Engineering Mathematics (4320002) - Winter 2023 Solution

Complete solution guide for Engineering Mathematics (4320002) Winter 2023 exam

study-materialsolutionsengineering-mathematics43200022023wintergtu

Q.1 Fill in the blanks [14 marks]

Q1.1 [1 mark]

Order of the matrix [2578]\begin{bmatrix} 2 & 5 \\ 7 & 8 \end{bmatrix} is _________

Answer: (d) 2×22 \times 2

Solution: The matrix has 2 rows and 2 columns, so its order is 2×22 \times 2.

Q1.2 [1 mark]

[4362]+[1558]=\begin{bmatrix} 4 & 3 \\ 6 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ 5 & 8 \end{bmatrix} = _________

Answer: (a) [581110]\begin{bmatrix} 5 & 8 \\ 11 & 10 \end{bmatrix}

Solution: [4362]+[1558]=[4+13+56+52+8]=[581110]\begin{bmatrix} 4 & 3 \\ 6 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 4+1 & 3+5 \\ 6+5 & 2+8 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 11 & 10 \end{bmatrix}

Q1.3 [1 mark]

Which of the following is a square matrix?

Answer: (c) [1354]\begin{bmatrix} 1 & 3 \\ 5 & 4 \end{bmatrix}

Solution: A square matrix has equal number of rows and columns. Only option (c) has 2×22 \times 2 dimensions.

Q1.4 [1 mark]

If A=[3]A = [3] and B=[4]B = [4] then A⋅B=A \cdot B = _________

Answer: (b) 12

Solution: A⋅B=[3]×[4]=[3×4]=[12]=12A \cdot B = [3] \times [4] = [3 \times 4] = [12] = 12

Q1.5 [1 mark]

ddxsin⁡x=\frac{d}{dx}\sin x = _________

Answer: (d) cos⁡x\cos x

Solution: The derivative of sin⁡x\sin x is cos⁡x\cos x.

Q1.6 [1 mark]

If f(x)=xexf(x) = xe^x then f′(0)=f'(0) = _________

Answer: (b) 1

Solution: Using product rule: f′(x)=ddx(xex)=ex+xex=ex(1+x)f'(x) = \frac{d}{dx}(xe^x) = e^x + xe^x = e^x(1 + x) f′(0)=e0(1+0)=1×1=1f'(0) = e^0(1 + 0) = 1 \times 1 = 1

Q1.7 [1 mark]

If y=x2y = x^2 then d2ydx2=\frac{d^2y}{dx^2} = _________

Answer: (b) 2

Solution: y=x2y = x^2 dydx=2x\frac{dy}{dx} = 2x d2ydx2=2\frac{d^2y}{dx^2} = 2

Q1.8 [1 mark]

∫cos⁡xdx=\int \cos x dx = _________ +c+ c

Answer: (a) sin⁡x\sin x

Solution: ∫cos⁡xdx=sin⁡x+c\int \cos x dx = \sin x + c

Q1.9 [1 mark]

∫01xdx=\int_0^1 x dx = _________

Answer: (c) 12\frac{1}{2}

Solution: ∫01xdx=[x22]01=122−022=12\int_0^1 x dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}

Q1.10 [1 mark]

∫11+x2dx=\int \frac{1}{1+x^2} dx = _________ +c+ c

Answer: (a) tan⁡−1x\tan^{-1} x

Solution: ∫11+x2dx=tan⁡−1x+c\int \frac{1}{1+x^2} dx = \tan^{-1} x + c

Q1.11 [1 mark]

Order of differential equation xsin⁡y+xy=xx\sin y + xy = x is _________

Answer: (b) 1

Solution: The equation can be written as dydx=1−xysin⁡y\frac{dy}{dx} = \frac{1-xy}{\sin y}. The highest order derivative is first order.

Q1.12 [1 mark]

Integration factor of dydx+y=x\frac{dy}{dx} + y = x is _________

Answer: (d) exe^x

Solution: For dydx+Py=Q\frac{dy}{dx} + Py = Q, integration factor =e∫Pdx=e∫1dx=ex= e^{\int P dx} = e^{\int 1 dx} = e^x

Q1.13 [1 mark]

i2=i^2 = _________

Answer: (b) -1

Solution: By definition, i2=−1i^2 = -1

Q1.14 [1 mark]

(2+3i)(2−3i)=(2+3i)(2-3i) = _________

Answer: (c) 13

Solution: (2+3i)(2−3i)=22−(3i)2=4−9i2=4−9(−1)=4+9=13(2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13

Q.2(A) Attempt any two [6 marks]

Q2.1(A)(1) [3 marks]

If A=[25−13]A = \begin{bmatrix} 2 & 5 \\ -1 & 3 \end{bmatrix}, B=[5846]B = \begin{bmatrix} 5 & 8 \\ 4 & 6 \end{bmatrix} and C=[4215]C = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix} then find 2A+3B−C2A + 3B - C

Solution: 2A=2[25−13]=[410−26]2A = 2\begin{bmatrix} 2 & 5 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 10 \\ -2 & 6 \end{bmatrix}

3B=3[5846]=[15241218]3B = 3\begin{bmatrix} 5 & 8 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 15 & 24 \\ 12 & 18 \end{bmatrix}

2A+3B=[410−26]+[15241218]=[19341024]2A + 3B = \begin{bmatrix} 4 & 10 \\ -2 & 6 \end{bmatrix} + \begin{bmatrix} 15 & 24 \\ 12 & 18 \end{bmatrix} = \begin{bmatrix} 19 & 34 \\ 10 & 24 \end{bmatrix}

2A+3B−C=[19341024]−[4215]=[1532919]2A + 3B - C = \begin{bmatrix} 19 & 34 \\ 10 & 24 \end{bmatrix} - \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 15 & 32 \\ 9 & 19 \end{bmatrix}

Q2.1(A)(2) [3 marks]

If M=[1437]M = \begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} and N=[6905]N = \begin{bmatrix} 6 & 9 \\ 0 & 5 \end{bmatrix} then prove that (M+N)T=MT+NT(M+N)^T = M^T + N^T

Solution: M+N=[1437]+[6905]=[713312]M + N = \begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} + \begin{bmatrix} 6 & 9 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 13 \\ 3 & 12 \end{bmatrix}

(M+N)T=[731312](M+N)^T = \begin{bmatrix} 7 & 3 \\ 13 & 12 \end{bmatrix}

MT=[1347]M^T = \begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}, NT=[6095]N^T = \begin{bmatrix} 6 & 0 \\ 9 & 5 \end{bmatrix}

MT+NT=[1347]+[6095]=[731312]M^T + N^T = \begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 9 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ 13 & 12 \end{bmatrix}

Hence, (M+N)T=MT+NT(M+N)^T = M^T + N^T is proved.

Q2.1(A)(3) [3 marks]

Solve differential equation: xdydx+y=xyx\frac{dy}{dx} + y = xy

Solution: xdydx+y=xyx\frac{dy}{dx} + y = xy dydx+yx=y\frac{dy}{dx} + \frac{y}{x} = y dydx=y−yx=y(1−1x)=y(x−1x)\frac{dy}{dx} = y - \frac{y}{x} = y\left(1 - \frac{1}{x}\right) = y\left(\frac{x-1}{x}\right)

Separating variables: dyy=x−1xdx\frac{dy}{y} = \frac{x-1}{x}dx

Integrating: ln⁡∣y∣=∫x−1xdx=∫(1−1x)dx=x−ln⁡∣x∣+C\ln|y| = \int\frac{x-1}{x}dx = \int\left(1 - \frac{1}{x}\right)dx = x - \ln|x| + C

y=Aex−ln⁡∣x∣=Aexxy = Ae^{x-\ln|x|} = A\frac{e^x}{x}

Q.2(B) Attempt any two [8 marks]

Q2.1(B)(1) [4 marks]

Solve equations 2x+3y=82x + 3y = 8, 3x+4y=113x + 4y = 11 using matrix method

Solution: Writing in matrix form: AX=BAX = B [2334][xy]=[811]\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 11 \end{bmatrix}

Finding A−1A^{-1}: ∣A∣=2(4)−3(3)=8−9=−1|A| = 2(4) - 3(3) = 8 - 9 = -1

A−1=1∣A∣[4−3−32]=[−433−2]A^{-1} = \frac{1}{|A|}\begin{bmatrix} 4 & -3 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix}

X=A−1B=[−433−2][811]=[−32+3324−22]=[12]X = A^{-1}B = \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} 8 \\ 11 \end{bmatrix} = \begin{bmatrix} -32+33 \\ 24-22 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

Therefore: x=1,y=2x = 1, y = 2

Q2.1(B)(2) [4 marks]

If A=[3214]A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} and B=[1201]B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} then prove that (AB)T=BTAT(AB)^T = B^T A^T

Solution: AB=[3214][1201]=[3816]AB = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 1 & 6 \end{bmatrix}

(AB)T=[3186](AB)^T = \begin{bmatrix} 3 & 1 \\ 8 & 6 \end{bmatrix}

AT=[3124]A^T = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}, BT=[1021]B^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}

BTAT=[1021][3124]=[3186]B^T A^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 8 & 6 \end{bmatrix}

Hence, (AB)T=BTAT(AB)^T = B^T A^T is proved.

Q2.1(B)(3) [4 marks]

If A=[23−12]A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} then prove that A2−4A+7I=OA^2 - 4A + 7I = O

Solution: A2=[23−12][23−12]=[112−41]A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix}

4A=4[23−12]=[812−48]4A = 4\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix}

7I=7[1001]=[7007]7I = 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

A2−4A+7I=[112−41]−[812−48]+[7007]=[0000]=OA^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

Hence proved.

Q.3(A) Attempt any two [6 marks]

Q3.1(A)(1) [3 marks]

Find derivative of f(x)=exf(x) = e^x using definition of differentiation

Solution: Using definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

f′(x)=lim⁡h→0ex+h−exh=lim⁡h→0ex⋅eh−exhf'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h}

=lim⁡h→0ex(eh−1)h=exlim⁡h→0eh−1h= \lim_{h \to 0} \frac{e^x(e^h - 1)}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}

Since lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1

Therefore: f′(x)=exf'(x) = e^x

Q3.1(A)(2) [3 marks]

If y=log⁡(sin⁡x)y = \log(\sin x) then find dydx\frac{dy}{dx}

Solution: y=log⁡(sin⁡x)y = \log(\sin x)

Using chain rule: dydx=1sin⁡x⋅ddx(sin⁡x)=1sin⁡x⋅cos⁡x=cos⁡xsin⁡x=cot⁡x\frac{dy}{dx} = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x

Q3.1(A)(3) [3 marks]

Evaluate: ∫(4x3+3x2+2x)dx\int\left(4x^3 + 3x^2 + \frac{2}{x}\right)dx

Solution: ∫(4x3+3x2+2x)dx\int\left(4x^3 + 3x^2 + \frac{2}{x}\right)dx

=∫4x3dx+∫3x2dx+∫2xdx= \int 4x^3 dx + \int 3x^2 dx + \int \frac{2}{x} dx

=4⋅x44+3⋅x33+2ln⁡∣x∣+C= 4 \cdot \frac{x^4}{4} + 3 \cdot \frac{x^3}{3} + 2\ln|x| + C

=x4+x3+2ln⁡∣x∣+C= x^4 + x^3 + 2\ln|x| + C

Q.3(B) Attempt any two [8 marks]

Q3.1(B)(1) [4 marks]

If y=etan⁡x+log⁡(sin⁡x)y = e^{\tan x} + \log(\sin x) then find dydx\frac{dy}{dx}

Solution: y=etan⁡x+log⁡(sin⁡x)y = e^{\tan x} + \log(\sin x)

dydx=ddx[etan⁡x]+ddx[log⁡(sin⁡x)]\frac{dy}{dx} = \frac{d}{dx}[e^{\tan x}] + \frac{d}{dx}[\log(\sin x)]

For first term: ddx[etan⁡x]=etan⁡x⋅sec⁡2x\frac{d}{dx}[e^{\tan x}] = e^{\tan x} \cdot \sec^2 x

For second term: ddx[log⁡(sin⁡x)]=1sin⁡x⋅cos⁡x=cot⁡x\frac{d}{dx}[\log(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x

Therefore: dydx=etan⁡xsec⁡2x+cot⁡x\frac{dy}{dx} = e^{\tan x} \sec^2 x + \cot x

Q3.1(B)(2) [4 marks]

The equation of motion of a particle is s=t4+3ts = t^4 + 3t. Find its velocity and acceleration at t=2t = 2 sec

Solution: Given: s=t4+3ts = t^4 + 3t

Velocity: v=dsdt=4t3+3v = \frac{ds}{dt} = 4t^3 + 3

At t=2t = 2: v=4(2)3+3=4(8)+3=32+3=35v = 4(2)^3 + 3 = 4(8) + 3 = 32 + 3 = 35 units/sec

Acceleration: a=dvdt=d2sdt2=12t2a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 12t^2

At t=2t = 2: a=12(2)2=12(4)=48a = 12(2)^2 = 12(4) = 48 units/sec²

Q3.1(B)(3) [4 marks]

Find the maximum and minimum value of the function f(x)=2x3−3x2−12x+5f(x) = 2x^3 - 3x^2 - 12x + 5

Solution: f(x)=2x3−3x2−12x+5f(x) = 2x^3 - 3x^2 - 12x + 5

f′(x)=6x2−6x−12=6(x2−x−2)=6(x−2)(x+1)f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)

For critical points: f′(x)=0f'(x) = 0 6(x−2)(x+1)=06(x-2)(x+1) = 0 x=2x = 2 or x=−1x = -1

f′′(x)=12x−6f''(x) = 12x - 6

At x=−1x = -1: f′′(−1)=12(−1)−6=−18<0f''(-1) = 12(-1) - 6 = -18 < 0 (Maximum) At x=2x = 2: f′′(2)=12(2)−6=18>0f''(2) = 12(2) - 6 = 18 > 0 (Minimum)

f(−1)=2(−1)3−3(−1)2−12(−1)+5=−2−3+12+5=12f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12 (Maximum) f(2)=2(8)−3(4)−12(2)+5=16−12−24+5=−15f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15 (Minimum)

Maximum value: 12 at x=−1x = -1 Minimum value: -15 at x=2x = 2

Q.4(A) Attempt any two [6 marks]

Q4.1(A)(1) [3 marks]

Evaluate: ∫xexdx\int xe^x dx

Solution: Using integration by parts: ∫udv=uv−∫vdu\int u dv = uv - \int v du

Let u=xu = x, dv=exdxdv = e^x dx Then du=dxdu = dx, v=exv = e^x

∫xexdx=x⋅ex−∫exdx=xex−ex+C=ex(x−1)+C\int xe^x dx = x \cdot e^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C

Q4.1(A)(2) [3 marks]

Evaluate: ∫dx9−4x2\int \frac{dx}{\sqrt{9-4x^2}}

Solution: ∫dx9−4x2=∫dx9(1−4x29)=∫dx31−(2x3)2\int \frac{dx}{\sqrt{9-4x^2}} = \int \frac{dx}{\sqrt{9(1-\frac{4x^2}{9})}} = \int \frac{dx}{3\sqrt{1-\left(\frac{2x}{3}\right)^2}}

Let 2x3=sin⁡θ\frac{2x}{3} = \sin \theta, then x=3sin⁡θ2x = \frac{3\sin \theta}{2}, dx=3cos⁡θ2dθdx = \frac{3\cos \theta}{2} d\theta

=∫3cos⁡θ2dθ31−sin⁡2θ=∫3cos⁡θ2dθ3cos⁡θ=∫12dθ=θ2+C= \int \frac{\frac{3\cos \theta}{2} d\theta}{3\sqrt{1-\sin^2 \theta}} = \int \frac{\frac{3\cos \theta}{2} d\theta}{3\cos \theta} = \int \frac{1}{2} d\theta = \frac{\theta}{2} + C

=12sin⁡−1(2x3)+C= \frac{1}{2}\sin^{-1}\left(\frac{2x}{3}\right) + C

Q4.1(A)(3) [3 marks]

Find complex conjugate of 1−i1+i\frac{1-i}{1+i}

Solution: 1−i1+i=(1−i)(1−i)(1+i)(1−i)=(1−i)21−i2=1−2i+i21−(−1)=1−2i−12=−2i2=−i\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{(1-i)^2}{1-i^2} = \frac{1-2i+i^2}{1-(-1)} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i

Complex conjugate of −i-i is −i‾=i\overline{-i} = i

Q.4(B) Attempt any two [8 marks]

Q4.1(B)(1) [4 marks]

Evaluate: ∫0π/2cos⁡xcos⁡x+sin⁡xdx\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx

Solution: Let I=∫0π/2cos⁡xcos⁡x+sin⁡xdxI = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx

Using property: ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x)dx = \int_0^a f(a-x)dx

I=∫0π/2cos⁡(π/2−x)cos⁡(π/2−x)+sin⁡(π/2−x)dx=∫0π/2sin⁡xsin⁡x+cos⁡xdxI = \int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2-x)}}{\sqrt{\cos(\pi/2-x)} + \sqrt{\sin(\pi/2-x)}} dx = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx

Adding both expressions: 2I=∫0π/2cos⁡x+sin⁡xcos⁡x+sin⁡xdx=∫0π/21dx=π22I = \int_0^{\pi/2} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}

Therefore: I=π4I = \frac{\pi}{4}

Q4.1(B)(2) [4 marks]

Find the area of circle x2+y2=a2x^2 + y^2 = a^2 using integration

Solution: For circle x2+y2=a2x^2 + y^2 = a^2, we have y=±a2−x2y = \pm\sqrt{a^2-x^2}

Area of circle = 4×4 \times Area in first quadrant =4∫0aa2−x2dx= 4\int_0^a \sqrt{a^2-x^2} dx

Let x=asin⁡θx = a\sin \theta, dx=acos⁡θdθdx = a\cos \theta d\theta When x=0x = 0, θ=0\theta = 0; when x=ax = a, θ=π/2\theta = \pi/2

=4∫0π/2a2−a2sin⁡2θ⋅acos⁡θdθ= 4\int_0^{\pi/2} \sqrt{a^2-a^2\sin^2 \theta} \cdot a\cos \theta d\theta =4∫0π/2acos⁡θ⋅acos⁡θdθ= 4\int_0^{\pi/2} a\cos \theta \cdot a\cos \theta d\theta =4a2∫0π/2cos⁡2θdθ= 4a^2\int_0^{\pi/2} \cos^2 \theta d\theta =4a2⋅π4=πa2= 4a^2 \cdot \frac{\pi}{4} = \pi a^2

Q4.1(B)(3) [4 marks]

Simplify: (cos⁡3θ+isin⁡3θ)4⋅(cos⁡θ−isin⁡θ)5(cos⁡2θ−isin⁡2θ)3⋅(cos⁡12θ+isin⁡12θ)\frac{(\cos 3\theta + i\sin 3\theta)^4 \cdot (\cos \theta - i\sin \theta)^5}{(\cos 2\theta - i\sin 2\theta)^3 \cdot (\cos 12\theta + i\sin 12\theta)}

Solution: Using De Moivre's theorem: (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta

Numerator: (cos⁡3θ+isin⁡3θ)4⋅(cos⁡θ−isin⁡θ)5(\cos 3\theta + i\sin 3\theta)^4 \cdot (\cos \theta - i\sin \theta)^5 =(cos⁡12θ+isin⁡12θ)⋅(cos⁡(−5θ)+isin⁡(−5θ))= (\cos 12\theta + i\sin 12\theta) \cdot (\cos(-5\theta) + i\sin(-5\theta)) =cos⁡(12θ−5θ)+isin⁡(12θ−5θ)= \cos(12\theta - 5\theta) + i\sin(12\theta - 5\theta) =cos⁡7θ+isin⁡7θ= \cos 7\theta + i\sin 7\theta

Denominator: (cos⁡2θ−isin⁡2θ)3⋅(cos⁡12θ+isin⁡12θ)(\cos 2\theta - i\sin 2\theta)^3 \cdot (\cos 12\theta + i\sin 12\theta) =(cos⁡(−6θ)+isin⁡(−6θ))⋅(cos⁡12θ+isin⁡12θ)= (\cos(-6\theta) + i\sin(-6\theta)) \cdot (\cos 12\theta + i\sin 12\theta) =cos⁡(−6θ+12θ)+isin⁡(−6θ+12θ)= \cos(-6\theta + 12\theta) + i\sin(-6\theta + 12\theta) =cos⁡6θ+isin⁡6θ= \cos 6\theta + i\sin 6\theta

Result: cos⁡7θ+isin⁡7θcos⁡6θ+isin⁡6θ=cos⁡(7θ−6θ)+isin⁡(7θ−6θ)=cos⁡θ+isin⁡θ\frac{\cos 7\theta + i\sin 7\theta}{\cos 6\theta + i\sin 6\theta} = \cos(7\theta - 6\theta) + i\sin(7\theta - 6\theta) = \cos \theta + i\sin \theta

Q.5(A) Attempt any two [6 marks]

Q5.1(A)(1) [3 marks]

If (3x−7)+2iy=5y+(5+x)i(3x - 7) + 2iy = 5y + (5 + x)i then find value of x and y

Solution: (3x−7)+2iy=5y+(5+x)i(3x - 7) + 2iy = 5y + (5 + x)i

Comparing real and imaginary parts: Real parts: 3x−7=5y3x - 7 = 5y ... (1) Imaginary parts: 2y=5+x2y = 5 + x ... (2)

From equation (2): x=2y−5x = 2y - 5 ... (3)

Substituting (3) in (1): 3(2y−5)−7=5y3(2y - 5) - 7 = 5y 6y−15−7=5y6y - 15 - 7 = 5y 6y−22=5y6y - 22 = 5y y=22y = 22

From (3): x=2(22)−5=44−5=39x = 2(22) - 5 = 44 - 5 = 39

Therefore: x=39,y=22x = 39, y = 22

Q5.1(A)(2) [3 marks]

Convert z=1+3iz = 1 + \sqrt{3}i into polar form

Solution: z=1+3iz = 1 + \sqrt{3}i

Modulus: ∣z∣=12+(3)2=1+3=4=2|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2

Argument: arg⁡(z)=tan⁡−1(31)=tan⁡−1(3)=π3\arg(z) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

Polar form: z=∣z∣(cos⁡θ+isin⁡θ)=2(cos⁡π3+isin⁡π3)z = |z|(\cos \theta + i\sin \theta) = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)

Q5.1(A)(3) [3 marks]

Express 4+2i(3+2i)(5−3i)\frac{4 + 2i}{(3 + 2i)(5 - 3i)} in a+iba + ib form

Solution: First, simplify denominator: (3+2i)(5−3i)=15−9i+10i−6i2=15+i−6(−1)=15+i+6=21+i(3 + 2i)(5 - 3i) = 15 - 9i + 10i - 6i^2 = 15 + i - 6(-1) = 15 + i + 6 = 21 + i

4+2i21+i=(4+2i)(21−i)(21+i)(21−i)=84−4i+42i−2i2212−i2=84+38i+2441+1=86+38i442\frac{4 + 2i}{21 + i} = \frac{(4 + 2i)(21 - i)}{(21 + i)(21 - i)} = \frac{84 - 4i + 42i - 2i^2}{21^2 - i^2} = \frac{84 + 38i + 2}{441 + 1} = \frac{86 + 38i}{442}

=86442+38442i=43221+19221i= \frac{86}{442} + \frac{38}{442}i = \frac{43}{221} + \frac{19}{221}i

Q.5(B) Attempt any two [8 marks]

Q5.1(B)(1) [4 marks]

Solve differential equation: dydx+2y=3ex\frac{dy}{dx} + 2y = 3e^x

Solution: This is a first-order linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q

Here: P=2P = 2, Q=3exQ = 3e^x

Integration factor: μ=e∫Pdx=e∫2dx=e2x\mu = e^{\int P dx} = e^{\int 2 dx} = e^{2x}

Multiplying equation by μ\mu: e2xdydx+2e2xy=3e2x⋅ex=3e3xe^{2x}\frac{dy}{dx} + 2e^{2x}y = 3e^{2x} \cdot e^x = 3e^{3x}

This gives: ddx(ye2x)=3e3x\frac{d}{dx}(ye^{2x}) = 3e^{3x}

Integrating both sides: ye2x=∫3e3xdx=3⋅e3x3+C=e3x+Cye^{2x} = \int 3e^{3x} dx = 3 \cdot \frac{e^{3x}}{3} + C = e^{3x} + C

Therefore: y=e3x+Ce2x=ex+Ce−2xy = \frac{e^{3x} + C}{e^{2x}} = e^x + Ce^{-2x}

Q5.1(B)(2) [4 marks]

Solve differential equation: dydx=(x+y)2\frac{dy}{dx} = (x + y)^2

Solution: Let v=x+yv = x + y, then dvdx=1+dydx\frac{dv}{dx} = 1 + \frac{dy}{dx}

So dydx=dvdx−1\frac{dy}{dx} = \frac{dv}{dx} - 1

Substituting in the original equation: dvdx−1=v2\frac{dv}{dx} - 1 = v^2 dvdx=v2+1\frac{dv}{dx} = v^2 + 1

Separating variables: dvv2+1=dx\frac{dv}{v^2 + 1} = dx

Integrating both sides: ∫dvv2+1=∫dx\int \frac{dv}{v^2 + 1} = \int dx tan⁡−1(v)=x+C\tan^{-1}(v) = x + C v=tan⁡(x+C)v = \tan(x + C)

Substituting back: x+y=tan⁡(x+C)x + y = \tan(x + C) Therefore: y=tan⁡(x+C)−xy = \tan(x + C) - x

Q5.1(B)(3) [4 marks]

Solve differential equation: dydx+yx=ex\frac{dy}{dx} + \frac{y}{x} = e^x, y(0)=2y(0) = 2

Solution: This is a first-order linear differential equation: dydx+yx=ex\frac{dy}{dx} + \frac{y}{x} = e^x

Here: P=1xP = \frac{1}{x}, Q=exQ = e^x

Integration factor: μ=e∫1xdx=eln⁡∣x∣=∣x∣=x\mu = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x| = x (for x>0x > 0)

Multiplying equation by μ=x\mu = x: xdydx+y=xexx\frac{dy}{dx} + y = xe^x

This gives: ddx(xy)=xex\frac{d}{dx}(xy) = xe^x

Integrating both sides using integration by parts: xy=∫xexdxxy = \int xe^x dx

For ∫xexdx\int xe^x dx: Let u=xu = x, dv=exdxdv = e^x dx Then du=dxdu = dx, v=exv = e^x ∫xexdx=xex−∫exdx=xex−ex=ex(x−1)\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x = e^x(x-1)

So: xy=ex(x−1)+Cxy = e^x(x-1) + C y=ex(x−1)+Cxy = \frac{e^x(x-1) + C}{x}

Using initial condition y(0)=2y(0) = 2: This presents a problem as we have division by zero. The equation needs to be solved more carefully near x=0x = 0.

For the general solution: y=ex(1−1x)+Cxy = e^x\left(1 - \frac{1}{x}\right) + \frac{C}{x}

Formula Cheat Sheet

Matrix Operations

  • Matrix addition: (A+B)ij=Aij+Bij(A + B)_{ij} = A_{ij} + B_{ij}
  • Matrix multiplication: (AB)ij=∑kAikBkj(AB)_{ij} = \sum_{k} A_{ik}B_{kj}
  • Transpose: (AT)ij=Aji(A^T)_{ij} = A_{ji}
  • Inverse of 2×2 matrix: A−1=1∣A∣[d−b−ca]A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} where A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

Differentiation Formulas

  • ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}
  • ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
  • ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x
  • ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x
  • ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x
  • ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}
  • Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'
  • Chain rule: ddxf(g(x))=f′(g(x))⋅g′(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

Integration Formulas

  • ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C (for n≠−1n \neq -1)
  • ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C
  • ∫exdx=ex+C\int e^x dx = e^x + C
  • ∫sin⁡xdx=−cos⁡x+C\int \sin x dx = -\cos x + C
  • ∫cos⁡xdx=sin⁡x+C\int \cos x dx = \sin x + C
  • ∫sec⁡2xdx=tan⁡x+C\int \sec^2 x dx = \tan x + C
  • ∫11+x2dx=tan⁡−1x+C\int \frac{1}{1+x^2} dx = \tan^{-1} x + C
  • ∫11−x2dx=sin⁡−1x+C\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C

Differential Equations

  • First-order linear: dydx+Py=Q\frac{dy}{dx} + Py = Q
  • Integration factor: μ=e∫Pdx\mu = e^{\int P dx}
  • Solution: y=1μ[∫μQdx+C]y = \frac{1}{\mu}\left[\int \mu Q dx + C\right]
  • Variable separable: dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y) → dyg(y)=f(x)dx\frac{dy}{g(y)} = f(x)dx

Complex Numbers

  • i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1
  • Modulus: ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}
  • Argument: arg⁡(a+bi)=tan⁡−1(ba)\arg(a + bi) = \tan^{-1}\left(\frac{b}{a}\right)
  • Polar form: z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta)
  • De Moivre's theorem: (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta

Problem-Solving Strategies

Matrix Problems

  1. Always check dimensions before performing operations
  2. For matrix equations: Use inverse method X=A−1BX = A^{-1}B
  3. For transpose properties: Use (AB)T=BTAT(AB)^T = B^T A^T
  4. For matrix powers: Calculate step by step, look for patterns

Differentiation Problems

  1. Identify the type: Product, quotient, chain rule, or implicit
  2. For complex functions: Break down using appropriate rules
  3. For applications: Remember v=dsdtv = \frac{ds}{dt}, a=dvdta = \frac{dv}{dt}
  4. For maxima/minima: Find critical points where f′(x)=0f'(x) = 0

Integration Problems

  1. Recognize standard forms first
  2. For substitution: Look for f′(x)f'(x) when f(x)f(x) appears
  3. For integration by parts: Choose uu as LIATE (Log, Inverse trig, Algebraic, Trig, Exponential)
  4. For definite integrals: Use fundamental theorem or properties

Differential Equations

  1. Identify the type: Linear, separable, or exact
  2. For linear equations: Find integration factor systematically
  3. For separable equations: Separate variables completely before integrating
  4. Always check initial conditions if given

Complex Numbers

  1. For operations: Convert to a+bia + bi form first
  2. For polar form: Calculate modulus and argument carefully
  3. For powers: Use De Moivre's theorem
  4. For division: Multiply by conjugate of denominator

Common Mistakes to Avoid

Matrix Operations

  • ❌ Don't assume AB=BAAB = BA (matrix multiplication is not commutative)
  • ❌ Don't forget to check if matrices can be multiplied (inner dimensions must match)
  • ❌ Don't confuse transpose with inverse

Differentiation

  • ❌ Don't forget the chain rule for composite functions
  • ❌ Don't mix up ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x and ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x
  • ❌ Don't forget to use product rule when multiplying functions

Integration

  • ❌ Don't forget the constant of integration +C+C
  • ❌ Don't confuse indefinite and definite integrals
  • ❌ Don't forget to substitute limits properly in definite integrals

Complex Numbers

  • ❌ Don't forget i2=−1i^2 = -1 when expanding
  • ❌ Don't confuse modulus with real part
  • ❌ Don't forget to rationalize denominators with complex numbers

Exam Tips

Time Management

  • Spend 2-3 minutes reading the entire paper first
  • Attempt easier questions first to build confidence
  • Reserve 15 minutes at the end for review

Writing Strategy

  • Show all steps clearly - partial marks are often awarded
  • Draw diagrams where helpful - especially for geometry problems
  • Write final answers clearly and box them if possible

Calculation Tips

  • Double-check arithmetic - many marks are lost due to calculation errors
  • Use calculator efficiently but don't become dependent on it
  • Cross-verify answers using different methods when possible

Question Selection

  • In OR questions, choose the one you're most confident about
  • Don't spend too much time on any single question
  • If stuck, move on and return later with fresh perspective

Good luck with your exam preparation!