### SECTION A
1. Differentiate y = cos ec e^√x [04marks]
2. Evaluate lim(x→1) ((1 − x³)/(1 − x)) [04marks]
3. Show that the vectors **i** − **j** − **k** and **i** + 2**j** − **k** are perpendicular [04marks]
4. Solve the equation dy/dx = sin x(1 + y). [04marks]
5. Evaluate ∫₀^(π/4) (3sec²x)/(4 + 3tan x) dx. [04marks]
---
### SECTION B: CALCULUS
6. (a). Find dy/dx and d²y/dx² at (1,1) if x³ + y³ + 3gx + 3fy + c = 0 [10marks]
(b). If y = e^(−ax) cos 2bx, show that y'' + 2ay' + (a² + 4b²)y = 0. [10marks]
7. (a). Differentiate from first principle y = cos²3x. [10marks]
(b). Find ∫ dx/((x−1)(x−2)(x+3)). [10marks]
---
8. (a). If y = cosh(k cos⁻¹ x), show that (1 − x²)y'' − xy' − k²y = 0. [06marks]
(b). Using integration by parts, evaluate the following:
(i) ∫₁² x² log x dx, (ii) ∫₀¹ cos⁻¹ x dx. [14marks]
---
### SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
9. (a). Solve the equation (x² + y²)dx − 3xy dy = 0. [12marks]
(b). The vertices of △PQR are represented by the vectors **i** + 3**j** + **k**, 3**i** − **j** + **k** and **i** + **j** − 3**k**, respectively. Calculate the cosine of the angles of △PQR and obtain its area. [08marks]
10. (a). Calculate the unit vector in direction of 3**a** + 2**b**, given that the vectors **a** = **i** − 2**j** + **k** and **b** = **i** + 2**j** − **k**. Hence obtain the angle between it and vectors **b**. [10marks]
(b). Solve the equation y dy/dx = sec²x e^(2tan x − 10y²). [10marks]
11. (a). Solve the equation cos x dy/dx − y sin x = cos²x − 1. [10marks]
(b). Given that **a** = **i** − 3**j** + **k** and **b** = 2**i** − **j** + **k**, find the projection of the vector 3**a** + **b** on **a** − 3**b**. [10marks]
---
# SOLUTIONS
## SECTION A
### Q1. Differentiate y = cosec(e^√x)
Let u = e^√x, so y = cosec(u)
**dy/du** = −cosec(u)cot(u)
**du/dx:** Let v = √x = x^(1/2)
dv/dx = 1/(2√x)
du/dx = e^√x · 1/(2√x)
**By chain rule:**
dy/dx = −cosec(e^√x) · cot(e^√x) · e^√x/(2√x)
$$\boxed{\frac{dy}{dx} = \frac{-e^{\sqrt{x}}\operatorname{cosec}(e^{\sqrt{x}})\cot(e^{\sqrt{x}})}{2\sqrt{x}}}$$
---
### Q2. Evaluate lim(x→1) [(1−x³)/(1−x)]
**Factor numerator:**
1 − x³ = (1−x)(1 + x + x²)
Therefore:
(1−x³)/(1−x) = (1−x)(1+x+x²)/(1−x) = 1 + x + x²
**Taking limit:**
lim(x→1) (1 + x + x²) = 1 + 1 + 1
$$\boxed{= 3}$$
---
### Q3. Show that i − j − k and i + 2j − k are Perpendicular
Two vectors are perpendicular if their **dot product = 0**.
Let **a** = **i** − **j** − **k** = (1, −1, −1)
Let **b** = **i** + 2**j** − **k** = (1, 2, −1)
**a · b** = (1)(1) + (−1)(2) + (−1)(−1)
= 1 − 2 + 1
= **0** ✓
Since **a · b = 0**, the vectors are **perpendicular**. □
---
### Q4. Solve dy/dx = sin x(1 + y)
**Separating variables:**
dy/(1+y) = sin x dx
**Integrating both sides:**
∫ dy/(1+y) = ∫ sin x dx
ln|1 + y| = −cos x + C
|1 + y| = e^(−cos x + C) = Ae^(−cos x)
$$\boxed{1 + y = Ae^{-\cos x}}$$
or **y = Ae^(−cos x) − 1**
---
### Q5. Evaluate ∫₀^(π/4) [3sec²x/(4 + 3tan x)] dx
**Substitution:** Let u = 4 + 3tan x
du/dx = 3sec²x → du = 3sec²x dx
**Limits:**
- x = 0: u = 4 + 3(0) = 4
- x = π/4: u = 4 + 3(1) = 7
**Integral becomes:**
∫₄⁷ du/u = [ln|u|]₄⁷
= ln 7 − ln 4
$$\boxed{= \ln\frac{7}{4}}$$
---
## SECTION B: CALCULUS
### Q6(a). Find dy/dx and d²y/dx² at (1,1) for x³ + y³ + 3gx + 3fy + c = 0
**Implicit differentiation:**
3x² + 3y²(dy/dx) + 3g + 3f(dy/dx) = 0
dy/dx(3y² + 3f) = −3x² − 3g
$$\frac{dy}{dx} = \frac{-(x^2 + g)}{y^2 + f}$$
**At (1,1):**
$$\frac{dy}{dx}\bigg|_{(1,1)} = \frac{-(1+g)}{1+f}$$
---
**Second derivative — differentiating dy/dx implicitly:**
Let p = dy/dx = −(x² + g)/(y² + f)
Using quotient rule:
$$\frac{d^2y}{dx^2} = \frac{-(2x)(y^2+f) - (-(x^2+g))(2y\frac{dy}{dx})}{(y^2+f)^2}$$
$$= \frac{-2x(y^2+f) + 2y(x^2+g)\frac{dy}{dx}}{(y^2+f)^2}$$
**Substituting dy/dx = −(x²+g)/(y²+f) at (1,1):**
$$\frac{d^2y}{dx^2}\bigg|_{(1,1)} = \frac{-2(1+f) + 2\cdot\frac{-(1+g)^2}{(1+f)}}{(1+f)^2}$$
$$= \frac{-2(1+f)^2 - 2(1+g)^2}{(1+f)^3}$$
$$\boxed{\frac{d^2y}{dx^2}\bigg|_{(1,1)} = \frac{-2[(1+f)^2 + (1+g)^2]}{(1+f)^3}}$$
---
### Q6(b). Show that y'' + 2ay' + (a² + 4b²)y = 0 for y = e^(−ax)cos 2bx
**First derivative:**
y' = −ae^(−ax)cos 2bx − 2be^(−ax)sin 2bx
y' = e^(−ax)(−a cos 2bx − 2b sin 2bx)
**Second derivative:**
y'' = −ae^(−ax)(−a cos 2bx − 2b sin 2bx) + e^(−ax)(2ab sin 2bx − 4b² cos 2bx)
y'' = e^(−ax)[a² cos 2bx + 2ab sin 2bx + 2ab sin 2bx − 4b² cos 2bx]
y'' = e^(−ax)[(a² − 4b²)cos 2bx + 4ab sin 2bx]
**Now compute y'' + 2ay' + (a² + 4b²)y:**
y'' + 2ay' = e^(−ax)[(a²−4b²)cos2bx + 4ab sin2bx]
+ 2a·e^(−ax)[−a cos2bx − 2b sin2bx]
= e^(−ax)[(a²−4b²)cos2bx + 4ab sin2bx − 2a²cos2bx − 4ab sin2bx]
= e^(−ax)[(a²−4b²−2a²)cos2bx]
= e^(−ax)[−(a²+4b²)cos2bx]
**Adding (a²+4b²)y:**
= e^(−ax)[−(a²+4b²)cos2bx] + (a²+4b²)·e^(−ax)cos2bx
= e^(−ax)cos2bx[−(a²+4b²) + (a²+4b²)]
**= 0** ✓ □
---
### Q7(a). Differentiate y = cos²3x from First Principles
**Definition:** dy/dx = lim(h→0) [f(x+h) − f(x)]/h
f(x) = cos²3x
f(x+h) = cos²3(x+h) = cos²(3x+3h)
**f(x+h) − f(x)** = cos²(3x+3h) − cos²3x
Using identity: cos²A − cos²B = −sin(A+B)sin(A−B):
= −sin(3x+3h+3x)·sin(3x+3h−3x)
= −sin(6x+3h)·sin(3h)
**Therefore:**
dy/dx = lim(h→0) [−sin(6x+3h)·sin(3h)]/h
= lim(h→0) −sin(6x+3h) · [sin(3h)/h]
= lim(h→0) −sin(6x+3h) · 3·[sin(3h)/(3h)]
As h→0: sin(3h)/(3h) → 1 and sin(6x+3h) → sin(6x)
$$\boxed{\frac{dy}{dx} = -3\sin 6x}$$
*(Note: this equals −2·3sin3x·cos3x = −3sin6x using double angle)* ✓
---
### Q7(b). Find ∫ dx/[(x−1)(x−2)(x+3)]
**Partial fractions:**
1/[(x−1)(x−2)(x+3)] = A/(x−1) + B/(x−2) + C/(x+3)
1 = A(x−2)(x+3) + B(x−1)(x+3) + C(x−1)(x−2)
**x = 1:** 1 = A(−1)(4) → A = −1/4
**x = 2:** 1 = B(1)(5) → B = 1/5
**x = −3:** 1 = C(−4)(−5) → C = 1/20
**Therefore:**
∫ dx/[(x−1)(x−2)(x+3)]
= ∫[−1/4·1/(x−1) + 1/5·1/(x−2) + 1/20·1/(x+3)]dx
$$\boxed{= -\frac{1}{4}\ln|x-1| + \frac{1}{5}\ln|x-2| + \frac{1}{20}\ln|x+3| + C}$$
---
### Q8(a). Show that (1−x²)y'' − xy' − k²y = 0 for y = cosh(k cos⁻¹x)
Let u = k cos⁻¹x
**y' = sinh(k cos⁻¹x) · k·(−1/√(1−x²))**
y' = −k sinh(u)/√(1−x²)
**Rearranging:** √(1−x²)·y' = −k sinh(u)
**Squaring:** (1−x²)(y')² = k² sinh²(u) = k²(cosh²u − 1) = k²(y² − 1)
**Differentiating both sides w.r.t. x:**
−2x(y')² + (1−x²)·2y'·y'' = 2k²y·y'
**Dividing by 2y'** (y' ≠ 0):
−x·y' + (1−x²)y'' = k²y
**(1−x²)y'' − xy' − k²y = 0** ✓ □
---
### Q8(b)(i). ∫₁² x² log x dx (Integration by Parts)
**Let:** u = log x → du = 1/x dx
**Let:** dv = x² dx → v = x³/3
$$\int x^2 \log x\, dx = \frac{x^3}{3}\log x - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx$$
$$= \frac{x^3}{3}\log x - \frac{1}{3}\int x^2\,dx$$
$$= \frac{x^3}{3}\log x - \frac{x^3}{9} + C$$
**Evaluating from 1 to 2:**
$$= \left[\frac{x^3}{3}\log x - \frac{x^3}{9}\right]_1^2$$
At x=2: 8/3·log2 − 8/9
At x=1: 1/3·log1 − 1/9 = 0 − 1/9 = −1/9
$$= \frac{8\log 2}{3} - \frac{8}{9} + \frac{1}{9}$$
$$\boxed{= \frac{8\log 2}{3} - \frac{7}{9}}$$
*(Using log base 10; if natural log: = 8ln2/3 − 7/9)*
---
### Q8(b)(ii). ∫₀¹ cos⁻¹x dx (Integration by Parts)
**Let:** u = cos⁻¹x → du = −1/√(1−x²) dx
**Let:** dv = dx → v = x
$$\int \cos^{-1}x\,dx = x\cos^{-1}x - \int x\cdot\frac{-1}{\sqrt{1-x^2}}\,dx$$
$$= x\cos^{-1}x + \int \frac{x}{\sqrt{1-x^2}}\,dx$$
**For the remaining integral:** let w = 1−x², dw = −2x dx
$$\int \frac{x}{\sqrt{1-x^2}}dx = -\sqrt{1-x^2}$$
**Therefore:**
$$\int \cos^{-1}x\,dx = x\cos^{-1}x - \sqrt{1-x^2} + C$$
**Evaluating from 0 to 1:**
At x=1: 1·cos⁻¹(1) − √0 = 1·0 − 0 = 0
At x=0: 0·cos⁻¹(0) − √1 = 0 − 1 = −1
$$\boxed{= 0-(-1) = 1}$$
---
---
## SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
### Q9(a). Solve (x² + y²)dx − 3xy dy = 0
**Rearranging:**
dy/dx = (x² + y²)/(3xy)
**This is homogeneous.** Let y = vx → dy/dx = v + x dv/dx
dy/dx = (x² + v²x²)/(3x·vx) = (1 + v²)/(3v)
**Substituting:**
v + x dv/dx = (1 + v²)/(3v)
x dv/dx = (1 + v²)/(3v) − v = (1 + v² − 3v²)/(3v) = (1 − 2v²)/(3v)
**Separating variables:**
3v dv/(1 − 2v²) = dx/x
**Integrating:**
Left side: let w = 1 − 2v², dw = −4v dv
∫3v/(1−2v²)dv = −3/4 ∫dw/w = −3/4 ln|1−2v²|
∫dx/x = ln|x|
**Therefore:**
−3/4 ln|1 − 2v²| = ln|x| + C
ln|1 − 2v²|^(−3/4) = ln|x| + C
**Substituting back v = y/x:**
$$\left(1 - \frac{2y^2}{x^2}\right)^{-3/4} = Ax$$
$$\boxed{\left(\frac{x^2-2y^2}{x^2}\right)^{3/4} = \frac{K}{x}}$$
or equivalently: **(x² − 2y²)³ = Kx²** (after simplification)
---
### Q9(b). Triangle PQR — Cosines of Angles and Area
**Position vectors:**
- P = **i** + 3**j** + **k** = (1, 3, 1)
- Q = 3**i** − **j** + **k** = (3, −1, 1)
- R = **i** + **j** − 3**k** = (1, 1, −3)
**Side vectors:**
PQ = Q − P = (2, −4, 0), |PQ| = √(4+16+0) = √20 = 2√5
QR = R − Q = (−2, 2, −4), |QR| = √(4+4+16) = √24 = 2√6
PR = R − P = (0, −2, −4), |PR| = √(0+4+16) = √20 = 2√5
---
**Cosine of angle at P (between PQ and PR):**
cos P = (PQ · PR)/(|PQ||PR|)
PQ · PR = (2)(0) + (−4)(−2) + (0)(−4) = 0 + 8 + 0 = 8
cos P = 8/(2√5 · 2√5) = 8/20 = **2/5**
---
**Cosine of angle at Q (between QP and QR):**
QP = −PQ = (−2, 4, 0)
QR = (−2, 2, −4)
QP · QR = (−2)(−2) + (4)(2) + (0)(−4) = 4 + 8 + 0 = 12
cos Q = 12/(2√5 · 2√6) = 12/(4√30) = **3/√30**
---
**Cosine of angle at R (between RP and RQ):**
RP = −PR = (0, 2, 4)
RQ = −QR = (2, −2, 4)
RP · RQ = (0)(2) + (2)(−2) + (4)(4) = 0 − 4 + 16 = 12
cos R = 12/(2√5 · 2√6) = 12/(4√30) = **3/√30**
---
**Area of △PQR:**
Using cross product: Area = ½|PQ × PR|
$$PQ \times PR = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -4 & 0 \\ 0 & -2 & -4 \end{vmatrix}$$
= **i**[(−4)(−4) − (0)(−2)] − **j**[(2)(−4) − (0)(0)] + **k**[(2)(−2) − (−4)(0)]
= **i**[16] − **j**[−8] + **k**[−4]
= (16, 8, −4)
|PQ × PR| = √(256 + 64 + 16) = √336 = 4√21
$$\boxed{\text{Area} = \frac{1}{2} \times 4\sqrt{21} = 2\sqrt{21} \text{ sq. units}}$$
---
### Q10(a). Unit Vector in Direction of 3a + 2b and Angle with b
**Given:**
**a** = (1, −2, 1), **b** = (1, 2, −1)
**3a + 2b:**
3**a** = (3, −6, 3)
2**b** = (2, 4, −2)
3**a** + 2**b** = (5, −2, 1)
|3**a** + 2**b**| = √(25 + 4 + 1) = √30
**Unit vector:**
$$\hat{u} = \frac{1}{\sqrt{30}}(5\mathbf{i} - 2\mathbf{j} + \mathbf{k})$$
---
**Angle between û and b:**
|**b**| = √(1 + 4 + 1) = √6
û · **b** = (5/√30)(1) + (−2/√30)(2) + (1/√30)(−1)
= (5 − 4 − 1)/√30 = 0/√30 = 0
$$\boxed{\theta = 90°}$$
The unit vector 3**a** + 2**b** is **perpendicular** to **b**.
---
### Q10(b). Solve y dy/dx = sec²x · e^(2tan x − 10y²)
**Rewriting:**
y dy/dx = sec²x · e^(2tan x) · e^(−10y²)
**Separating variables:**
y · e^(10y²) dy = sec²x · e^(2tan x) dx
**Integrating left side:**
Let u = 10y², du = 20y dy → y dy = du/20
∫y e^(10y²) dy = (1/20)∫e^u du = e^(10y²)/20
**Integrating right side:**
Let v = 2tan x, dv = 2sec²x dx → sec²x dx = dv/2
∫sec²x · e^(2tan x) dx = (1/2)∫e^v dv = e^(2tan x)/2
**Therefore:**
$$\frac{e^{10y^2}}{20} = \frac{e^{2\tan x}}{2} + C$$
$$\boxed{e^{10y^2} = 10e^{2\tan x} + K}$$
---
### Q11(a). Solve cos x dy/dx − y sin x = cos²x − 1
**Rewriting in standard form:**
dy/dx − y tan x = cos x − sec x
**This is a first-order linear ODE:** dy/dx + P(x)y = Q(x)
where P(x) = −tan x, Q(x) = cos x − sec x
**Integrating factor:**
μ = e^(∫−tan x dx) = e^(ln|cos x|) = **cos x**
**Multiplying through by cos x:**
d/dx[y cos x] = cos x(cos x − sec x)
= cos²x − 1
= −sin²x
**Integrating:**
y cos x = ∫−sin²x dx = ∫−(1 − cos2x)/2 dx
= −x/2 + sin2x/4 + C
$$\boxed{y\cos x = -\frac{x}{2} + \frac{\sin 2x}{4} + C}$$
---
### Q11(b). Projection of 3a + b on a − 3b
**Given:**
**a** = (1, −3, 1), **b** = (2, −1, 1)
**Compute 3a + b:**
3**a** = (3, −9, 3)
**b** = (2, −1, 1)
3**a** + **b** = (5, −10, 4)
**Compute a − 3b:**
**a** = (1, −3, 1)
3**b** = (6, −3, 3)
**a** − 3**b** = (−5, 0, −2)
|**a** − 3**b**| = √(25 + 0 + 4) = √29
**Projection formula:**
proj = (3**a** + **b**) · (**a** − 3**b**) / |**a** − 3**b**|
**(3a+b)·(a−3b):**
= (5)(−5) + (−10)(0) + (4)(−2)
= −25 + 0 − 8
= −33
$$\boxed{\text{Projection} = \frac{-33}{\sqrt{29}} = \frac{-33\sqrt{29}}{29}}$$