Engineering Mathematics Test 1

Concept:

Eigen vector (X) that corresponding to Eigen value (λ) satisfies the equation AX = λX.

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 3&4\\ 2&t \end{array}} \right]\)

\(X = \left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right]\)

Let λ is an Eigen value,

AX = λX

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 3&4\\ 2&t \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 8\\ {8 - t} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4\lambda }\\ { - \lambda } \end{array}} \right]\)

By comparing both the sides,

8 = 4λ ⇒ λ = 2

8 – t = -λ ⇒ 8 – t = -2 ⇒ t = 10 

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 1&4&2\\ 2&6&5 \end{array}} \right]\)

Performing row transformation: R₃ → R₃ – R₁, we get:

\(A \approx \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 1&4&2\\ 1&4&2 \end{array}} \right]\)

Now, performing row transformation: R₃ → R₃ – R₂, we get:

\(A \approx \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 1&4&2\\ 0&0&0 \end{array}} \right]\)

∴ The determinant of the matrix is 0, i.e.

|A| = 0

Also, since:

\(\left| {\begin{array}{*{20}{c}} 1&2\\ 1&4 \end{array}} \right| \ne 0\)

Rank (A) = 2

Concept:

For a given complex function with poles, the complex integral \(I = \mathop \oint \limits_C f\left( z \right)dz\) is given by

Residue theorem as;

\(I = \mathop \oint \limits_C f\left( z \right)dz = 2\pi \;i \times \left\{ {Sum\;of\;residue\;of\;poles\;inside\;or\;on\;C} \right\}\)

Calculation:

\(f\left( z \right) = \oint \frac{{{z^3} - 2z + 3}}{{z - 2}}dz\)

Contour: |z|= 3

Simple pole, z = 2 and it is lies inside the contour.

Residue of f(z) at z = 2 is,

\(\mathop {{\rm{lt}}}\limits_{z \to 2} \left( {z - 2} \right)\frac{{\left( {{z^3} - 2z + 3} \right)}}{{\left( {z - 2} \right)}} = {2^3} - 2\left( 2 \right) + 3 = 7\)

f(z) = 2πi(7) = 14πi

\(P\left( {an\;Ace} \right) = p = \frac{1}{6}\)

The probability of not getting an ace will be:

\(q = 1 - p = \frac{5}{6}n = 4\)

The probability of getting 0 aces will be:

\(P\left( {0\;Ace} \right) = {\left( {\frac{5}{6}} \right)^4}\)

\(P\left( {1\;Ace} \right) = {\;^4}{C_1}\;{\left( {\frac{5}{6}} \right)^3} \cdot \left( {\frac{1}{6}} \right)\)

\(P\left( {2\;Ace} \right) = \;{\;^4}{C_2}{\left( {\frac{5}{6}} \right)^2} \cdot {\left( {\frac{1}{6}} \right)^2}\)

\(P\left( {3\;Ace} \right) = \;{\;^4}{C_3}\left( {\frac{5}{6}} \right) \cdot {\left( {\frac{1}{6}} \right)^3}\)

\(P\left( {4\;Ace} \right) = \;{\;^4}{C_4}{\left( {\frac{5}{6}} \right)^0} \cdot {\left( {\frac{1}{6}} \right)^4}\)

\(P\left( {4\;Ace} \right) = {\left( {\frac{1}{6}} \right)^4}\)

Concept:

M dx + N dy = 0, will be exact differential equation if:

\(\frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}} = \frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}}\)

Where M and N are functions of x and y.

Calculation:

M = α xy³ + y cos x

N = x²y² + β sin x

\(\frac{{\partial M}}{{\partial y}} = 3\;\alpha \;x{y^2} + \cos x\)

\(\frac{{\partial N}}{{\partial x}} = 2x{y^2} + \beta \cos x\)

The differential equation to be exact,

3 α xy² + cos x = 2xy² + β cos x

\( \Rightarrow 3\alpha = 2 \Rightarrow \alpha = \frac{2}{3}\)

Concept:

Gradient / Normal vector of surface / Slope of surface / Rate of maximum increase or decrease:

Let ϕ(x, y, z) = C represents any scalar point function, then its gradient is defined as

\(grad\;\phi = \nabla \phi = \hat i\frac{{\partial \phi }}{{\partial x}} + \hat j\frac{{\partial \phi }}{{\partial y}} + \hat k\frac{{\partial \phi }}{{\partial z}}\)

Calculation:

We have T = x² + y² – 2z²

\(\nabla T = \frac{{\partial T}}{{\partial x}}i + \frac{{\partial T}}{{\partial y}}j + \frac{{\partial T}}{{\partial z}}k\)

= 2xi + 2yj + 4zkj

∴ ∇T is the direction of the maximum rate of change and the mosquito should also fly in the direction of the gradient vector to get away from heat as soon as possible.

At P(2, 2, 1) will be:

∇T = 4i + 4j – 4k

\(\mathop \smallint \limits_{ - \pi }^\pi \left| x \right|\cos nxdx\)

The given function f(x) = |x| cos nx is even function. So, the given integral can be reduced to

\( = 2\mathop \smallint \limits_0^\pi \left| x \right|\cos nxdx\)

\( = 2\mathop \smallint \limits_0^\pi x\cos nxdx\)

\(\smallint x\cos nxdx = x\smallint \cos nxdx - \smallint \frac{{\sin nx}}{n}dx\)

\( = x\left( {\frac{{\sin nx}}{n}} \right) + \frac{1}{{{n^2}}}\left( {\cos nx} \right)\)

\( = 2\left[ {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}} \right]_0^\pi \)

\( = 2\left[ {\frac{{\cos n\pi - 1}}{{{n^2}}}} \right]\)

\( = \frac{{ - 2}}{{{n^2}}}\left( {1 - {{\left( { - 1} \right)}^n}} \right)\)

= 0 when n is even

\( = - \frac{4}{{{n^2}}}\) when n is odd

Concept:

For eigen values of a matrix A,

I A - λI I = 0

Analysis:

Given :

\(U = {\left[ {\begin{array}{*{20}{c}} 1\\ 2 \end{array}} \right]_{2 \times 1}}\;\)

\(V = {\left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]_{2 \times 1}}\)

\({V^T} = {\left[ {\begin{array}{*{20}{c}} 1&1 \end{array}} \right]_{1 \times 2}}\)

\(A = U{V^T}\)

\( = {\left[ {\begin{array}{*{20}{c}} 1\\ 2 \end{array}} \right]_{2 \times 1}}{\left[ {\begin{array}{*{20}{c}} 1&1 \end{array}} \right]_{1 \times 2}}\)

\( = {\left[ {\begin{array}{*{20}{c}} 1&1\\ 2&2 \end{array}} \right]_{2 \times 2}}\)

For eigen values of matrix A

I A - λI I = 0

\(\left| {\begin{array}{*{20}{c}} {1 - λ }&1\\ 2&{2 - λ } \end{array}} \right|\) = 0

(1 - λ) (2 - λ) - 2 = 0

λ² - 3λ + 2 - 2 = 0

λ² - 3 λ = 0

λ (λ - 3) = 0

λ = 0 and 3

So, the largest eigen value is 3.

We have f(x) = 2x² + 3x – m log x

Differntiating it w.r.t. x, we get:

\(f'\left( x \right) = 4x + 3 - \frac{m}{x}\) 

\(= \frac{{4{x^2} + 3x - m}}{x}\) 

The function f(x) is monotonic decreasing if f’(x) < 0, i.e.

\(\frac{{4{x^2} + 3x - m}}{x} < 0\) 

m > 4x² + 3x

For the open interval (0, 1), the maximum value of m will be obtained for x = 1.

F(x, y) = x³ + y³ + 3xy – 1

\( \frac{{dy}}{{dx}} = \frac{{\frac{{ - \partial f}}{{\partial x}}}}{{\frac{{\partial f}}{{\partial y}}}} \)

\(= \frac{{ - 3{x^2} + 3y}}{{3{y^2} + 3x}} = - \frac{{{x^2} + y}}{{{y^2} + x}}\)

\(\\ \frac{{du}}{{dx}} = \frac{{\partial u}}{{\partial x}} + \frac{{\partial u}}{{\partial y}}.\frac{{dy}}{{dx}} \)

\(= \left( {1.\log xy + x.\frac{1}{x}} \right) + \left( {\frac{x}{y}} \right).\frac{{dy}}{{dx}}\)

\(\Rightarrow \frac{{du}}{{dx}} = 1 + \log xy - \frac{x}{y}\frac{{\left( {{x^2} + y} \right)}}{{\left( {{y^2} + x} \right)}}\)

∇ ⋅ (∇ × f) = 0

The divergence of the curl is zero.

∇ × (∇f) = 0

Curl of a gradient is the zero vector.

\(\nabla \times \left( {\nabla \times \vec f} \right) = \nabla \left( {\nabla \cdot \vec f} \right) - {\nabla ^2}\vec f\) 

\(\nabla \times (\log r\;\vec r) = (\nabla \log r) \times \vec r + \log r\left( {\nabla \times \vec r} \right)\;\)

\(= \frac{1}{r}\bar r \times \bar r + 0 = 0\) 

y’’ + 9y = 0

y’’ + 9y = 0

Differential Eq. is of 2^nd order and Homogeneous.

Differential Eq. → D² + 9 = 0

Auxiliary Eq. →

∴ m² + 9 = 0 ⇒ m² = -9 = 9 i²

m = ± 3i

On comparing α ± iβ on root of above eqn.

i.e. Root of above eq. is imaginary α = 0, β = 3

So, Solution above eq.

y = C.F = e^αx [C₁ cos β_x + C₂ sin β_x]

∴ y = C₁ cos 3x + C₂ sin 3x

From given boundary condition: y (0) = 0

At x = 0, y = 0

⇒ C₁ = 0

y(π/2) = √2

\(\sqrt 2 = {C_1}\cos 3\left( {\frac{\pi }{2}} \right) + {C_2}\sin \frac{{3\pi }}{2} = {C_2}\sin \frac{{3\pi }}{2} = - {C_2}\)

⇒ C₂ = -√2

∴ y = -√2 sin 3x

y(π/4) = ?

\(y = - \sqrt 2 \sin \frac{{3\pi }}{4} = - \sqrt 2 \times \frac{1}{{\sqrt 2 }} = - 1\)

Concept:

Laurent series of f(z) is given by:

\(f\left( z \right) = \mathop \sum \limits_{n = - 1}^\infty {a_n}{z^n}\)

\( = a^{- 1}z^{-1} + {a_0} + {a_1}z + {a_2}{z^2} + {a_3}{z^3} + \ldots \)

Calculation:

Given:

\(f\left( z \right) = \frac{1}{{\left( {z - 1} \right)\left( {z - 2} \right)}} = \frac{1}{{z - 2}} - \frac{1}{{z - 1}}\)

The given region is 1 < |z| < 2

\(1 < \left| z \right| \Rightarrow \frac{1}{{\left| z \right|}} < 1\)

\(\left| z \right| < 2 \Rightarrow \frac{{\left| z \right|}}{2} < 1\;\)

\( \Rightarrow f\left( z \right) = \frac{1}{{2\left( {\frac{z}{2} - 1} \right)}} - \frac{1}{{z\left( {1 - \frac{1}{z}} \right)}}\)

\( = \frac{{ - 1}}{{2\left( {1 - \frac{z}{2}} \right)}} - \frac{1}{{z\left( {1 - \frac{1}{z}} \right)}}\)

\( = \frac{{ - 1}}{2}{\left( {1 - \frac{z}{2}} \right)^{ - 1}} - \frac{1}{z}{\left( {1 - \frac{1}{z}} \right)^{ - 1}}\)

\( = \frac{{ - 1}}{2}\left[ {1 + \frac{z}{2} + {{\left( {\frac{z}{2}} \right)}^2} + \ldots } \right] - \frac{1}{z}\left[ {1 + \frac{1}{z} + {{\left( {\frac{1}{z}} \right)}^2} + \ldots } \right]\)

\( = \left[ {\frac{{ - 1}}{2} - \frac{z}{4} - \frac{{{z^2}}}{8} + \ldots } \right] - \left[ {\frac{1}{z} + \frac{1}{{{z^2}}} + \frac{1}{{{z^3}}} + \ldots } \right]\)

Co-efficient of \(\frac{1}{{{z^2}}}\;\)is -1

Given vector  \(\vec v = \frac{1}{3}\left( { - {y^3}\hat i + {x^3}\hat j + {z^3}\hat k} \right)\)

Bounded by open surface of hemisphere x² + y² + z² = 1

Such that z ≥ 0, and closed curve (x² + y² = 1)

We have to find

\(I = \smallint \left( {\nabla \times u} \right).\hat nds\)

⇒ from stokes theorem, we have

\(\mathop \int\!\!\!\int \nolimits_s \left( {\nabla \times u} \right).\hat nds = \mathop \oint \nolimits_c u.dr\)

\( \Rightarrow I = \frac{1}{3}\left\{ { - \oint {y^3}dx + \oint {x^3}dy} \right\}\)

Now, changing the integral with substitution

x = cos θ ⇒ dx = -sin θ dθ

y = sin θ ⇒ dy = cos θ dθ

\(I = \frac{1}{3}\mathop \smallint \nolimits_0^{2\pi } {\sin ^4}\theta d\theta + \mathop \smallint \nolimits_0^{2\pi } \frac{1}{3}{\cos ^4}\theta d\theta \)

\(A = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}\\ { - 1}&2&0\\ 0&0&{ - 2} \end{array}} \right]\)

B = A³ – A² - 4A + 5I

|A - λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&0&{ - 1}\\ { - 1}&{2 - \lambda }&0\\ 0&0&{ - 2 - \lambda } \end{array}} \right| = 0\)

⇒ (1- λ) (2 - λ) (-2 - λ) – 1 (0) = 0

⇒ λ = 1, λ = 2, λ = -2

If λ is an Eigen value pf A, then λⁿ will be an Eigen value pf Aⁿ.

If λ is an Eigen value of A, then A, then k λ will be an Eigen value of kA where k is a scalar.

Eigen values of A are 1, 2, -2

Eigen values of A² are 1, 4, 4

Eigen values of A³ are 1, 8, -8

Eigen values of 4A are 4, 8, -8

Eigen values of 5I are 5, 5, 5

B = A3 – A2 - 4A + 5I

λ_1B = λ1³– λ12– 4λ1 + 5 = 1 - 1 - 4 + 5 = 1

λ2B = λ23– λ22– 4λ2 + 5 = 8 - 4 - 8 + 5 = 1

λ3B = λ33– λ32– 4λ3 + 5 = -8 - 4 + 8 + 5 = 1

Eigen values of B are 1, 1, 1

Since x, y, z are in A.P with common difference ‘d’, we can write:

x = x,

y = x + d,

z = x + 2d

The given matrix can now be written as:

\(\left[ A \right] = \left[ {\begin{array}{*{20}{c}} 4&5&x\\ 5&6&{x + d}\\ 6&k&{x + 2d} \end{array}} \right]\)

Applying R₂ → R₂ – R₁, we get the equivalent matrix as:

\(\left[ A \right] \approx \left[ {\begin{array}{*{20}{c}} 4&5&x\\ 1&1&d\\ 1&{k - 6}&d \end{array}} \right]\)

Applying R₃ → R₃ – R₂, we get:

\(\left[ A \right] \approx \left[ {\begin{array}{*{20}{c}} 4&5&x\\ 1&1&d\\ 0&{k - 7}&0 \end{array}} \right]\)

For the rank of the matrix to be 2, the determinant of the above must be 0, i.e.

|A| = 0

Solving for the determinant, we get:

(k - 7) (4d - x) = 0

k = 7 or x = 4d

∴ For k = 7, and x = 4d

We have y” + λy = 0

The auxillary equation will be:

D² + λ = 0

\(D = \pm \sqrt \lambda \;i\) 

∴ The solution will be:

\({y_1} = {C_1}\cos \sqrt k \;\lambda x + {C_2}\sin \sqrt \lambda \;x\) 

\(y' = - {C_1}\sqrt \lambda \sin \sqrt \lambda \;x + {C_2}\sqrt \lambda \cos \sqrt \lambda \;x\) 

From y’(0) = 0, we get:

\(0 = {C_2}\sqrt \lambda \) 

C₂ = 0

For y’(π) = 0, we get:

\(0 = - {C_1}\sqrt \lambda \sin \sqrt \lambda \;\pi\) 

\(\sin \sqrt \lambda \;\pi = \sin n\pi \) 

\(\sqrt \lambda = n\) 

λ = n²

Concept:

Taylor Series expansion for f(x) about x = a is given as:

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right) + {\rm{f'}}\left( {\rm{a}} \right)\left( {{\rm{x}} - {\rm{a}}} \right) + \frac{{{\rm{f''}}\left( {\rm{a}} \right)}}{{2!}}{\left( {{\rm{x}} - {\rm{a}}} \right)^2} + \frac{{{\rm{f'''}}\left( {\rm{a}} \right)}}{{3!}}{\left( {{\rm{x}} - {\rm{a}}} \right)^3} + \ldots \)

Calculation:

Let f(x) = e^x

\({\rm{f}}\left( {\rm{x}} \right) = {{\rm{e}}^{\rm{\pi }}} + {{\rm{e}}^{\rm{\pi }}}\left( {{\rm{x}} - {\rm{\pi }}} \right) + \frac{{{{\rm{e}}^{\rm{\pi }}}}}{{2!}}{\left( {{\rm{x}} - {\rm{\pi }}} \right)^2} + \frac{{{{\rm{e}}^{\rm{\pi }}}}}{{3!}}{\left( {{\rm{x}} - {\rm{\pi }}} \right)^3} + \ldots \)

Coefficient of (x - π)² = 0.5 e^π

Let f(x) = sin x

\({\rm{f}}\left( {\rm{x}} \right) = \sin {\rm{\pi }} + \cos {\rm{\pi }}\left( {{\rm{x}} - {\rm{\pi }}} \right) - \frac{{\sin {\rm{\pi }}}}{{2!}}{\left( {{\rm{x}} - {\rm{\pi }}} \right)^2} + \frac{{\cos {\rm{\pi }}}}{{3!}}{\left( {{\rm{x}} - {\rm{\pi }}} \right)^3} + \ldots {\rm{\;}}\)

\( = 0 - \left( {{\rm{x}} - {\rm{\pi }}} \right) - 0 - \frac{1}{{3!}}{\left( {{\rm{x}} - {\rm{\pi }}} \right)^3} + \ldots \)

Coefficient of (x - π)² = 0

The given system has 4 equations and 3 unknowns.

Hence it is a over-determined system of equations.

The equations are;

x₁ + 3x₂ + 2x₃ = 1       --(1)

2x₁ + 2x₂ - 3x₃ = 1     ---(2)

4x₁ + 4x₂ - 6x₃ = 2     ---(3)

2x₁ + 5x₂ + 2x₃ = 1     ---(4)

Note that equation (2) and (3) are linearly dependent on each other and equation 3 is twice that of equation (2).

Hence considering equation 1, 2 and 4 -

\(\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 2 & -3 \\ 2 & 5 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right]\)

R₂ → R₂ - 2R₁ and R₃ → R₃ - 2R₁

\(\left[ \begin{matrix} 1 & 3 & 2 \\ 0 & -4 & -7 \\ 0 & -1 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} \right]\)

\({{R}_{1}}\to {{R}_{3}}-\frac{{{R}_{2}}}{4}\)

\(\left[ \begin{matrix} 1 & 3 & 2 \\ 0 & -4 & -7 \\ 0 & 0 & -\frac{1}{4} \\ \end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\ -1 \\ -\frac{3}{4} \\ \end{matrix} \right]\)

\(\Rightarrow 0.{{x}_{1}}+0.{{x}_{2}}+\left( -\frac{1}{4} \right)\times {{x}_{3}}=\frac{-3}{4}\)

⇒ x₃ = 3

Important point:

(1) Under-determined system: Number of equations < Number of unknowns

(2) Over-Determined system: Number of equation > Number of unknowns

(3) Equally Determined system: Number of equation = Number of unknowns

Concept:

\(\int f\left( z \right)dz=2\pi \left[ sum~of~residue \right]\) 

Calculation:

Residue

\(\begin{matrix} lt \\ z\to 0 \\\end{matrix}~z\left( \frac{{{e}^{kz}}}{z} \right)=1\) 

\(\mathop{\oint }_{\left| z \right|=1}^{{}}{{e}^{\frac{kz}{z}}}dz=2\pi i\left[ 1 \right]=2\pi i\) 

Substitute

z = e^iθ

z = cos θ + i sin θ

dz = i e^iθ

\(2\pi i=\underset{z=1}{\overset{{}}{\mathop \oint }}\,\frac{{{e}^{kz}}}{z}dz=\underset{0}{\overset{2\pi }{\mathop \int }}\,\frac{{{e}^{k\left( \cos \theta +i\sin \theta \right)}}}{{{e}^{i\theta }}}i{{e}^{i\theta }}d\theta\) 

\(=i\underset{0}{\overset{2\pi }{\mathop \int }}\,{{e}^{k\cos \theta }}\left[ \cos \left( k\sin \theta \right)+i\left( k\sin \theta \right) \right]d\theta \) 

Equating the real and imaginary part

\(0=\underset{0}{\overset{2\pi }{\mathop \int }}\,{{e}^{k\cos \theta }}\sin \left( k\sin \theta \right)d\theta \)