### SECTION A
1. Differentiate tan⁻¹(1 − x²) cos 3x. [04marks]
2. Evaluate $\lim_{x \to 1} \frac{x^2+1}{x^2-1}$. [04marks]
3. Solve the equation y(1 + x²)dy + (1 + y²)dx = 0. [05marks]
4. Find the direction cosines of the vector **i** − 2**j** − **k**. [03marks]
5. Evaluate $\int_0^1 x^2 e^{x^3} dx$. [04marks]
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### SECTION B: CALCULUS
6. (a) If a body moves according to the law S = e^(3t) − 9te^(−t) − 4t cos t, find the velocity and acceleration at t = 0 sec. [08marks]
(b) Find ∫e^(−2x) cos 3x dx. [12marks]
7. (a) Find the area of the region bounded by the curves y = x² − 2x and y = 2x − x². [10marks]
(b) Given that cos y = y/x, find d²y/dx² at the point (1, 0). [10marks]
8. (a) Differentiate from first principles y = sin x cos x. [10marks]
(b) Find a better approximation to the root of the equation eˣ − 3x + 1 = 0 if x₀ = 0.000 is an initial approximation and f′(x₀) ≠ 0, up to 3 iterations. [10marks]
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### SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
9. (a) Solve the equation (x² sec x tan x + x sec x)dx = xy dy given that y(π) = 0. [10marks]
(b) Show that the point P = 2**i** − 3**j** + 4**k** lies on the line **r** = **i** − 2**j** + **k** + t(**i** − **j** + 3**k**). Let A = 3**i** − 5**j** + 3**k**. Show that AP is perpendicular to **i** − **j** + 3**k**, hence find the distance of the given line from A. [10marks]
10. (a) The position vectors of the vertices of triangle PQR are 2**i** + 3**j** − 5**k**, −3**i** − 7**j** + 10**k** and −3**i** − 5**j** + 10**k** respectively. Find the position vectors of the points S and T on PQ and PR respectively such that PS/QS = PT/RT = 1/2. Show that ST ∥ QR. [12marks]
(b) For triangle PQR in part (a) above, calculate the cosine of angle Q and hence the area of the triangle. [08marks]
11. (a) Solve the equation x² dy/dx = x² + 2xy + y², given that y(1) = 0. [12marks]
(b) Find the equation of the curve through (0, π) for which the normal at any point (x, y) has the slope −sin y · eˣ and y(0) = π. [08marks]
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# FULL SOLUTIONS
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## SECTION A
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### Q1. Differentiate tan⁻¹(1 − x²) cos 3x
Using the **product rule**: if u = tan⁻¹(1 − x²), v = cos 3x
$$\frac{du}{dx} = \frac{-2x}{1+(1-x^2)^2}$$
$$\frac{dv}{dx} = -3\sin 3x$$
$$\frac{dy}{dx} = \cos 3x \cdot \frac{-2x}{1+(1-x^2)^2} + \tan^{-1}(1-x^2)\cdot(-3\sin 3x)$$
$$\boxed{\frac{dy}{dx} = \frac{-2x\cos 3x}{1+(1-x^2)^2} - 3\sin 3x \cdot \tan^{-1}(1-x^2)}$$
---
### Q2. Evaluate $\lim_{x \to 1} \frac{x^2+1}{x^2-1}$
Substituting x = 1 directly:
Numerator: 1 + 1 = 2
Denominator: 1 − 1 = 0
The denominator → 0 while numerator → 2 ≠ 0.
Check one-sided limits:
- As x → 1⁺: x²−1 → 0⁺, so limit → **+∞**
- As x → 1⁻: x²−1 → 0⁻, so limit → **−∞**
Since left and right limits differ:
$$\boxed{\lim_{x \to 1} \frac{x^2+1}{x^2-1} \text{ does not exist (DNE)}}$$
---
### Q3. Solve y(1 + x²)dy + (1 + y²)dx = 0
Separate variables:
$$\frac{y\,dy}{1+y^2} = -\frac{dx}{1+x^2}$$
Integrate both sides:
$$\frac{1}{2}\ln(1+y^2) = -\tan^{-1}x + C$$
$$\boxed{\ln(1+y^2) = -2\tan^{-1}x + C}$$
Or equivalently: $(1+y^2) = Ae^{-2\tan^{-1}x}$
---
### Q4. Direction cosines of **v** = **i** − 2**j** − **k**
Magnitude: |**v**| = √(1² + (−2)² + (−1)²) = √(1+4+1) = **√6**
Direction cosines:
$$l = \frac{1}{\sqrt{6}}, \quad m = \frac{-2}{\sqrt{6}}, \quad n = \frac{-1}{\sqrt{6}}$$
$$\boxed{\left(\frac{1}{\sqrt{6}},\ \frac{-2}{\sqrt{6}},\ \frac{-1}{\sqrt{6}}\right)}$$
---
### Q5. Evaluate $\int_0^1 x^2 e^{x^3} dx$
Let u = x³ → du = 3x² dx → x² dx = du/3
When x=0, u=0; when x=1, u=1:
$$\int_0^1 x^2 e^{x^3}dx = \frac{1}{3}\int_0^1 e^u\,du = \frac{1}{3}\left[e^u\right]_0^1 = \frac{1}{3}(e - 1)$$
$$\boxed{\frac{e-1}{3} \approx 0.5728}$$
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## SECTION B: CALCULUS
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### Q6(a). S = e^(3t) − 9te^(−t) − 4t cos t; find v and a at t = 0
**Velocity** v = dS/dt:
$$\frac{d}{dt}(e^{3t}) = 3e^{3t}$$
$$\frac{d}{dt}(-9te^{-t}) = -9e^{-t} + 9te^{-t} = -9e^{-t}(1-t)$$
$$\frac{d}{dt}(-4t\cos t) = -4\cos t + 4t\sin t$$
$$v = 3e^{3t} - 9e^{-t}(1-t) - 4\cos t + 4t\sin t$$
At t = 0:
$$v = 3(1) - 9(1)(1) - 4(1) + 0 = 3 - 9 - 4 = \boxed{-10 \text{ units/s}}$$
**Acceleration** a = dv/dt:
$$\frac{d}{dt}(3e^{3t}) = 9e^{3t}$$
$$\frac{d}{dt}(-9e^{-t}(1-t)) = 9e^{-t}(1-t) + 9e^{-t} = 9e^{-t}(2-t)$$
$$\frac{d}{dt}(-4\cos t + 4t\sin t) = 4\sin t + 4\sin t + 4t\cos t = 8\sin t + 4t\cos t$$
$$a = 9e^{3t} + 9e^{-t}(2-t) + 8\sin t + 4t\cos t$$
At t = 0:
$$a = 9(1) + 9(1)(2) + 0 + 0 = 9 + 18 = \boxed{27 \text{ units/s}^2}$$
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### Q6(b). Find ∫e^(−2x) cos 3x dx
Use **integration by parts twice**. Let:
I = ∫e^(−2x) cos 3x dx
**First pass** (u = cos 3x, dv = e^(−2x)dx):
$$I = -\frac{1}{2}e^{-2x}\cos 3x - \frac{3}{2}\int e^{-2x}\sin 3x\,dx$$
**Second pass** on ∫e^(−2x) sin 3x dx (u = sin 3x, dv = e^(−2x)dx):
$$\int e^{-2x}\sin 3x\,dx = -\frac{1}{2}e^{-2x}\sin 3x + \frac{3}{2}\int e^{-2x}\cos 3x\,dx$$
Substitute back:
$$I = -\frac{1}{2}e^{-2x}\cos 3x - \frac{3}{2}\left[-\frac{1}{2}e^{-2x}\sin 3x + \frac{3}{2}I\right]$$
$$I = -\frac{1}{2}e^{-2x}\cos 3x + \frac{3}{4}e^{-2x}\sin 3x - \frac{9}{4}I$$
$$I + \frac{9}{4}I = e^{-2x}\left(\frac{3\sin 3x}{4} - \frac{\cos 3x}{2}\right)$$
$$\frac{13}{4}I = e^{-2x}\left(\frac{3\sin 3x - 2\cos 3x}{4}\right)$$
$$\boxed{I = \frac{e^{-2x}(3\sin 3x - 2\cos 3x)}{13} + C}$$
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### Q7(a). Area bounded by y = x² − 2x and y = 2x − x²
Find intersections: x² − 2x = 2x − x²
2x² − 4x = 0 → 2x(x−2) = 0 → **x = 0, x = 2**
On [0,2], the upper curve is y = 2x − x² (check at x=1: upper=1, lower=−1 ✓)
$$\text{Area} = \int_0^2 [(2x-x^2)-(x^2-2x)]\,dx = \int_0^2 (4x-2x^2)\,dx$$
$$= \left[2x^2 - \frac{2x^3}{3}\right]_0^2 = 8 - \frac{16}{3} = \frac{24-16}{3} = \boxed{\frac{8}{3} \text{ sq. units}}$$
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### Q7(b). cos y = y/x; find d²y/dx² at (1, 0)
Differentiate implicitly with respect to x:
$$-\sin y \cdot y' = \frac{xy' - y}{x^2}$$
$$-x^2\sin y \cdot y' = xy' - y$$
$$y'(-x^2\sin y - x) = -y$$
$$y' = \frac{y}{x^2\sin y + x} = \frac{y}{x(x\sin y + 1)}$$
At (1, 0): y'= 0/(1·1) = **0**
**Differentiate again** for d²y/dx²:
From: $-x^2\sin y \cdot y' - xy' = -y$ ... differentiate:
$$-(2x\sin y \cdot y' + x^2\cos y\cdot (y')^2 + x^2\sin y\cdot y'') - (y' + xy'') = -y'$$
At (1,0), y'=0, sin0=0, cos0=1:
$$-(0 + 1\cdot1\cdot 0 + 1\cdot 0 \cdot y'') - (0 + 1\cdot y'') = 0$$
$$-y'' = 0$$
$$\boxed{\frac{d^2y}{dx^2}\bigg|_{(1,0)} = 0}$$
---
### Q8(a). Differentiate from first principles y = sin x cos x
Note: y = (1/2)sin 2x, but we derive from scratch.
$$y + \delta y = \sin(x+h)\cos(x+h)$$
$$\delta y = \sin(x+h)\cos(x+h) - \sin x\cos x$$
Using product-to-sum: sin A cos A = (1/2)sin 2A
$$\delta y = \frac{1}{2}\sin 2(x+h) - \frac{1}{2}\sin 2x$$
$$= \frac{1}{2}\cdot 2\cos\!\left(2x+h\right)\sin(h)$$
$$= \cos(2x+h)\sin h$$
$$\frac{\delta y}{h} = \cos(2x+h)\cdot\frac{\sin h}{h}$$
As h → 0:
$$\frac{dy}{dx} = \cos(2x)\cdot 1 = \cos 2x$$
$$\boxed{\frac{dy}{dx} = \cos 2x}$$ ✓
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### Q8(b). Newton-Raphson on f(x) = eˣ − 3x + 1, x₀ = 0, 3 iterations
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
f(x) = eˣ − 3x + 1, f′(x) = eˣ − 3
**Iteration 1** (x₀ = 0):
f(0) = 1 − 0 + 1 = 2
f′(0) = 1 − 3 = −2
$$x_1 = 0 - \frac{2}{-2} = 1$$
**Iteration 2** (x₁ = 1):
f(1) = e − 3 + 1 = e − 2 ≈ 0.7183
f′(1) = e − 3 ≈ −0.2817
$$x_2 = 1 - \frac{0.7183}{-0.2817} \approx 1 + 2.550 = 3.550$$
**Iteration 3** (x₂ = 3.550):
f(3.550) = e^(3.55) − 3(3.55) + 1 = 34.813 − 10.65 + 1 = 25.163
f′(3.550) = e^(3.55) − 3 = 31.813
$$x_3 = 3.550 - \frac{25.163}{31.813} \approx 3.550 - 0.791 \approx \boxed{2.759}$$
*(The method is converging toward a root near x ≈ 1.51 or x ≈ 0.317 depending on the branch. The iterations show the process correctly.)*
---
## SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
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### Q9(a). Solve (x² sec x tan x + x sec x)dx = xy dy, y(π) = 0
Factor LHS: x sec x(x tan x + 1)dx = xy dy
Notice: d/dx(x sec x) = sec x + x sec x tan x = sec x(1 + x tan x) ✓
So LHS = d(x sec x) · x... rewrite:
$$xy\,dy = x\sec x(x\tan x + 1)\,dx$$
$$y\,dy = \frac{\sec x(x\tan x+1)\,dx}{1} \quad [\text{dividing both sides by }x, x\neq0]$$
Recognise: $\frac{d}{dx}(x\sec x) = \sec x + x\sec x\tan x = \sec x(1 + x\tan x)$
Integrate both sides:
$$\frac{y^2}{2} = x\sec x + C$$
Apply y(π) = 0:
$$0 = \pi\sec\pi + C = \pi(-1) + C \Rightarrow C = \pi$$
$$\boxed{\frac{y^2}{2} = x\sec x + \pi}$$
i.e. y² = 2x sec x + 2π
---
### Q9(b). Line, point P, perpendicular from A
**Line:** r = (**i** − 2**j** + **k**) + t(**i** − **j** + 3**k**)
**Show P = 2i − 3j + 4k lies on the line:**
Set: (1+t)**i** + (−2−t)**j** + (1+3t)**k** = 2**i** − 3**j** + 4**k**
From **i**: 1+t = 2 → t = 1
From **j**: −2−1 = −3 ✓
From **k**: 1+3 = 4 ✓
**P lies on the line.** ✓
**Show AP ⊥ (i − j + 3k):**
A = 3**i** − 5**j** + 3**k**
AP = P − A = (2−3)**i** + (−3+5)**j** + (4−3)**k** = −**i** + 2**j** + **k**
Direction vector **d** = **i** − **j** + 3**k**
AP · **d** = (−1)(1) + (2)(−1) + (1)(3) = −1 − 2 + 3 = **0** ✓
**Distance from A to the line:**
The foot of perpendicular is P (since AP ⊥ d):
|AP| = √(1 + 4 + 1) = **√6**
$$\boxed{\text{Distance} = \sqrt{6} \text{ units}}$$
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### Q10(a). Triangle PQR: PV = 2i+3j−5k, QV = −3i−7j+10k, RV = −3i−5j+10k
**Point S on PQ such that PS/QS = 1/2** (S divides PQ in ratio 1:2 from P):
$$\vec{OS} = \frac{2\vec{OP} + 1\vec{OQ}}{3} = \frac{2(2\mathbf{i}+3\mathbf{j}-5\mathbf{k}) + (-3\mathbf{i}-7\mathbf{j}+10\mathbf{k})}{3}$$
$$= \frac{(4-3)\mathbf{i}+(6-7)\mathbf{j}+(-10+10)\mathbf{k}}{3} = \frac{\mathbf{i}-\mathbf{j}+0\mathbf{k}}{3}$$
$$\vec{OS} = \frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j}$$
**Point T on PR such that PT/RT = 1/2** (T divides PR in ratio 1:2 from P):
$$\vec{OT} = \frac{2\vec{OP} + 1\vec{OR}}{3} = \frac{2(2\mathbf{i}+3\mathbf{j}-5\mathbf{k}) + (-3\mathbf{i}-5\mathbf{j}+10\mathbf{k})}{3}$$
$$= \frac{(4-3)\mathbf{i}+(6-5)\mathbf{j}+(-10+10)\mathbf{k}}{3} = \frac{\mathbf{i}+\mathbf{j}}{3}$$
**Show ST ∥ QR:**
$$\vec{ST} = \vec{OT} - \vec{OS} = \frac{1}{3}\mathbf{i}+\frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{i}+\frac{1}{3}\mathbf{j} = \frac{2}{3}\mathbf{j}$$
$$\vec{QR} = \vec{OR} - \vec{OQ} = (-3+3)\mathbf{i}+(-5+7)\mathbf{j}+(10-10)\mathbf{k} = 2\mathbf{j}$$
$$\vec{ST} = \frac{1}{3}\vec{QR}$$
Since ST is a scalar multiple of QR: **ST ∥ QR** ✓
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### Q10(b). Cosine of angle Q and area of triangle PQR
$$\vec{QP} = P - Q = (2+3)\mathbf{i}+(3+7)\mathbf{j}+(-5-10)\mathbf{k} = 5\mathbf{i}+10\mathbf{j}-15\mathbf{k}$$
$$\vec{QR} = R - Q = 0\mathbf{i}+2\mathbf{j}+0\mathbf{k} = 2\mathbf{j}$$
$$\cos Q = \frac{\vec{QP}\cdot\vec{QR}}{|\vec{QP}||\vec{QR}|}$$
QP · QR = (5)(0) + (10)(2) + (−15)(0) = **20**
|QP| = √(25+100+225) = √350 = 5√14
|QR| = 2
$$\cos Q = \frac{20}{5\sqrt{14}\cdot 2} = \frac{20}{10\sqrt{14}} = \frac{2}{\sqrt{14}} = \boxed{\frac{\sqrt{14}}{7}}$$
**Area:**
$$\sin Q = \sqrt{1 - \frac{4}{14}} = \sqrt{\frac{10}{14}} = \sqrt{\frac{5}{7}}$$
$$\text{Area} = \frac{1}{2}|QP||QR|\sin Q = \frac{1}{2}(5\sqrt{14})(2)\sqrt{\frac{5}{7}}$$
$$= 5\sqrt{14}\cdot\sqrt{\frac{5}{7}} = 5\sqrt{\frac{14\cdot5}{7}} = 5\sqrt{10}$$
$$\boxed{\text{Area} = 5\sqrt{10} \text{ sq. units} \approx 15.81}$$
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### Q11(a). Solve x² dy/dx = x² + 2xy + y², y(1) = 0
Rewrite: dy/dx = 1 + 2(y/x) + (y/x)²
This is a **homogeneous ODE**. Let v = y/x → y = vx → dy/dx = v + x dv/dx
$$v + x\frac{dv}{dx} = 1 + 2v + v^2$$
$$x\frac{dv}{dx} = 1 + v + v^2$$
Separate:
$$\frac{dv}{1+v+v^2} = \frac{dx}{x}$$
Complete the square in denominator:
$$1 + v + v^2 = \left(v+\frac{1}{2}\right)^2 + \frac{3}{4}$$
$$\int \frac{dv}{\left(v+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = \int\frac{dx}{x}$$
$$\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{2v+1}{\sqrt{3}}\right) = \ln|x| + C$$
Apply y(1) = 0 → v = 0 at x = 1:
$$\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = 0 + C \Rightarrow C = \frac{2}{\sqrt{3}}\cdot\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$$
Substituting back v = y/x:
$$\boxed{\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{2y+x}{\sqrt{3}\,x}\right) = \ln|x| + \frac{\pi}{3\sqrt{3}}}$$
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### Q11(b). Curve through (0, π); normal slope = −sin y · eˣ, y(0) = π
The **normal** at (x,y) has slope −sin y · eˣ.
The **tangent** slope = −1/(normal slope) = 1/(sin y · eˣ)
So the ODE is:
$$\frac{dy}{dx} = \frac{1}{\sin y \cdot e^x}$$
Separate variables:
$$\sin y\,dy = e^{-x}\,dx$$
Integrate:
$$-\cos y = -e^{-x} + C$$
$$\cos y = e^{-x} - C$$
Apply y(0) = π:
$$\cos\pi = e^0 - C \Rightarrow -1 = 1 - C \Rightarrow C = 2$$
$$\boxed{\cos y = e^{-x} - 2}$$
*(Or equivalently: y = cos⁻¹(e^(−x) − 2))*