2025 IJMB mathematics paper II

1. Differentiate cos⁻¹(2 + x) sin 5x

2. Evaluate $\lim_{y \to 3} \dfrac{\sqrt{y-3}}{y^2 - 8y - 9}$

3. Solve the equation $\dfrac{dy}{dx} = \dfrac{\sin x}{\cos y}$

4. Find the direction cosines of the vectors $\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ and $\mathbf{i} + 2\mathbf{j} - \mathbf{k}$, hence find the cosine of the angle between them.

5. Find $\displaystyle\int 5x^2 e^{x^3} dx$

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SECTION B: CALCULUS

1. (a) A curve C has parametric equations: $x = t + t^{-1}$ and $y = t^m + t^{-m}$

Show that $(x^2 - 4)\left(\dfrac{dy}{dx}\right)^2 = m^2(y^2 - 4)$

(b) Find the indefinite integral $\displaystyle\int \dfrac{e^t}{(e^t - 3)(e^t + 3)} \, dt$ (You may use $x = e^t$) [08marks]

2. (a) Given that $I_n = \displaystyle\int \tan^n x \, dx$, show that $(n-1)I_n = \tan^{n-1}x - (n-1)I_{n-2}$, using integration by parts. Hence obtain $I_3$. [10marks]

(b) Differentiate $\cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$ and express in its simplest form. [10marks]

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

3. (a) Find the maxima and minima of $f(x) = x(1-x^2)$ and points at which they occur, sketch the curve for $x \in \mathbb{R}$. Hence, obtain the area bounded by the curve and the x-axis. [12marks]

(b) Differentiate from first principles $y = \cos 3x$ [08marks]

9. (a) If $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + k$, $\mathbf{b} = \mathbf{i} - 5\mathbf{j} + \mathbf{k}$ and $\mathbf{c} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$, show that $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$ [08marks]

(b) Solve the equation $\dfrac{dv}{dx} - xy = 3x$, for $y(0) = 3$ [12marks]

10. (a) Find the particular solution of $2x^3 \dfrac{dy}{dx} = y^2 - 3xy$, given that $y(1) = 1$ [12marks]

(b) Given vectors $\mathbf{a} = \mathbf{i} - \mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$, compute the projection of $(2\mathbf{a} + \mathbf{b})$ on $(\mathbf{a} - 2\mathbf{b})$ [08marks]

11. (a) Find the equation of the curve satisfying $b^2 \dfrac{dv}{y} - a^2 \dfrac{dv}{v} = 0$, passing through $(0, b)$ [10marks]

(b) Find the area of triangle ABC whose position vectors of A, B, C are $\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, $\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$, and $\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$ respectively. [10marks]

ANSWERS

SECTION A

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Q1. Differentiate $y = \cos^{-1}(2+x)\sin 5x$

Using product rule: $\dfrac{dy}{dx} = u'v + uv'$

• $u = \cos^{-1}(2+x)$, so $u' = \dfrac{-1}{\sqrt{1-(2+x)^2}}$

• $v = \sin 5x$, so $v' = 5\cos 5x$

$$\boxed{\dfrac{dy}{dx} = \dfrac{-\sin 5x}{\sqrt{1-(2+x)^2}} + 5\cos 5x \cdot \cos^{-1}(2+x)}$$

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Q2. Evaluate $\lim_{y\to 3} \dfrac{\sqrt{y-3}}{y^2-8y-9}$

Factor denominator: $y^2 - 8y - 9 = (y-9)(y+1)$

Hmm — at $y=3$: numerator $= 0$, denominator $= (3-9)(4) = -24 \neq 0$

$$\lim_{y \to 3} \dfrac{\sqrt{y-3}}{(y-9)(y+1)} = \dfrac{\sqrt{0}}{(-6)(4)} = \dfrac{0}{-24} = \boxed{0}$$

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Q3. Solve $\dfrac{dy}{dx} = \dfrac{\sin x}{\cos y}$

Separating variables:

$$\cos y \, dy = \sin x \, dx$$

$$\sin y = -\cos x + C$$

$$\boxed{\sin y + \cos x = C}$$

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Q4. Direction cosines of $\mathbf{p} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}$ and $\mathbf{q} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$

$|\mathbf{p}| = \sqrt{1+4+1} = \sqrt{6}$, $|\mathbf{q}| = \sqrt{6}$

Direction cosines of p: $\left(\dfrac{1}{\sqrt6},\ \dfrac{-2}{\sqrt6},\ \dfrac{1}{\sqrt6}\right)$

Direction cosines of q: $\left(\dfrac{1}{\sqrt6},\ \dfrac{2}{\sqrt6},\ \dfrac{-1}{\sqrt6}\right)$

Angle between them:

$$\cos\theta = \dfrac{\mathbf{p}\cdot\mathbf{q}}{|\mathbf{p}||\mathbf{q}|} = \dfrac{1-4-1}{6} = \dfrac{-4}{6} = -\dfrac{2}{3}$$

$$\boxed{\cos\theta = -\tfrac{2}{3}}$$

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Q5. $\displaystyle\int 5x^2 e^{x^3} dx$

Let $u = x^3$, $du = 3x^2\,dx$, so $x^2\,dx = \tfrac{du}{3}$

$$= 5 \cdot \dfrac{1}{3}\int e^u\,du = \dfrac{5}{3}e^{x^3} + C$$

$$\boxed{\dfrac{5}{3}e^{x^3} + C}$$

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SECTION B

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B1(b). $\displaystyle\int \dfrac{e^t}{(e^t-3)(e^t+3)}\,dt$

Let $x = e^t$, $dx = e^t\,dt$:

$$= \int \dfrac{dx}{(x-3)(x+3)} = \int \dfrac{dx}{x^2-9}$$

Partial fractions: $\dfrac{1}{(x-3)(x+3)} = \dfrac{1}{6}\left(\dfrac{1}{x-3} - \dfrac{1}{x+3}\right)$

$$= \dfrac{1}{6}\ln\left|\dfrac{x-3}{x+3}\right| + C = \boxed{\dfrac{1}{6}\ln\left|\dfrac{e^t-3}{e^t+3}\right| + C}$$

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B2(b). Differentiate $\cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$

Let $x = \tan\theta$, so $\theta = \tan^{-1}x$

$$\dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$$

So $\cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x$

$$\dfrac{d}{dx}\left[2\tan^{-1}x\right] = \dfrac{2}{1+x^2}$$

$$\boxed{\dfrac{2}{1+x^2}}$$

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B2(a). $I_3$ (using $(n-1)I_n = \tan^{n-1}x - (n-1)I_{n-2}$)

For $n=3$: $2I_3 = \tan^2 x - 2I_1$

$I_1 = \displaystyle\int \tan x\,dx = \ln|\sec x|$

$$\boxed{I_3 = \dfrac{\tan^2 x}{2} - \ln|\sec x| + C}$$

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SECTION C

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Q3(a). $f(x) = x(1-x^2) = x - x^3$

$f'(x) = 1 - 3x^2 = 0 \Rightarrow x = \pm\dfrac{1}{\sqrt{3}}$

$f''(x) = -6x$

• At $x = \dfrac{1}{\sqrt3}$: $f''< 0$ → Maximum, $f = \dfrac{2}{3\sqrt3} = \dfrac{2\sqrt3}{9}$

• At $x = -\dfrac{1}{\sqrt3}$: $f'' > 0$ → Minimum, $f = -\dfrac{2\sqrt3}{9}$

Area (curve crosses x-axis at $x = 0, \pm1$):

$$A = 2\int_0^1 (x - x^3)\,dx = 2\left[\dfrac{x^2}{2} - \dfrac{x^4}{4}\right]_0^1 = 2\left(\dfrac{1}{2}-\dfrac{1}{4}\right) = 2\cdot\dfrac{1}{4}$$

$$\boxed{A = \dfrac{1}{2} \text{ sq. units}}$$

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Q3(b). Differentiate $y = \cos 3x$ from first principles

$$\dfrac{dy}{dx} = \lim_{h\to 0}\dfrac{\cos 3(x+h) - \cos 3x}{h}$$

Using sum-to-product: $\cos A - \cos B = -2\sin\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)$

$$= \lim_{h\to 0}\dfrac{-2\sin(3x + \tfrac{3h}{2})\sin(\tfrac{3h}{2})}{h}$$

$$= \lim_{h\to 0} -2\sin\!\left(3x+\tfrac{3h}{2}\right)\cdot\dfrac{\sin(\tfrac{3h}{2})}{\tfrac{3h}{2}}\cdot\dfrac{3}{2}$$

$$= -2\sin 3x \cdot 1 \cdot \dfrac{3}{2} = \boxed{-3\sin 3x}$$

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Q9(a). Show $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}$

$\mathbf{b}+\mathbf{c} = (1+2)\mathbf{i}+(-5+1)\mathbf{j}+(1+2)\mathbf{k} = 3\mathbf{i}-4\mathbf{j}+3\mathbf{k}$

$\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = (3)(3)+(-2)(-4)+(1)(3) = 9+8+3 = 20$

$\mathbf{a}\cdot\mathbf{b} = 3-(-2)(-5)+1 = 3-10+1 = -6$…

Wait, let me recompute: $\mathbf{a}\cdot\mathbf{b} = (3)(1)+(-2)(-5)+(1)(1) = 3+10+1 = 14$

$\mathbf{a}\cdot\mathbf{c} = (3)(2)+(-2)(1)+(1)(2) = 6-2+2 = 6$

$\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c} = 14+6 = \mathbf{20}$ ✓ Shown.

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Q9(b). $\dfrac{dv}{dx} - xv = 3x$, $y(0)=3$

(Reading as $\dfrac{dy}{dx} - xy = 3x$)

Linear ODE. Integrating factor: $\mu = e^{-\int x\,dx} = e^{-x^2/2}$

$$\dfrac{d}{dx}\left(ye^{-x^2/2}\right) = 3xe^{-x^2/2}$$

$$ye^{-x^2/2} = \int 3xe^{-x^2/2}\,dx = -3e^{-x^2/2} + C$$

$$y = -3 + Ce^{x^2/2}$$

At $y(0)=3$: $3 = -3 + C \Rightarrow C = 6$

$$\boxed{y = -3 + 6e^{x^2/2}}$$

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Q10(b). Projection of $(2\mathbf{a}+\mathbf{b})$ on $(\mathbf{a}-2\mathbf{b})$

$\mathbf{a} = \mathbf{i}-\mathbf{j}-\mathbf{k}$, $\mathbf{b} = 2\mathbf{i}-3\mathbf{j}+\mathbf{k}$

$2\mathbf{a}+\mathbf{b} = (2+2)\mathbf{i}+(-2-3)\mathbf{j}+(-2+1)\mathbf{k} = 4\mathbf{i}-5\mathbf{j}-\mathbf{k}$

$\mathbf{a}-2\mathbf{b} = (1-4)\mathbf{i}+(-1+6)\mathbf{j}+(-1-2)\mathbf{k} = -3\mathbf{i}+5\mathbf{j}-3\mathbf{k}$

$|\mathbf{a}-2\mathbf{b}| = \sqrt{9+25+9} = \sqrt{43}$

Projection $= \dfrac{(2\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-2\mathbf{b})}{|\mathbf{a}-2\mathbf{b}|}$

$= \dfrac{(4)(-3)+(-5)(5)+(-1)(-3)}{\sqrt{43}} = \dfrac{-12-25+3}{\sqrt{43}} = \dfrac{-34}{\sqrt{43}}$

$$\boxed{-\dfrac{34}{\sqrt{43}} = -\dfrac{34\sqrt{43}}{43}}$$

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Q11(b). Area of triangle ABC

$A = \mathbf{i}-\mathbf{j}+2\mathbf{k}$, $B = \mathbf{i}+4\mathbf{j}-3\mathbf{k}$, $C = \mathbf{i}-3\mathbf{j}+2\mathbf{k}$

$\overrightarrow{AB} = 0\mathbf{i}+5\mathbf{j}-5\mathbf{k}$

$\overrightarrow{AC} = 0\mathbf{i}-2\mathbf{j}+0\mathbf{k}$

$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&5&-5\\0&-2&0\end{vmatrix}$

$= \mathbf{i}(5\cdot0-(-5)(-2)) - \mathbf{j}(0-0) + \mathbf{k}(0-0)$

$= \mathbf{i}(0-10) = -10\mathbf{i}$

$|\overrightarrow{AB}\times\overrightarrow{AC}| = 10$

$$\text{Area} = \dfrac{1}{2}\times 10 = \boxed{5 \text{ sq. units}}$$