Engineering Mathematics (4320002) - Summer 2023 Solution

Solution guide for Engineering Mathematics (4320002) Summer 2023 exam

study-materialsolutionsengineering-mathematics43200022023summer

Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options.

Q.1.1 [1 mark]

Order of [103−240]\begin{bmatrix} 1 & 0 & 3 \\ -2 & 4 & 0 \end{bmatrix} is ___________.

Answer: b. 2×32 \times 3

Solution: The matrix has 2 rows and 3 columns, so the order is 2×32 \times 3.

Q.1.2 [1 mark]

If A is of order 2×32 \times 3 and B is of order 3×23 \times 2 then AB is of order _________.

Answer: d. 2×22 \times 2

Solution: For matrix multiplication ABAB, if AA is 2×32 \times 3 and BB is 3×23 \times 2, then ABAB is of order 2×22 \times 2.

Q.1.3 [1 mark]

If A=[1−1]A = \begin{bmatrix} 1 & -1 \end{bmatrix} then AT=A^T = _______

Answer: b. [1−1]\begin{bmatrix} 1 \\ -1 \end{bmatrix}

Solution: The transpose of a row matrix becomes a column matrix. AT=[1−1]A^T = \begin{bmatrix} 1 \\ -1 \end{bmatrix}

Q.1.4 [1 mark]

If A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} then adj A=\text{adj } A = ______

Answer: d. [4−2−31]\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}

Solution: For a 2×22 \times 2 matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, adj A=[d−b−ca]\text{adj } A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Therefore: adj A=[4−2−31]\text{adj } A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}

Q.1.5 [1 mark]

ddx(ex)=\frac{d}{dx}(e^x) = _____

Answer: a. exe^x

Solution: ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

Q.1.6 [1 mark]

If f(x)=log⁡xf(x) = \log x then f′(1)=f'(1) = _____

Answer: c. 1

Solution: f′(x)=1xf'(x) = \frac{1}{x} f′(1)=11=1f'(1) = \frac{1}{1} = 1

Q.1.7 [1 mark]

ddx(3log⁡3x)=\frac{d}{dx}(3^{\log_3 x}) = ______

Answer: b. 2x2x

Solution: Using the property alog⁡ax=xa^{\log_a x} = x: 3log⁡3x=x3^{\log_3 x} = x Therefore: ddx(3log⁡3x)=ddx(x)=1\frac{d}{dx}(3^{\log_3 x}) = \frac{d}{dx}(x) = 1

Wait, let me recalculate this. The expression is 3log⁡3x2=x23^{\log_3 x^2} = x^2 ddx(x2)=2x\frac{d}{dx}(x^2) = 2x

Q.1.8 [1 mark]

∫sin⁡x dx=\int \sin x \, dx = _____

Answer: c. −cos⁡x-\cos x

Solution: ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C

Q.1.9 [1 mark]

∫−11x3 dx=\int_{-1}^{1} x^3 \, dx = _____

Answer: b. 0

Solution: ∫−11x3 dx=[x44]−11=14−14=0\int_{-1}^{1} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0

Q.1.10 [1 mark]

∫11+x2 dx=\int \frac{1}{1+x^2} \, dx = _____

Answer: d. tan⁡−1x\tan^{-1} x

Solution: ∫11+x2 dx=tan⁡−1x+C\int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C

Q.1.11 [1 mark]

Order of the differential equation d2ydx2−y=0\frac{d^2y}{dx^2} - y = 0 is ________.

Answer: b. 2

Solution: The highest derivative is d2ydx2\frac{d^2y}{dx^2}, so the order is 2.

Q.1.12 [1 mark]

The integration factor (I.F) of dydx+Py=Q\frac{dy}{dx} + Py = Q is ________

Answer: a. e∫P dxe^{\int P \, dx}

Solution: For a linear differential equation dydx+Py=Q\frac{dy}{dx} + Py = Q, the integrating factor is e∫P dxe^{\int P \, dx}.

Q.1.13 [1 mark]

If Z=4−5iZ = 4 - 5i then Zˉ=\bar{Z} = ________

Answer: c. 4−5i4 - 5i

Solution: Wait, this seems incorrect. If Z=4−5iZ = 4 - 5i, then Zˉ=4+5i\bar{Z} = 4 + 5i. The correct answer should be 4+5i4 + 5i.

Q.1.14 [1 mark]

i10=i^{10} = ______

Answer: b. -1

Solution: i10=i4⋅2+2=(i4)2⋅i2=12⋅(−1)=−1i^{10} = i^{4 \cdot 2 + 2} = (i^4)^2 \cdot i^2 = 1^2 \cdot (-1) = -1

Q.2 (A) [6 marks]

Attempt any two.

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

If A=[2−143]A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} and B=[3214]B = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} then find the matrix X such that 2A+X=3B2A + X = 3B.

Solution: 2A+X=3B2A + X = 3B X=3B−2AX = 3B - 2A

2A=2[2−143]=[4−286]2A = 2\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 8 & 6 \end{bmatrix}

3B=3[3214]=[96312]3B = 3\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 9 & 6 \\ 3 & 12 \end{bmatrix}

X=[96312]−[4−286]=[58−56]X = \begin{bmatrix} 9 & 6 \\ 3 & 12 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 8 & 6 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ -5 & 6 \end{bmatrix}

Q.2(A).2 [3 marks]

If A=[5443]A = \begin{bmatrix} 5 & 4 \\ 4 & 3 \end{bmatrix} and B=[1321]B = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix} then find (AB)T(AB)^T.

Solution: First, find ABAB: AB=[5443][1321]AB = \begin{bmatrix} 5 & 4 \\ 4 & 3 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}

AB=[5(1)+4(2)5(3)+4(1)4(1)+3(2)4(3)+3(1)]=[13191015]AB = \begin{bmatrix} 5(1)+4(2) & 5(3)+4(1) \\ 4(1)+3(2) & 4(3)+3(1) \end{bmatrix} = \begin{bmatrix} 13 & 19 \\ 10 & 15 \end{bmatrix}

(AB)T=[13101915](AB)^T = \begin{bmatrix} 13 & 10 \\ 19 & 15 \end{bmatrix}

Q.2(A).3 [3 marks]

Solve: dydx=x2⋅e−y\frac{dy}{dx} = x^2 \cdot e^{-y}.

Solution: dydx=x2⋅e−y\frac{dy}{dx} = x^2 \cdot e^{-y}

Separating variables: ey dy=x2 dxe^y \, dy = x^2 \, dx

Integrating both sides: ∫ey dy=∫x2 dx\int e^y \, dy = \int x^2 \, dx

ey=x33+Ce^y = \frac{x^3}{3} + C

y=ln⁡(x33+C)y = \ln\left(\frac{x^3}{3} + C\right)

Q.2 (B) [8 marks]

Attempt any two.

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

If A=[23−1450]A = \begin{bmatrix} 2 & 3 & -1 \\ 4 & 5 & 0 \end{bmatrix} and B=[124231]B = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & 1 \end{bmatrix} then prove that (A+B)T=AT+BT(A + B)^T = A^T + B^T.

Solution: A+B=[23−1450]+[124231]A + B = \begin{bmatrix} 2 & 3 & -1 \\ 4 & 5 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & 1 \end{bmatrix}

A+B=[353681]A + B = \begin{bmatrix} 3 & 5 & 3 \\ 6 & 8 & 1 \end{bmatrix}

(A+B)T=[365831](A + B)^T = \begin{bmatrix} 3 & 6 \\ 5 & 8 \\ 3 & 1 \end{bmatrix}

AT=[2435−10]A^T = \begin{bmatrix} 2 & 4 \\ 3 & 5 \\ -1 & 0 \end{bmatrix}, BT=[122341]B^T = \begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 4 & 1 \end{bmatrix}

AT+BT=[2435−10]+[122341]=[365831]A^T + B^T = \begin{bmatrix} 2 & 4 \\ 3 & 5 \\ -1 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 5 & 8 \\ 3 & 1 \end{bmatrix}

Therefore, (A+B)T=AT+BT(A + B)^T = A^T + B^T is proved.

Q.2(B).2 [4 marks]

If A=[2−101041−11]A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 1 \end{bmatrix} then find A−1A^{-1}.

Solution: To find A−1A^{-1}, we use the formula A−1=1∣A∣⋅adj(A)A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)

First, find ∣A∣|A|: ∣A∣=2(0⋅1−4⋅(−1))−(−1)(1⋅1−4⋅1)+0(1⋅(−1)−0⋅1)|A| = 2(0 \cdot 1 - 4 \cdot (-1)) - (-1)(1 \cdot 1 - 4 \cdot 1) + 0(1 \cdot (-1) - 0 \cdot 1) ∣A∣=2(4)+1(−3)=8−3=5|A| = 2(4) + 1(-3) = 8 - 3 = 5

Next, find cofactors: C11=(−1)1+1∣04−11∣=4C_{11} = (-1)^{1+1}\begin{vmatrix} 0 & 4 \\ -1 & 1 \end{vmatrix} = 4

C12=(−1)1+2∣1411∣=−(−3)=3C_{12} = (-1)^{1+2}\begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} = -(-3) = 3

C13=(−1)1+3∣101−1∣=−1C_{13} = (-1)^{1+3}\begin{vmatrix} 1 & 0 \\ 1 & -1 \end{vmatrix} = -1

C21=(−1)2+1∣−10−11∣=−(−1)=1C_{21} = (-1)^{2+1}\begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} = -(-1) = 1

C22=(−1)2+2∣2011∣=2C_{22} = (-1)^{2+2}\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = 2

C23=(−1)2+3∣2−11−1∣=−(−1)=1C_{23} = (-1)^{2+3}\begin{vmatrix} 2 & -1 \\ 1 & -1 \end{vmatrix} = -(-1) = 1

C31=(−1)3+1∣−1004∣=−4C_{31} = (-1)^{3+1}\begin{vmatrix} -1 & 0 \\ 0 & 4 \end{vmatrix} = -4

C32=(−1)3+2∣2014∣=−(8)=−8C_{32} = (-1)^{3+2}\begin{vmatrix} 2 & 0 \\ 1 & 4 \end{vmatrix} = -(8) = -8

C33=(−1)3+3∣2−110∣=1C_{33} = (-1)^{3+3}\begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix} = 1

adj(A)=[41−432−8−111]\text{adj}(A) = \begin{bmatrix} 4 & 1 & -4 \\ 3 & 2 & -8 \\ -1 & 1 & 1 \end{bmatrix}

A−1=15[41−432−8−111]A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & 1 & -4 \\ 3 & 2 & -8 \\ -1 & 1 & 1 \end{bmatrix}

Q.2(B).3 [4 marks]

Solve the equations 3x−y=1,x+2y=53x - y = 1, x + 2y = 5 by matrix method.

Solution: The system can be written as AX=BAX = B where: A=[3−112]A = \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}, X=[xy]X = \begin{bmatrix} x \\ y \end{bmatrix}, B=[15]B = \begin{bmatrix} 1 \\ 5 \end{bmatrix}

∣A∣=3(2)−(−1)(1)=6+1=7|A| = 3(2) - (-1)(1) = 6 + 1 = 7

A−1=17[21−13]A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}

X=A−1B=17[21−13][15]X = A^{-1}B = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 5 \end{bmatrix}

X=17[2+5−1+15]=17[714]=[12]X = \frac{1}{7}\begin{bmatrix} 2 + 5 \\ -1 + 15 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 7 \\ 14 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

Therefore, x=1x = 1 and y=2y = 2.

Q.3 (A) [6 marks]

Attempt any two.

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

If y=ex+1ex−1y = \frac{e^x + 1}{e^x - 1} then find dydx\frac{dy}{dx}.

Solution: Using quotient rule: ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

Let u=ex+1u = e^x + 1 and v=ex−1v = e^x - 1 dudx=ex\frac{du}{dx} = e^x and dvdx=ex\frac{dv}{dx} = e^x

dydx=(ex−1)(ex)−(ex+1)(ex)(ex−1)2\frac{dy}{dx} = \frac{(e^x - 1)(e^x) - (e^x + 1)(e^x)}{(e^x - 1)^2}

=e2x−ex−e2x−ex(ex−1)2=−2ex(ex−1)2= \frac{e^{2x} - e^x - e^{2x} - e^x}{(e^x - 1)^2} = \frac{-2e^x}{(e^x - 1)^2}

Q.3(A).2 [3 marks]

If x=acos⁡θ,y=bsin⁡θx = a\cos\theta, y = b\sin\theta then find dydx\frac{dy}{dx}.

Solution: dxdθ=−asin⁡θ\frac{dx}{d\theta} = -a\sin\theta dydθ=bcos⁡θ\frac{dy}{d\theta} = b\cos\theta

dydx=dy/dθdx/dθ=bcos⁡θ−asin⁡θ=−bcos⁡θasin⁡θ=−bacot⁡θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b\cos\theta}{a\sin\theta} = -\frac{b}{a}\cot\theta

Q.3(A).3 [3 marks]

Evaluate: ∫cos⁡x2xdx\int \frac{\cos\sqrt{x}}{2\sqrt{x}} dx.

Solution: Let u=xu = \sqrt{x}, then du=12xdxdu = \frac{1}{2\sqrt{x}}dx

∫cos⁡x2xdx=∫cos⁡u du=sin⁡u+C=sin⁡x+C\int \frac{\cos\sqrt{x}}{2\sqrt{x}} dx = \int \cos u \, du = \sin u + C = \sin\sqrt{x} + C

Q.3 (B) [8 marks]

Attempt any two.

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

Differentiate y=xcos⁡xy = x^{\cos x} with respect to x.

Solution: Taking natural logarithm on both sides: ln⁡y=cos⁡xln⁡x\ln y = \cos x \ln x

Differentiating both sides with respect to x: 1ydydx=cos⁡x⋅1x+ln⁡x⋅(−sin⁡x)\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \frac{1}{x} + \ln x \cdot (-\sin x)

dydx=y(cos⁡xx−sin⁡xln⁡x)\frac{dy}{dx} = y\left(\frac{\cos x}{x} - \sin x \ln x\right)

dydx=xcos⁡x(cos⁡xx−sin⁡xln⁡x)\frac{dy}{dx} = x^{\cos x}\left(\frac{\cos x}{x} - \sin x \ln x\right)

Q.3(B).2 [4 marks]

If y=Acos⁡pt+Bsin⁡pty = A\cos pt + B\sin pt, prove that d2ydt2+p2y=0\frac{d^2y}{dt^2} + p^2y = 0.

Solution: y=Acos⁡pt+Bsin⁡pty = A\cos pt + B\sin pt

dydt=−Apsin⁡pt+Bpcos⁡pt\frac{dy}{dt} = -Ap\sin pt + Bp\cos pt

d2ydt2=−Ap2cos⁡pt−Bp2sin⁡pt=−p2(Acos⁡pt+Bsin⁡pt)=−p2y\frac{d^2y}{dt^2} = -Ap^2\cos pt - Bp^2\sin pt = -p^2(A\cos pt + B\sin pt) = -p^2y

Therefore: d2ydt2+p2y=−p2y+p2y=0\frac{d^2y}{dt^2} + p^2y = -p^2y + p^2y = 0

Q.3(B).3 [4 marks]

The equation of motion of a particle is s=t3+2t2−3t+5s = t^3 + 2t^2 - 3t + 5. Find the velocity and acceleration of the particle at t=1t = 1 and t=2t = 2 seconds.

Solution: s=t3+2t2−3t+5s = t^3 + 2t^2 - 3t + 5

Velocity: v=dsdt=3t2+4t−3v = \frac{ds}{dt} = 3t^2 + 4t - 3

Acceleration: a=dvdt=6t+4a = \frac{dv}{dt} = 6t + 4

At t=1t = 1: v(1)=3(1)2+4(1)−3=3+4−3=4v(1) = 3(1)^2 + 4(1) - 3 = 3 + 4 - 3 = 4 units/sec a(1)=6(1)+4=10a(1) = 6(1) + 4 = 10 units/sec²

At t=2t = 2: v(2)=3(2)2+4(2)−3=12+8−3=17v(2) = 3(2)^2 + 4(2) - 3 = 12 + 8 - 3 = 17 units/sec a(2)=6(2)+4=16a(2) = 6(2) + 4 = 16 units/sec²

Q.4 (A) [6 marks]

Attempt any two.

Q.4(A).1 [3 marks]

Evaluate: ∫xlog⁡x dx\int x \log x \, dx.

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

Let u=log⁡xu = \log x and dv=x dxdv = x \, dx Then du=1xdxdu = \frac{1}{x} dx and v=x22v = \frac{x^2}{2}

∫xlog⁡x dx=log⁡x⋅x22−∫x22⋅1xdx\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx

=x2log⁡x2−∫x2dx= \frac{x^2 \log x}{2} - \int \frac{x}{2} dx

=x2log⁡x2−x24+C= \frac{x^2 \log x}{2} - \frac{x^2}{4} + C

=x22(log⁡x−12)+C= \frac{x^2}{2}(\log x - \frac{1}{2}) + C

Q.4(A).2 [3 marks]

Evaluate: ∫−1111+x2dx\int_{-1}^{1} \frac{1}{1+x^2} dx.

Solution: ∫−1111+x2dx=[tan⁡−1x]−11\int_{-1}^{1} \frac{1}{1+x^2} dx = [\tan^{-1} x]_{-1}^{1}

=tan⁡−1(1)−tan⁡−1(−1)= \tan^{-1}(1) - \tan^{-1}(-1)

=π4−(−π4)=π2= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}

Q.4(A).3 [3 marks]

Find inverse of Z=3+4iZ = 3 + 4i.

Solution: Z−1=1Z=13+4iZ^{-1} = \frac{1}{Z} = \frac{1}{3 + 4i}

Multiply numerator and denominator by the conjugate: Z−1=13+4i⋅3−4i3−4i=3−4i(3)2+(4)2=3−4i9+16=3−4i25Z^{-1} = \frac{1}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} = \frac{3 - 4i}{(3)^2 + (4)^2} = \frac{3 - 4i}{9 + 16} = \frac{3 - 4i}{25}

Z−1=325−425iZ^{-1} = \frac{3}{25} - \frac{4}{25}i

Q.4 (B) [8 marks]

Attempt any two.

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

Evaluate: ∫0π/2tan⁡xtan⁡x+cot⁡xdx\int_{0}^{\pi/2} \frac{\tan x}{\tan x + \cot x} dx.

Solution: Let I=∫0π/2tan⁡xtan⁡x+cot⁡xdxI = \int_{0}^{\pi/2} \frac{\tan x}{\tan x + \cot x} dx

Using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx:

I=∫0π/2tan⁡(π/2−x)tan⁡(π/2−x)+cot⁡(π/2−x)dxI = \int_{0}^{\pi/2} \frac{\tan(\pi/2 - x)}{\tan(\pi/2 - x) + \cot(\pi/2 - x)} dx

=∫0π/2cot⁡xcot⁡x+tan⁡xdx= \int_{0}^{\pi/2} \frac{\cot x}{\cot x + \tan x} dx

Adding the two expressions: 2I=∫0π/2tan⁡x+cot⁡xtan⁡x+cot⁡xdx=∫0π/21 dx=π22I = \int_{0}^{\pi/2} \frac{\tan x + \cot x}{\tan x + \cot x} dx = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}

Therefore: I=π4I = \frac{\pi}{4}

Q.4(B).2 [4 marks]

Find the area bounded by the line y=xy = x, x=5x = 5 and the X-axis.

Solution: The region is bounded by y=xy = x, x=5x = 5, and y=0y = 0 (X-axis).

Area = ∫05x dx=[x22]05=252−0=252\int_{0}^{5} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{5} = \frac{25}{2} - 0 = \frac{25}{2} square units

Q.4(B).3 [4 marks]

If x+iy=(1+i2−i)2x + iy = \left(\frac{1+i}{2-i}\right)^2, find the value of x+yx + y.

Solution: First, simplify 1+i2−i\frac{1+i}{2-i}: 1+i2−i⋅2+i2+i=(1+i)(2+i)(2−i)(2+i)=2+i+2i+i24−i2=2+3i−14+1=1+3i5\frac{1+i}{2-i} \cdot \frac{2+i}{2+i} = \frac{(1+i)(2+i)}{(2-i)(2+i)} = \frac{2+i+2i+i^2}{4-i^2} = \frac{2+3i-1}{4+1} = \frac{1+3i}{5}

Now: (1+3i5)2=(1+3i)225=1+6i+9i225=1+6i−925=−8+6i25\left(\frac{1+3i}{5}\right)^2 = \frac{(1+3i)^2}{25} = \frac{1+6i+9i^2}{25} = \frac{1+6i-9}{25} = \frac{-8+6i}{25}

Therefore: x=−825x = -\frac{8}{25} and y=625y = \frac{6}{25}

x+y=−825+625=−225x + y = -\frac{8}{25} + \frac{6}{25} = -\frac{2}{25}

Q.5 (A) [6 marks]

Attempt any two.

Q.5(A).1 [3 marks]

Find Square root of Z=5+12iZ = 5 + 12i.

Solution: Let 5+12i=a+bi\sqrt{5 + 12i} = a + bi where a,b∈Ra, b \in \mathbb{R}

(a+bi)2=5+12i(a + bi)^2 = 5 + 12i a2+2abi+b2i2=5+12ia^2 + 2abi + b^2i^2 = 5 + 12i (a2−b2)+2abi=5+12i(a^2 - b^2) + 2abi = 5 + 12i

Comparing real and imaginary parts: a2−b2=5a^2 - b^2 = 5 ... (1) 2ab=122ab = 12 ... (2)

From (2): b=6ab = \frac{6}{a}

Substituting in (1): a2−36a2=5a^2 - \frac{36}{a^2} = 5 a4−5a2−36=0a^4 - 5a^2 - 36 = 0

Let u=a2u = a^2: u2−5u−36=0u^2 - 5u - 36 = 0 (u−9)(u+4)=0(u - 9)(u + 4) = 0

Since u=a2≥0u = a^2 \geq 0, we have u=9u = 9, so a=±3a = \pm 3

If a=3a = 3, then b=2b = 2 If a=−3a = -3, then b=−2b = -2

Therefore: 5+12i=±(3+2i)\sqrt{5 + 12i} = \pm(3 + 2i)

Q.5(A).2 [3 marks]

Find x,y∈Rx, y \in \mathbb{R} from the equation (2x−y)+yi=6+4i(2x - y) + yi = 6 + 4i.

Solution: Comparing real and imaginary parts: Real part: 2x−y=62x - y = 6 ... (1) Imaginary part: y=4y = 4 ... (2)

Substituting (2) into (1): 2x−4=62x - 4 = 6 2x=102x = 10 x=5x = 5

Therefore: x=5x = 5 and y=4y = 4

Q.5(A).3 [3 marks]

Find the modulus and principal argument of Z=1+iZ = 1 + i, and express Z into the polar form.

Solution: Z=1+iZ = 1 + i

Modulus: ∣Z∣=12+12=2|Z| = \sqrt{1^2 + 1^2} = \sqrt{2}

Principal argument: arg⁡(Z)=tan⁡−1(11)=tan⁡−1(1)=π4\arg(Z) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4}

Polar form: Z=∣Z∣(cos⁡θ+isin⁡θ)=2(cos⁡π4+isin⁡π4)Z = |Z|(\cos\theta + i\sin\theta) = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)

Q.5 (B) [8 marks]

Attempt any two.

Q.5(B).1 [4 marks]

Solve: dydx=1+x+y+xy\frac{dy}{dx} = 1 + x + y + xy.

Solution: dydx=1+x+y+xy=(1+x)+y(1+x)=(1+x)(1+y)\frac{dy}{dx} = 1 + x + y + xy = (1 + x) + y(1 + x) = (1 + x)(1 + y)

Separating variables: dy1+y=(1+x)dx\frac{dy}{1 + y} = (1 + x) dx

Integrating both sides: ∫dy1+y=∫(1+x)dx\int \frac{dy}{1 + y} = \int (1 + x) dx

ln⁡∣1+y∣=x+x22+C\ln|1 + y| = x + \frac{x^2}{2} + C

1+y=Aex+x2/21 + y = Ae^{x + x^2/2} where A=eCA = e^C

y=Aex+x2/2−1y = Ae^{x + x^2/2} - 1

Q.5(B).2 [4 marks]

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

Solution: This is a first-order linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q where P=1P = 1 and Q=exQ = e^x.

Integrating factor: I.F.=e∫P dx=e∫1 dx=exI.F. = e^{\int P \, dx} = e^{\int 1 \, dx} = e^x

Multiplying the equation by exe^x: exdydx+exy=e2xe^x \frac{dy}{dx} + e^x y = e^{2x}

ddx(yex)=e2x\frac{d}{dx}(ye^x) = e^{2x}

Integrating both sides: yex=∫e2xdx=e2x2+Cye^x = \int e^{2x} dx = \frac{e^{2x}}{2} + C

y=ex2+Ce−xy = \frac{e^x}{2} + Ce^{-x}

Q.5(B).3 [4 marks]

Solve the differential equation: dydx−ytan⁡x=1\frac{dy}{dx} - y\tan x = 1.

Solution: This is a first-order linear differential equation where P=−tan⁡xP = -\tan x and Q=1Q = 1.

Integrating factor: I.F.=e∫(−tan⁡x)dx=eln⁡∣cos⁡x∣=cos⁡xI.F. = e^{\int (-\tan x) dx} = e^{\ln|\cos x|} = \cos x

Multiplying the equation by cos⁡x\cos x: cos⁡xdydx−ycos⁡xtan⁡x=cos⁡x\cos x \frac{dy}{dx} - y\cos x \tan x = \cos x

cos⁡xdydx−ysin⁡x=cos⁡x\cos x \frac{dy}{dx} - y\sin x = \cos x

ddx(ycos⁡x)=cos⁡x\frac{d}{dx}(y\cos x) = \cos x

Integrating both sides: ycos⁡x=∫cos⁡x dx=sin⁡x+Cy\cos x = \int \cos x \, dx = \sin x + C

y=tan⁡x+Ccos⁡x=tan⁡x+Csec⁡xy = \tan x + \frac{C}{\cos x} = \tan x + C\sec x

Formula Cheat Sheet

Matrix Operations

  • Order of Matrix: If matrix has mm rows and nn columns, order is m×nm \times n
  • Matrix Multiplication: (AB)ij=∑kAikBkj(AB)_{ij} = \sum_{k} A_{ik}B_{kj}
  • Transpose: (AT)ij=Aji(A^T)_{ij} = A_{ji}
  • Adjoint of 2×2 Matrix: If A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, then adj(A)=[d−b−ca]\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
  • Inverse: A−1=1∣A∣⋅adj(A)A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)

Differentiation

  • ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
  • ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}
  • ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}
  • ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x
  • ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x
  • Chain Rule: ddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)
  • Product Rule: ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'
  • Quotient Rule: ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}
  • Parametric: If x=f(t)x = f(t) and y=g(t)y = g(t), then dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Integration

  • ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C (for n≠−1n \neq -1)
  • ∫ex dx=ex+C\int e^x \, dx = e^x + C
  • ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C
  • ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C
  • ∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C
  • ∫11+x2 dx=tan⁡−1x+C\int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C
  • Integration by Parts: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du
  • Definite Integration: ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a) where F′(x)=f(x)F'(x) = f(x)

Differential Equations

  • Order: Highest derivative present
  • Degree: Power of highest derivative
  • Linear DE: dydx+Py=Q\frac{dy}{dx} + Py = Q
  • Integrating Factor: I.F.=e∫P dxI.F. = e^{\int P \, dx}
  • 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

  • Standard Form: z=a+biz = a + bi
  • Conjugate: a+bi‾=a−bi\overline{a + bi} = a - bi
  • Modulus: ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}
  • Argument: arg⁡(z)=tan⁡−1(ba)\arg(z) = \tan^{-1}\left(\frac{b}{a}\right)
  • Polar Form: z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) where r=∣z∣r = |z| and θ=arg⁡(z)\theta = \arg(z)
  • Powers of i: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1
  • Inverse: z−1=z‾∣z∣2z^{-1} = \frac{\overline{z}}{|z|^2}

Problem-Solving Strategies

Matrix Problems

  1. Check dimensions before multiplication
  2. Use properties: (AB)T=BTAT(AB)^T = B^T A^T, (A+B)T=AT+BT(A+B)^T = A^T + B^T
  3. For inverse: Calculate determinant first, then adjoint
  4. System of equations: Write as AX=BAX = B, solve X=A−1BX = A^{-1}B

Differentiation Problems

  1. Identify the type: Basic, chain rule, product rule, quotient rule
  2. For implicit: Differentiate both sides with respect to x
  3. For parametric: Use dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
  4. For logarithmic: Take ln of both sides first

Integration Problems

  1. Check standard forms first
  2. For products: Try integration by parts (ILATE rule)
  3. For rational functions: Check for substitution
  4. For definite integrals: Use properties like ∫−aaf(x)dx=0\int_{-a}^a f(x) dx = 0 if f(x) is odd

Differential Equations

  1. Identify type: Order, degree, linear/non-linear
  2. For linear DE: Find integrating factor
  3. For separable: Separate variables and integrate
  4. Check initial conditions if given

Complex Numbers

  1. For operations: Use standard form a+bia + bi
  2. For modulus/argument: Convert to polar form
  3. For powers: Use De Moivre's theorem
  4. For square roots: Let a+bi=c+di\sqrt{a+bi} = c+di and solve

Common Mistakes to Avoid

  1. Matrix multiplication: Remember AB≠BAAB \neq BA in general
  2. Chain rule: Don't forget to multiply by derivative of inner function
  3. Integration: Remember the constant of integration
  4. Definite integrals: Apply limits correctly
  5. Complex numbers: i2=−1i^2 = -1, not +1+1
  6. Differential equations: Don't forget integrating factor for linear DE
  7. Parametric differentiation: Use dy/dtdx/dt\frac{dy/dt}{dx/dt}, not dt/dydt/dx\frac{dt/dy}{dt/dx}

Exam Tips

Time Management

  • Q.1 (MCQs): Spend 15-20 minutes maximum
  • Short answers: 3-4 minutes per question
  • Long answers: 8-10 minutes per question
  • Keep 10 minutes for final review

Strategy

  1. Read all questions first to identify easy ones
  2. Attempt easy questions first to build confidence
  3. Show all steps clearly for partial marks
  4. Check units in application problems
  5. Verify answers where possible (especially in matrix problems)

During Exam

  • Write clearly and organize solutions
  • Draw diagrams where helpful
  • State formulas before using them
  • Don't panic if stuck on one question - move to next
  • Use remaining time to review and check calculations

Good Luck with your exams!