### SECTION A
1. Differentiate (1 + sin x)/(1 + cos x). [04marks]
2. Find the direction cosines of the vector 2**i** + 2**j** − **k**. [04marks]
3. Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3 - x^4}$. [04marks]
4. If y = eˣ sin x, show that y'' − 2y' + 2y = 0. [04marks]
5. Show that $\int_0^1 \frac{dx}{(1+x)[2+\log(1+x)]} = \log\frac{\log 2e^2}{2}$. [04marks]
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### SECTION B: CALCULUS
6. (a) If y = xeʸ, show that:
- (i) (1 − y)y' = eʸ
- (ii) (1 − y)y'' = (2 − y)(y')²
[10marks]
(b) Evaluate $\int_3^{10} \frac{1}{x\sqrt{x+2}}\,dx$. [10marks]
7. (a) Differentiate from first principles y = √(7x − 5). [10marks]
(b) Find the first four terms of the series expansion of the function f(x) = 1/(3x − 1) for |x| < 1, ascending powers of x. [10marks]
8. (a) Sketch the graph of the function y = x³ − 2x² − 5x + 6 and obtain the area bounded by the curve and x-axis. [14marks]
(b) Differentiate tan⁻¹(√(1 + x²) + x). [06marks]
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### SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
9. (a) Solve the equation dy/dx − y tan x = sin²x, y(π/2) = 0. [10marks]
(b) Find the angles the vector **a** = 3**i** + **j** − 2**k** makes with the coordinate axes. [10marks]
10. (a) Solve the equation yx dx − (1 + x²)dy = 0, y(0) = 1. [10marks]
(b) Given the vectors **a** = 3**i** + **j** − **k** and **b** = 2**i** − **j** − **k**, compute the projection of (3**a** − 2**b**) on the vector (2**a** + 3**b**). [10marks]
11. (a) Solve the differential equation:
$$\frac{dy}{dx} = \frac{x^2 - 3xy + 4y^2}{x^2}$$
if y(1) = 0. [15marks]
(b) Determine the value of y if the vectors y**i** − 2**j**, **j** − **k** and **k** − **i** are linearly dependent. [05marks]
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# FULL SOLUTIONS
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## SECTION A
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### Q1. Differentiate (1 + sin x)/(1 + cos x)
Using the **quotient rule**: u = 1 + sin x, v = 1 + cos x
$$\frac{dy}{dx} = \frac{(1+\cos x)\cos x - (1+\sin x)(-\sin x)}{(1+\cos x)^2}$$
$$= \frac{\cos x + \cos^2 x + \sin x + \sin^2 x}{(1+\cos x)^2}$$
$$= \frac{\cos x + \sin x + 1}{(1+\cos x)^2}$$
$$\boxed{\frac{dy}{dx} = \frac{1 + \sin x + \cos x}{(1+\cos x)^2}}$$
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### Q2. Direction cosines of **v** = 2**i** + 2**j** − **k**
$$|\mathbf{v}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$
$$\boxed{l = \frac{2}{3},\quad m = \frac{2}{3},\quad n = \frac{-1}{3}}$$
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### Q3. Evaluate $\lim_{x \to 0}\frac{x - \sin x}{x^3 - x^4}$
Factor denominator: x³ − x⁴ = x³(1 − x)
$$\lim_{x\to 0}\frac{x-\sin x}{x^3(1-x)}$$
Using the standard result $\lim_{x\to 0}\frac{x-\sin x}{x^3} = \frac{1}{6}$ (via L'Hôpital or Taylor series):
sin x = x − x³/6 + ... → x − sin x = x³/6 + ...
$$\lim_{x\to 0}\frac{x^3/6 + \ldots}{x^3(1-x)} = \frac{1/6}{1} = \boxed{\frac{1}{6}}$$
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### Q4. If y = eˣ sin x, show y'' − 2y' + 2y = 0
y = eˣ sin x
y' = eˣ sin x + eˣ cos x = eˣ(sin x + cos x)
y'' = eˣ(sin x + cos x) + eˣ(cos x − sin x) = 2eˣ cos x
Now compute y'' − 2y' + 2y:
$$= 2e^x\cos x - 2e^x(\sin x + \cos x) + 2e^x\sin x$$
$$= 2e^x\cos x - 2e^x\sin x - 2e^x\cos x + 2e^x\sin x$$
$$= 0 \quad ✓$$
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### Q5. Show $\int_0^1\frac{dx}{(1+x)[2+\log(1+x)]} = \log\frac{\log 2e^2}{2}$
Let u = 2 + log(1+x) → du = 1/(1+x) dx
When x = 0: u = 2 + log 1 = 2
When x = 1: u = 2 + log 2
$$\int_2^{2+\log 2}\frac{du}{u} = \Big[\log u\Big]_2^{2+\log 2}$$
$$= \log(2+\log 2) - \log 2 = \log\frac{2+\log 2}{2}$$
Now 2 + log 2 = log e² + log 2 = log 2e² (using log e² = 2 with natural log interpretation):
$$= \log\frac{\log 2e^2}{2} \quad ✓$$
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## SECTION B: CALCULUS
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### Q6(a). y = xeʸ; show (i) and (ii)
**Differentiate** y = xeʸ implicitly:
$$y' = e^y + xe^y y' = e^y(1 + xy')$$
Since y = xeʸ → eʸ = y/x:
$$y' = e^y + yy'$$
$$y' - yy' = e^y \Rightarrow y'(1-y) = e^y$$
$$\boxed{(1-y)y' = e^y} \quad ✓ \text{ (i)}$$
**Differentiate again** for (ii):
Differentiate (1−y)y' = eʸ:
$$-y'\cdot y' + (1-y)y'' = e^y\cdot y'$$
$$(1-y)y'' = e^y y' + (y')^2$$
From (i): eʸ = (1−y)y', substitute:
$$(1-y)y'' = (1-y)y'\cdot y' + (y')^2 = (y')^2[(1-y)+1] = (y')^2(2-y)$$
$$\boxed{(1-y)y'' = (2-y)(y')^2} \quad ✓ \text{ (ii)}$$
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### Q6(b). Evaluate $\int_3^{10}\frac{1}{x\sqrt{x+2}}\,dx$
Let u = √(x+2) → u² = x+2 → x = u²−2 → dx = 2u du
When x=3: u=√5; when x=10: u=√12 = 2√3
$$\int_{\sqrt{5}}^{2\sqrt{3}}\frac{2u\,du}{(u^2-2)\cdot u} = \int_{\sqrt{5}}^{2\sqrt{3}}\frac{2\,du}{u^2-2}$$
$$= \int_{\sqrt{5}}^{2\sqrt{3}}\frac{2\,du}{(u-\sqrt{2})(u+\sqrt{2})}$$
Using partial fractions: $\frac{2}{u^2-2} = \frac{1}{\sqrt{2}}\left(\frac{1}{u-\sqrt{2}}-\frac{1}{u+\sqrt{2}}\right)$
$$= \frac{1}{\sqrt{2}}\Big[\ln|u-\sqrt{2}| - \ln|u+\sqrt{2}|\Big]_{\sqrt{5}}^{2\sqrt{3}}$$
$$= \frac{1}{\sqrt{2}}\left[\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|\right]_{\sqrt{5}}^{2\sqrt{3}}$$
At u = 2√3: $\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}+\sqrt{2}}$
At u = √5: $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$
$$= \frac{1}{\sqrt{2}}\left[\ln\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}+\sqrt{2}} - \ln\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\right]$$
Numerically:
$\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}+\sqrt{2}} = \frac{3.464-1.414}{3.464+1.414} = \frac{2.050}{4.878} \approx 0.4203$
$\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} = \frac{2.236-1.414}{2.236+1.414} = \frac{0.822}{3.650} \approx 0.2252$
$$= \frac{1}{\sqrt{2}}[\ln 0.4203 - \ln 0.2252] = \frac{1}{\sqrt{2}}\ln\frac{0.4203}{0.2252} = \frac{1}{\sqrt{2}}\ln(1.866)$$
$$= \frac{0.6245}{\sqrt{2}} \approx \boxed{0.4414}$$
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### Q7(a). First principles: y = √(7x − 5)
$$y + \delta y = \sqrt{7(x+h)-5} = \sqrt{7x-5+7h}$$
$$\delta y = \sqrt{7x+7h-5} - \sqrt{7x-5}$$
Rationalise:
$$\frac{\delta y}{h} = \frac{7h}{h\left(\sqrt{7x+7h-5}+\sqrt{7x-5}\right)} = \frac{7}{\sqrt{7x+7h-5}+\sqrt{7x-5}}$$
As h → 0:
$$\frac{dy}{dx} = \frac{7}{2\sqrt{7x-5}}$$
$$\boxed{\frac{dy}{dx} = \frac{7}{2\sqrt{7x-5}}}$$
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### Q7(b). First four terms of f(x) = 1/(3x − 1), |x| < 1
$$f(x) = \frac{1}{3x-1} = \frac{-1}{1-3x} = -(1-3x)^{-1}$$
Using geometric series $(1-u)^{-1} = 1+u+u^2+u^3+\ldots$ with u = 3x:
$$f(x) = -(1 + 3x + 9x^2 + 27x^3 + \ldots)$$
$$\boxed{f(x) = -1 - 3x - 9x^2 - 27x^3 - \ldots}$$
Valid for |3x| < 1, i.e. |x| < 1/3.
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### Q8(a). Sketch y = x³ − 2x² − 5x + 6; find area bounded by curve and x-axis
**Find roots** (x-intercepts):
Test x = 1: 1 − 2 − 5 + 6 = 0 ✓
Factor: (x−1)(x²−x−6) = (x−1)(x−3)(x+2)
**Roots: x = −2, 1, 3**
**Sketch description:**
- Cubic, positive leading coefficient
- Crosses x-axis at x = −2, 1, 3
- y-intercept: (0, 6)
- Local max between x = −2 and x = 1; local min between x = 1 and x = 3
**Area:**
$$\text{Area} = \left|\int_{-2}^{1}(x^3-2x^2-5x+6)\,dx\right| + \left|\int_{1}^{3}(x^3-2x^2-5x+6)\,dx\right|$$
Let F(x) = x⁴/4 − 2x³/3 − 5x²/2 + 6x
**F(1):** 1/4 − 2/3 − 5/2 + 6 = 3/12 − 8/12 − 30/12 + 72/12 = **37/12**
**F(−2):** 16/4 − 2(−8)/3 − 5(4)/2 + 6(−2) = 4 + 16/3 − 10 − 12 = −18 + 16/3 = **−38/3**
**F(3):** 81/4 − 18 − 45/2 + 18 = 81/4 − 45/2 = 81/4 − 90/4 = **−9/4**
$$\int_{-2}^{1} = F(1) - F(-2) = \frac{37}{12} + \frac{38}{3} = \frac{37}{12} + \frac{152}{12} = \frac{189}{12} = \frac{63}{4}$$
$$\int_{1}^{3} = F(3) - F(1) = -\frac{9}{4} - \frac{37}{12} = -\frac{27}{12} - \frac{37}{12} = -\frac{64}{12} = -\frac{16}{3}$$
$$\text{Total Area} = \frac{63}{4} + \frac{16}{3} = \frac{189}{12} + \frac{64}{12} = \boxed{\frac{253}{12} \approx 21.08 \text{ sq. units}}$$
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### Q8(b). Differentiate tan⁻¹(√(1 + x²) + x)
Let u = √(1+x²) + x
$$\frac{du}{dx} = \frac{x}{\sqrt{1+x^2}} + 1 = \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}}$$
$$\frac{d}{dx}\tan^{-1}(u) = \frac{1}{1+u^2}\cdot\frac{du}{dx}$$
$$u^2 = (\sqrt{1+x^2}+x)^2 = 1+x^2+2x\sqrt{1+x^2}+x^2 = 1+2x^2+2x\sqrt{1+x^2}$$
$$1+u^2 = 2+2x^2+2x\sqrt{1+x^2} = 2(1+x^2+x\sqrt{1+x^2})$$
$$= 2\sqrt{1+x^2}(\sqrt{1+x^2}+x)$$
Therefore:
$$\frac{dy}{dx} = \frac{1}{2\sqrt{1+x^2}(\sqrt{1+x^2}+x)}\cdot\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}$$
$$\boxed{\frac{dy}{dx} = \frac{1}{2(1+x^2)}}$$
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## SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
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### Q9(a). dy/dx − y tan x = sin²x, y(π/2) = 0
This is a **linear first-order ODE**.
Integrating factor: $\mu = e^{-\int\tan x\,dx} = e^{\ln|\cos x|} = \cos x$
Multiply through:
$$\frac{d}{dx}(y\cos x) = \sin^2 x\cos x$$
Integrate:
$$y\cos x = \int\sin^2 x\cos x\,dx = \frac{\sin^3 x}{3} + C$$
Apply y(π/2) = 0:
$$0\cdot\cos(\pi/2) = \frac{\sin^3(\pi/2)}{3} + C \Rightarrow 0 = \frac{1}{3} + C \Rightarrow C = -\frac{1}{3}$$
$$\boxed{y\cos x = \frac{\sin^3 x - 1}{3}}$$
i.e. $y = \frac{\sin^3 x - 1}{3\cos x}$
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### Q9(b). Angles vector a = 3i + j − 2k makes with coordinate axes
$$|\mathbf{a}| = \sqrt{9+1+4} = \sqrt{14}$$
$$\alpha = \cos^{-1}\!\left(\frac{3}{\sqrt{14}}\right) \approx \cos^{-1}(0.8018) \approx \boxed{36.7°}\ \text{(with x-axis)}$$
$$\beta = \cos^{-1}\!\left(\frac{1}{\sqrt{14}}\right) \approx \cos^{-1}(0.2673) \approx \boxed{74.5°}\ \text{(with y-axis)}$$
$$\gamma = \cos^{-1}\!\left(\frac{-2}{\sqrt{14}}\right) \approx \cos^{-1}(-0.5345) \approx \boxed{122.3°}\ \text{(with z-axis)}$$
**Verify:** cos²α + cos²β + cos²γ = 9/14 + 1/14 + 4/14 = 1 ✓
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### Q10(a). yx dx − (1 + x²)dy = 0, y(0) = 1
Separate variables:
$$\frac{x\,dx}{1+x^2} = \frac{dy}{y}$$
Integrate:
$$\frac{1}{2}\ln(1+x^2) = \ln|y| + C$$
$$\ln(1+x^2)^{1/2} = \ln|y| + C$$
$$\sqrt{1+x^2} = Ay$$
Apply y(0) = 1: √1 = A·1 → **A = 1**
$$\boxed{y = \sqrt{1+x^2}}$$
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### Q10(b). Projection of (3a − 2b) onto (2a + 3b)
**a** = 3**i** + **j** − **k**, **b** = 2**i** − **j** − **k**
**3a − 2b:**
= (9−4)**i** + (3+2)**j** + (−3+2)**k** = **5i + 5j − k**
**2a + 3b:**
= (6+6)**i** + (2−3)**j** + (−2−3)**k** = **12i − j − 5k**
|**2a + 3b**| = √(144 + 1 + 25) = √170
**Dot product:**
(5**i**+5**j**−**k**)·(12**i**−**j**−5**k**) = 60 − 5 + 5 = **60**
**Projection:**
$$\text{proj} = \frac{(3\mathbf{a}-2\mathbf{b})\cdot(2\mathbf{a}+3\mathbf{b})}{|2\mathbf{a}+3\mathbf{b}|} = \frac{60}{\sqrt{170}}$$
$$\boxed{= \frac{60}{\sqrt{170}} = \frac{60\sqrt{170}}{170} = \frac{6\sqrt{170}}{17} \approx 4.603}$$
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### Q11(a). dy/dx = (x² − 3xy + 4y²)/x², y(1) = 0
Rewrite: dy/dx = 1 − 3(y/x) + 4(y/x)²
**Homogeneous ODE.** Let v = y/x → y = vx → dy/dx = v + x dv/dx:
$$v + x\frac{dv}{dx} = 1 - 3v + 4v^2$$
$$x\frac{dv}{dx} = 1 - 4v + 4v^2 = (1-2v)^2$$
Separate:
$$\frac{dv}{(1-2v)^2} = \frac{dx}{x}$$
Integrate LHS: let w = 1−2v, dw = −2dv:
$$\int\frac{dv}{(1-2v)^2} = \frac{-1}{2}\int\frac{dw}{w^2} = \frac{1}{2(1-2v)}$$
So:
$$\frac{1}{2(1-2v)} = \ln|x| + C$$
Apply y(1) = 0 → v = 0 at x = 1:
$$\frac{1}{2(1)} = 0 + C \Rightarrow C = \frac{1}{2}$$
$$\frac{1}{2(1-2v)} = \ln x + \frac{1}{2}$$
Substitute back v = y/x:
$$\frac{1}{2\left(1 - \frac{2y}{x}\right)} = \ln x + \frac{1}{2}$$
$$\frac{x}{2(x-2y)} = \ln x + \frac{1}{2}$$
$$\boxed{x = 2(x-2y)\!\left(\ln x + \frac{1}{2}\right)}$$
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### Q11(b). Find y if yi − 2j, j − k and k − i are linearly dependent
Three vectors are linearly dependent iff their determinant = 0:
$$\begin{vmatrix}y & -2 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{vmatrix} = 0$$
Expand along row 1:
$$y\begin{vmatrix}1&-1\\0&1\end{vmatrix} - (-2)\begin{vmatrix}0&-1\\-1&1\end{vmatrix} + 0 = 0$$
$$y(1-0) + 2(0-1) = 0$$
$$y - 2 = 0$$
$$\boxed{y = 2}$$