sum x_i{y}_i=(-5)(33)+(-1)(25)+(3)(17)+(10)(3)+(13)(-3)=-165-25+51+30-39=-148$$

$$\sum x_i^2=25+1+9+100+169=304,\sum y_i^2=1089+625+289+9+9=2021$$

Step 3: Compute $S_{xx}$, $S_{yy}$, $S_{xy}$:

$$S_{xy}=\sum x_i{y}_i-n\overline{x}\overline{y}=-148-5(4)(15)=-148-300=-448$$

$$S_{xx}=\sum x_i^2-n{\overline{x}}^2=304-5(16)=224$$

$$S_{yy}=\sum y_i^2-n{\overline{y}}^2=2021-5(225)=896$$

Step 4 : Pearson's correlation coefficient:

$$r=\frac{S_{xy}}{\sqrt{S_{xx}\cdot S_{yy}}}=\frac{-448}{\sqrt{224\times 896}}=\frac{-448}{\sqrt{200704}}=\frac{-448}{448}=-1$$

$$\boxed{{\displaystyle r=-1}}$$

Comment: $r=-1$ indicates a perfect negative linear correlation. As $x$ increases, $y$ decreases perfectly linearly — the strongest possible negative relationship.

Part (b): Regression Line $y$ on $x$

$$b=\frac{S_{xy}}{S_{xx}}=\frac{-448}{224}=-2$$

$$a=\overline{y}-b\overline{x}=15-(-2)(4)=23$$

$$\boxed{{\displaystyle y=23-2x}}$$

Estimate of $x$ when $y=16$:

$$16=23-2x⟹2x=7⟹\boxed{{\displaystyle x=3.5}}$$

Question 17 : Matrix Diagonalisation and Applications (15 marks)

$$A=\left(\begin{array}{cc}2& 2\\ 1& 3\end{array}\right)$$

Part (a): Eigenvalues and Eigenvectors (6 marks)

The characteristic equation is $det(A-\lambda I)=0$:

$$det\left(\begin{array}{cc}2-\lambda & 2\\ 1& 3-\lambda \end{array}\right)=(2-\lambda )(3-\lambda )-2={\lambda }^2-5\lambda +4=0$$

$$(\lambda -1)(\lambda -4)=0⟹\boxed{{\displaystyle {\lambda }_1=1,{\lambda }_2=4}}$$

Eigenvector for ${\lambda }_1=1$: Solve $(A-I)\mathbf{v}=\mathbf{0}$:

$$\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right)\mathbf{v}=\mathbf{0}⟹v_1+2v_2=0⟹v_1=-2v_2$$

$$\boxed{{\displaystyle {\mathbf{v}}_1=\left(\begin{array}{c}-2\\ 1\end{array}\right)}}$$

Eigenvector for ${\lambda }_2=4$: Solve $(A-4I)\mathbf{v}=\mathbf{0}$:

$$\left(\begin{array}{cc}-2& 2\\ 1& -1\end{array}\right)\mathbf{v}=\mathbf{0}⟹v_1=v_2$$

$$\boxed{{\displaystyle {\mathbf{v}}_2=\left(\begin{array}{c}1\\ 1\end{array}\right)}}$$

Part (b): Matrix P and its Inverse (2 marks)

$$P=\left(\begin{array}{cc}-2& 1\\ 1& 1\end{array}\right),D=P^{-1}AP=\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)$$

$$P^{-1}=\frac{1}{detP}\left(\begin{array}{cc}1& -1\\ -1& -2\end{array}\right)=\frac{1}{-3}\left(\begin{array}{cc}1& -1\\ -1& -2\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}\right)$$

Part (c): Finding $A^6$ and $f(A)$ (5 marks)

Since $A=PD{P}^{-1}$, then $A^6=P{D}^6{P}^{-1}$:

$$D^6=\left(\begin{array}{cc}{1}^6& 0\\ 0& 4^6\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 4096\end{array}\right)$$

$$A^6=\left(\begin{array}{cc}-2& 1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 4096\end{array}\right)\left(\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}\right)$$

$$\boxed{{\displaystyle A^6=\left(\begin{array}{cc}1366& 2730\\ 1365& 2731\end{array}\right)}}$$

For $f(A)$ where $f(t)=t^4-3t^3-6t^2+7t+3$:

By the Cayley-Hamilton theorem, $A$ satisfies its own characteristic equation ${\lambda }^2-5\lambda +4=0$, so $A^2=5A-4I$.

Divide $f(t)$ by the characteristic polynomial $p(t)=t^2-5t+4$:

$$f(t)=(t^2+2t)(t^2-5t+4)+(-t+3)$$

Since $p(A)=0$ (Cayley-Hamilton), the first term vanishes:

$$f(A)=-A+3I=-\left(\begin{array}{cc}2& 2\\ 1& 3\end{array}\right)+3\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\boxed{{\displaystyle \left(\begin{array}{cc}1& -2\\ -1& 0\end{array}\right)}}$$

Part (d):  Real Cube Root of A (2 marks)

We need $B$ such that $B^3=A$ with real eigenvalues. Since $A$ has eigenvalues 1 and 4:

$${\mu }_1=\sqrt[3]{1}=1,{\mu }_2=\sqrt[3]{4}$$

$$\boxed{{\displaystyle B=P\left(\begin{array}{cc}1& 0\\ 0& \sqrt[3]{4}\end{array}\right)P^{-1}}}$$

Question 18: Conics (15 marks)

Part (a) : Hyperbola from Locus Definition (7 marks)

Find the locus of all points $P(x,y)$ such that $|P{F}_1-P{F}_2|=2$ where $F_1(4,0)$ and $F_2(-4,0)$.

This is the definition of a hyperbola with:

• $2a=2⟹a=1$, so $a^2=1$
• $c=4$ (half the distance between foci), so $c^2=16$
• $b^2=c^2-a^2=16-1=15$

Since the foci are on the $x$-axis:

$$\boxed{{\displaystyle x^2-\frac{y^2}{15}=1}}$$

Part (b): Identifying the Ellipse (8 marks)

Given: $E:4x^2+9y^2-48x+72y=-144$.

Step 1: Complete the square in $x$ and $y$:

$$4(x^2-12x)+9(y^2+8y)=-144$$

$$4[(x-6{)}^2-36]+9[(y+4{)}^2-16]=-144$$

$$4(x-6{)}^2-144+9(y+4{)}^2-144=-144$$

$$4(x-6{)}^2+9(y+4{)}^2=144$$

Step 2: Divide by 144:

$$\frac{(x-6{)}^2}{36}+\frac{(y+4{)}^2}{16}=1$$

Step 3:  Identify the conic:

This is of the form ${\displaystyle \frac{(x-h{)}^2}{a^2}}+{\displaystyle \frac{(y-k{)}^2}{b^2}}=1$ with $a^2\ne b^2$, both positive — this is an ellipse.

$$\boxed{{\displaystyle \text{Ellipse with centre }(6,-4),a=6,b=4}}$$

The foci lie at distance $c=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$ from the centre along the major axis.

Question 19: Coupled Sequences (15 marks)

Given: $U_1=12$, $U_{n+1}={\displaystyle \frac{U_n+2V_n}{3}}$, $V_1=1$, $V_{n+1}={\displaystyle \frac{U_n+3V_n}{4}}$, $W_n=V_n-U_n$.

Part (a): Show $\{W_n\}$ is geometric (3 marks)

$$W_{n+1}=V_{n+1}-U_{n+1}=\frac{U_n+3V_n}{4}-\frac{U_n+2V_n}{3}$$

$$=\frac{3(U_n+3V_n)-4(U_n+2V_n)}{12}=\frac{3U_n+9V_n-4U_n-8V_n}{12}=\frac{V_n-U_n}{12}=\frac{W_n}{12}$$

Since ${\displaystyle \frac{W_{n+1}}{W_n}}={\displaystyle \frac{1}{12}}$ is constant, $\{W_n\}$ is a geometric sequence with common ratio $r={\displaystyle \frac{1}{12}}$.

Part (b): Express $W_n$ in terms of $n$ (4 marks)

$$W_1=V_1-U_1=1-12=-11$$

$$\boxed{{\displaystyle W_n=-11\cdot {\left(\frac{1}{12}\right)}^{n-1}}}$$

Part (c): Convergence and Limit of $\{W_n\}$ (3 marks)

Since $|r|={\displaystyle \frac{1}{12}}<1$, the geometric sequence converges.

$$\underset{n\to \mathrm{\infty }}{lim}{W}_n=\underset{n\to \mathrm{\infty }}{lim}-11\cdot {\left(\frac{1}{12}\right)}^{n-1}=0$$

As $n\to \mathrm{\infty }$, $W_n\to 0$, meaning $V_n-U_n\to 0$: the two sequences converge to the same limit.

Part (d): Show $U_n$ is Decreasing and $V_n$ is Increasing (5 marks)

For $\{U_n\}$:

$$U_{n+1}-U_n=\frac{U_n+2V_n}{3}-U_n=\frac{2V_n-2U_n}{3}=\frac{2W_n}{3}$$

Since $W_n=-11\cdot (1/12{)}^{n-1}<0$ for all $n$, we have $U_{n+1}-U_n<0$, so $\{U_n\}$ is decreasing.

For $\{V_n\}$:

$$V_{n+1}-V_n=\frac{U_n+3V_n}{4}-V_n=\frac{U_n-V_n}{4}=\frac{-W_n}{4}$$

Since $W_n<0$, we have $-W_n>0$, so $V_{n+1}-V_n>0$, meaning $\{V_n\}$ is increasing.

Question 20: Study of $f(x)={\displaystyle \frac{2\ln x}{x}}$ (15 marks)

Part (a): Domain (1 mark)

$$\boxed{{\displaystyle D_f=(0,+\mathrm{\infty })}}$$

Part (b): Limits at the Boundaries (2 marks)

As $x\to 0^{+}$:

$$\underset{x\to 0^{+}}{lim}\frac{2\ln x}{x}=-\mathrm{\infty }$$

As $x\to +\mathrm{\infty }$:

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{2\ln x}{x}=\underset{x\to +\mathrm{\infty }}{lim}\frac{2/x}{1}=0\text{(by L'Hôpital)}$$

Part (c): Asymptotes (2 marks)

• Vertical asymptote: $x=0$ (y-axis), since $f(x)\to -\mathrm{\infty }$ as $x\to 0^{+}$
• Horizontal asymptote: $y=0$ (x-axis), since $f(x)\to 0$ as $x\to +\mathrm{\infty }$

Part (d): First and Second Derivatives (2 marks)

$$f^{\prime }(x)=\frac{\frac{2}{x}\cdot x-2\ln x}{x^2}=\frac{2(1-\ln x)}{x^2}$$

$$f^{″}(x)=\frac{-\frac{2}{x}\cdot x^2-2(1-\ln x)\cdot 2x}{x^4}=\frac{2(2\ln x-3)}{x^3}$$

Part (e)(i): Increasing/Decreasing Intervals (2 marks)

$f^{\prime }(x)=0$ when $\ln x=1$, i.e. $x=e$.

• $f^{\prime }(x)>0$ for $x\in (0,e)$ — function is INCREASING
• $f^{\prime }(x)<0$ for $x\in (e,+\mathrm{\infty })$ — function is DECREASING
• Maximum at $x=e$: $f(e)={\displaystyle \frac{2}{e}}$

Part (e)(ii) — Concavity (2 marks)

$f^{″}(x)=0$ when $\ln x={\displaystyle \frac{3}{2}}$, i.e. $x=e^{3/2}$.

• Concave DOWNWARD on $(0,e^{3/2})$
• Concave UPWARD on $(e^{3/2},+\mathrm{\infty })$
• Inflection point at $(e^{3/2},{\displaystyle \frac{3}{e^{3/2}}})$

Part (f):Intersections with Axes (2 marks)

No y-intercept (domain excludes $x=0$).

x-intercept: $f(x)=0⟹\ln x=0⟹x=1$:

$$\boxed{{\displaystyle \text{x-intercept at }(1,0)}}$$

Part (g): Sketch Summary (2 marks)

Key features:

• Vertical asymptote: $x=0$; Horizontal asymptote: $y=0$
• Crosses x-axis at $(1,0)$
• Maximum at $(e,{\displaystyle \frac{2}{e}})\approx (2.72,0.74)$
• Inflection at $(e^{3/2},{\displaystyle \frac{3}{e^{3/2}}})\approx (4.48,0.67)$
• Curve rises from $-\mathrm{\infty }$ near the origin, crosses zero at $x=1$, reaches its peak at $x=e$, then slowly falls toward zero.

⚠️Important Disclaimer

The solutions on this page are prepared by the Mathrone Academy team for revision and learning purposes only. This is not an official NESA marking scheme or REB-approved answer guide. While every effort has been made to ensure accuracy, answers and explanations may differ from the official examiners' marking guide. Always refer to your school teacher or the official NESA publications for authoritative marking guidance. Mathrone Academy accepts no responsibility for any discrepancies between these solutions and official results. 

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