Engineering Mathematics Test 2

Concept:

The order of differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

The given differential equation is

\(\frac{{{d^4}y}}{{d{x^4}}} = \cos \left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)\)

The highest derivative in the given differential equation is 4. Hence the order is 4.

To observe the degree, the given differential equation must be expressed in a form free from radicals as the derivatives are concerned.

The above equation is expressed in terms of cos function. Hence degree is not defined. 

1. \({\vec F_1} = \left( {x + y} \right)\hat i + \left( {2y - 3z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_1} = 1 + 2 - 4 = - 1 \ne 0\) 

Therefore, \({\vec F_1}\) is not solenoidal.

2. \({\vec F_2} = \left( {x - y} \right)\hat i + \left( {2y - z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_2} = 1 + 2 - 4 = - 1 \ne 0\) 

Therefore, \({\vec F_2}\) is not solenoidal.

3. \({\vec F_3} = \left( {2x - y} \right)\hat i + \left( {y - 3z} \right)\hat j + \left( {3x + 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_3} = 2 + 1 + 4 = 8 \ne 0\) 

Therefore, \({\vec F_3}\) is not solenoidal.

4. \({\vec F_4} = \left( {2x + y} \right)\hat i + \left( {2y - 3z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_4} = 2 + 2 - 4 = 0\) 

Therefore, \({\vec F_4}\) is solenoidal.

Concept:

\(f = {f_x}\hat i + {f_y}\hat i + {f_2}\hat k\)

The divergence of the above vector field is defined as

\(\nabla \cdot f = \frac{d}{{dx}}{f_x} + \frac{{d{f_1}}}{{dy}} + \frac{{d{f_2}}}{{dz}}\)

Application:

\(A = 2{x^2}y\;\hat i + 3{x^2}{y^2}\hat j + 0\;\hat k\)

\(\nabla \cdot A = \frac{d}{{dx}}\left( {2{x^2}y} \right) + \frac{d}{{dy}}\left( {3{x^2}{y^2}} \right)\)

\(= 4xy + 6{x^2}y\)

Given (x, y, z) ∈ [1, 3], i.e. the cube coordinates lies in the range:

1 ≤ x ≤ 3

1 ≤ y ≤ 3

1 ≤ z ≤ 3

∴ Midpoint (M) of the cube will be:

\(M = \frac{{{x_2} + {x_1}}}{2},\frac{{{y_2} + {y_1}}}{2},\frac{{{z_2} + {z_1}}}{2}\)

Putting on the respective values, we get:

M = (2, 2, 2)

∴ The divergence of the vector A̅ at the given mid point (M) will be:

A_{(2, 2, 2)} = 4(2) (2) + 6(2²) (2)

= 16 + 48 = 64

Given f(x, y) = y^x

First, partially differentiate the function with respect to y, we get:

\(\frac{{\partial f}}{{\partial y}} = x{y^{x - 1}}\) 

Again differentiating it with respect to x, we get:

\(\frac{{{\partial ^2}f}}{{\partial x\partial y}} = {y^{x - 1}}\left( 1 \right) + x\left( {{y^{x - 1}}\log y} \right)\) 

\( = {y^{x - 1}}(x\log y + 1)\) 

At x = 2, y = 1

\(\frac{{{\partial ^2}f}}{{\partial x\partial y}} = {\left( 1 \right)^{2 - 1}}\left( {2\log 1 + 1} \right)\) 

= 1(2 × 0 + 1) = 1

For eigenvalues of the matrix:

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

From the above, we can write:

(-1 - λ) [(1 - λ)² - 4] = 0

The above can be written as:

(-1 - λ) [(1 - λ) + 2][(1 - λ) - 2] = 0

(λ + 1)( -1 - λ)(3 - λ) = 0

λ = -1, -1, 3

∴ All the Eigen values are in the range:

-1 ≤ λ ≤ 3

-2 ≤ λ -1 ≤ 2

|λ - 1| ≤ 2

Concept:

\({\rm{Standard\;deviation\;}}\left( {\rm{\sigma }} \right) = \sqrt {E\left( {{X^2}} \right) - {{\left( {E\left( X \right)} \right)}^2}} \)

\(E\left( X \right) = \mathop \sum \nolimits_{i = 1}^3 {X_i}.P\left( {{X_i}} \right)\)

Calculation:

E(X) = 1(0.3) + 2(0.5) + 3(0.2)

∴ E(X) = 1.9

Now,

\(E\left( {{X^2}} \right) = \mathop \sum \nolimits_{i = 1}^3 X_i^2.P\left( {{X_i}} \right)\)

E(X²) = 1² (0.3) + 2² (0.5) + 3² (0.2)

∴ E(X²) = 4.1

\(\therefore \sigma = \sqrt {4.1 - {{\left( {1.9} \right)}^2}} \)

∴ σ = 0.7 

\({\lambda ^3} + \;a{\lambda ^2} + 47\lambda \; - \;60 = 0\)

Since one eigenvalue of matrix is 4 it satisfies the characteristic equation

\({4^3} + \;a{4^2} + 47\left( 4 \right)\; - \;60 = 0\)

\(64 + \;16a + 188\; - \;60 = 0\)

\(a = \; - 12\)

\({\lambda ^3} - 12\;{\lambda ^2} + 47\lambda \; - \;60 = 0\)

Solving this will give

\(\left( {\lambda - \;3} \right)\left( {\lambda \; - \;4} \right)\left( {\lambda \; - \;5} \right) = 0\)

\(\lambda \; = \;3\;or\;\lambda \; = \;4\;or\;\lambda \; = \;5\)

y = f(x) = 5x² + 10x in the internal x = (1, 2)

Since, the function y is continuous in the interval (1, 2), as well as is differentiable at each point so, from Lagrange mean value theorem there exist at least a point where:

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\) 

Here, we have

a = 1, b = 2

So, for x = a = 1, we obtain:

y = f(a) = f(1) = 5(1)² + 10(1) = 15

and for x = b = 2

y = f(b) = f(2) = 5(2)² + 10(2) = 40

Therefore:

\(f'\left( c \right) = \frac{{40 - 15}}{{2 - 1}} = 25\) 

Concept:

Revolution about x-axis: The volume of the solid generated by the revolution about the x-axis, of the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is

\(V = \mathop \smallint \nolimits_a^b π {y^2}dx\)

Calculation:

\(V = \mathop \smallint \nolimits_a^b π {y^2}dx \)

\(= \mathop \smallint \nolimits_0^3 π {\left( 2 \right)^2}dx = \mathop \smallint \nolimits_0^3 4π dx \)

\(= \left[ {4π x} \right]_0^3 \)

= 12 π cubic meters

Let X be the number of successful hits.

Suppose n missiles are fired.

P(X ≥ 3) ≥ 0.95

From binominal distribution,

1 – P(X < 3) ≥ 0.95

P(X < 3) ≤ 0.05

⇒ P(X = 0) + P(X = 1) + P(X = 2) ≤ 0.05

\( \Rightarrow {n_{{c_0}}}{p^0}{q^n} + {n_{{c_1}}}p\;{q^{n - 1}} + {n_{{c_2}}}{p^2}{q^{n - 2}} \le 0.05\) 

\( \Rightarrow {\left( {\frac{1}{4}} \right)^n} + n \cdot \left( {\frac{3}{4}} \right){\left( {\frac{1}{4}} \right)^{n - 1}} + \left( {\frac{n}{2}} \right){\left( {\frac{3}{4}} \right)^2}{\left( {\frac{1}{4}} \right)^{n - 2}} \le 0.05\) 

⇒ 10 (9n² – 3n + 2) ≤ 4ⁿ.

We have:

\(\frac{{{d^2}u}}{{d{x^2}}} - k\frac{{du}}{{dx}} = 0\) 

(D² – kD) u = 0

The Auxillary Equation will be:

m² – km = 0

m (m – k) = 0

m = 0, k

Thus, the complete solution will be:

\(u = {C_1}{e^{0x}} + {C_2}{e^{kx}}\)

\(u = {C_1} + {C_2}{e^{kx}}\)

From the given conditions, we get:

u(0) = 0, i.e.

0 = C₁ + C₂

C₁ + C₂ = 0   ---(1)

And

u(L) = U, i.e.

\(U = {C_1} + {C_2}\;{e^{kL}}\)   ---(2)

Subtracting equation (1) from (2), we get:

\(U={C_2}\left( {{e^{kL}} - 1} \right)\)

\({C_2} = \frac{U}{{\left( {{e^{kL}} - 1} \right)}}\)

From equation (1), we have:

\({C_1} = - {C_2} = \frac{{ - U}}{{\left( {{e^{kL}} - 1} \right)}}\)

Substitute these values in the expression for u, we get:

\(u = \frac{{ - U}}{{\left( {{e^{kL}} - 1} \right)}} + \frac{U}{{\left( {{e^{kL}} - 1} \right)}}{e^{kx}}\)

\( u= U\left( {\frac{{1 - {e^{kx}}}}{{1 - {e^{kL}}}}} \right)\)

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\\ \ldots & \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 1&a&3\\ 1&{11}&a \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 1\\ 3\\ b \end{array}} \right]\)

The Augmented matrix can be written by

\([A|B] = \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 1&a&3\\ 1&{11}&a \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 3\\ b \end{array}} \right]\)

R₃ → R₃ – R₁

R₂ → R₂ – R₁

\(= \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 0&{a - 2}&1\\ 0&9&{a - 2} \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 2\\ {b - 1} \end{array}} \right]\)

\({R_3} \to {R_3} - \left( {\frac{9}{{a - 2}}} \right){R_2}\)

\(= \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 0&{a - 2}&1\\ 0&0&{\frac{{{{\left( {a - 2} \right)}^2} - 9}}{{a - 2}}} \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 2\\ {\frac{{\left( {a - 2} \right)\left( {b - 1} \right) - 18}}{{a - 2}}} \end{array}} \right]\)

The system ho have a unique solution, \(\frac{{{{\left( {a - 2} \right)}^2} - 9}}{{a - 2}} \ne 0\)

⇒ (a - 2)² – 9 ≠ 0

⇒ a – 2 ≠ ±3

⇒ a ≠ -1 or 5

The unit vector normal to the surface will be given by:

\(n = \frac{{\nabla \phi }}{{\left| {\nabla \phi } \right|}}\)

ϕ = x² + y² + (z - 2)² = 9

∇ϕ = 2xi + 2yj + 2(z - 2) k

\(n = \frac{{xi + 2yj + \left( {z - 2} \right)k}}{{\sqrt {{x^2} + {y^2} + {{\left( {z - 2} \right)}^2}} }}\)

\(F \cdot n = \left[ {xi + yj + \left( {z - 2} \right)k} \right]\frac{{\left[ {xi + yj + \left( {z - 2} \right)} \right]k}}{{\sqrt {{x^2} + {y^2} + {{\left( {z - 2} \right)}^2}} }}\)

\(F \cdot n = \frac{{{x^2} + {y^2} + {{\left( {z - 2} \right)}^2}}}{{\sqrt {{x^2} + {y^2} + {{\left( {z - 2} \right)}^2}} }}\)

\(= \frac{9}{{\sqrt 9 }} = \frac{9}{3} = 3\)

Thus:

\(\int\!\!\!\int \left( {F \cdot n} \right)dxdy=\int\!\!\!\int 3dxdy \)

\(= 3\int\!\!\!\int dxdy\)

3 × Area = 3 × π (3)²

= 27 π  

We have \(\frac{{dy}}{{dx}} = 2\left( {1 + y} \right)\sqrt y \) 

\(\frac{{dy}}{{2\left( {1 + y} \right)\sqrt y }} = dx\) 

\(\frac{{dt}}{{1 + {t^2}}} = dx\) 

Putting √y = t, the above expression becomes:

tan^-1 t = x + C₁

\({\tan ^{ - 1}}\sqrt y = x + {C_1}\) 

From y(0) = 0 we get:

tan^-1 (0) = 0 + C₁

C₁ = mπ

From y(x/2) = 1 we get:

\({\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{2} + C\) 

\(n\pi + \frac{\pi }{4} = \frac{\pi }{2} + {C_1}\) 

\({C_1} = n\pi - \frac{\pi }{4}\) 

Clearly, the solution exists on [0, 2π] 

Concept:

Laurentz Series is obtained by the arrangement and manipulation of standard series or expansions, i.e.

(1 - x)^-1 = 1 + x + x² + x³ + …… |x| < 1

(1 + x)^-1 = 1 – x + x² – x³ + ….. |x| < 1

(1 - x)^-2 = 1 + 2x + 3x² + ….. |x| < 1

(1 + x)^-2 1 – 2x + 3x² – 4x² + …. |x| < 1

Observe that in all the expansions; |x| should be less than 1.

 ∴ We need to manipulate the variable to satisfy the above condition.

Application:

Given region |z| < 2

⇒ \(\frac{2}{\left| z \right|}<1\) 

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

This can be interpreted as \(\frac{1}{\left| z \right|}<1\) 

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

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

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

\(\left| \frac{1}{z} \right|<1\Rightarrow {{\left( 1-\frac{1}{z} \right)}^{-1}}=1+\frac{1}{z}+\frac{1}{{{z}^{2}}}+\ldots \) 

Similarly, we can write:

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

\(\left| \frac{2}{z} \right|<1\Rightarrow {{\left( 1-\frac{2}{z} \right)}^{-1}}=1+\frac{2}{z}+{{\left( \frac{2}{z} \right)}^{2}}+\ldots \) 

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

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

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Properties of Eigenvalues:

The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A

The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

Calculation:

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

The given matrix is triangular.

Therefore, Eigenvalues = diagonal elements

= 1, 1

As the Eigenvalues are not unique, it is not diagonalizable.

Determinant of A = 1 ≠ 0

Therefore A is non-singular

To get Eigenvector,

Ax = λx

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = 1\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {{x_1} - {x_2}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

⇒ x₂ = 0 and x₁ can be any value.

The Eigenvector is \(\left[ {\begin{array}{*{20}{c}} k\\ 0 \end{array}} \right]\) 

Therefore, the given vector is Eigenvector.

Concept:

Directional derivative of a function f along the vector \(\hat u \) is given by:

\(DD = \nabla f.\frac{{\vec u}}{{\left| u \right|}}\)

where \(\frac{{\vec u}}{{\left| u \right|}}={\hat{u}}\)

grad f or ∇ f is defined by the equation,

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

Calculation:

Given:

z = y²e^2x

\(gradz\;= \nabla z \;=\; 2{y^2}{e^{2x}}\hat i + 2y{e^{2x}}\hat j\)

At (2, -1), \(\nabla z = 2{e^4}\hat i - 2{e^4}\hat j\)

Unit vector \(\vec b = \alpha \hat i + \beta \hat j\)

Directional derivative \( = \nabla z.\vec b = \left( {2{e^4}\hat i - 2{e^4}\hat j} \right).\alpha \hat i + \beta \hat j\) = 2e4α – 2e4β

Given that, the directional derivative is zero.

⇒ 2e⁴α – 2e⁴β = 0 ⇒ α = β

As b is a unit vector, \(\sqrt {{\alpha ^2} + {\beta ^2}} = 1\)

\( \Rightarrow \alpha = \beta = \frac{1}{{\sqrt 2 }}\)

\(\left| {\alpha + \beta } \right| = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2 \)

Concept:

f(z) = ϕ + iψ

ϕ = Real part, ψ = Imaginary part

If f(z) is analytic function

\(\left. {\begin{array}{*{20}{c}} {\frac{{\partial \phi }}{{\partial x}} = \frac{{\partial \psi }}{{\partial y}}}\\ {\frac{{\partial \psi }}{{\partial x}} = - \frac{{\partial \phi }}{{\partial y}}} \end{array}} \right\} \to C.R\;equation\)

\(\therefore d\psi = \frac{{\partial \psi }}{{\partial x}}dx + \frac{{\partial \psi }}{{\partial y}}dy\)

\(d\psi = \frac{{ - \partial \phi }}{{\partial y}}dx + \frac{{\partial \phi }}{{\partial x}} \ dy\)

Calculation:

Given, f(z) = x² – y² + i ψ (x, y)

\(\phi = {x^2} - {y^2}\)

∵ f(z) is analytic function

\(d\psi = \frac{{ - \partial \phi }}{{\partial y}}dx + \frac{{\partial \phi }}{{\partial x}} \cdot dy\)

\(\frac{{\partial \phi }}{{\partial y}} = \; - 2y,\;\frac{{\partial \phi }}{{\partial x}} = 2x\)

\(d\psi = 2y\;dx + 2x\;dy\)

\(\smallint d\psi = 2\smallint d\left( {xy} \right)\)

ψ = 2 xy

Given, z = 1 + i

Comparing it with z = x + iy 

∴ x = 1, y = 1

\(\therefore {\left( \psi \right)_{\left( {1,\;1} \right)}} = 2 \times 1 \times 1=2\)

Concept:

Binomial distribution:

Let p is the probability that an event will happen in a single trail (called the probability of success) and

q = 1 – p is the probability that an event will fail to happen (probability of failure)

The probability that the event will happen exactly r times in n trails (i.e. x successes and n – r failures will occur) is given by the probability function

\(f\left( x \right) = P\left( {X = r} \right) = {n_{{C_r}}}{p^r}{q^{n - r}}\)

where the random variable X denotes the number of successes in n trials and r = 0, 1, 2, … n

For Binomial distribution,

Mean = μ = np

Variance = σ2 = npq

Standard deviation = σ = √(npq)

Calculation:

Let V be the event that occurs in a trial with probability p. Mathematical expectation E of the number of trials to first occurrence of V in a sequence of trials is E = 1/p.

Similarly, the expected number of trials to get n^th success = n/p

Given that, the probability (p) = 1/5 = 0.2

The expected number of trials to get first success = 1/0.2 = 5

So, the expected number of failures preceding the first success is 4

The expected number of trials to get second success = 2/0.2 = 10

Expected number of successes in first 50 trials = np = 50 × 0.2 = 10

Let x² + y² = t

Differentiating with respect to x, we get:

\(2x + 2y\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}\) 

\(\frac{y}{x}\frac{{dy}}{{dx}} = \frac{1}{{2x}}\frac{{dt}}{{dx}} - 1\) 

Hence, the given equation becomes:

\(\frac{1}{{2x}}\frac{{dt}}{{dx}} - 1 + \frac{{2t - 1}}{{t + 1}} = 0\) 

\(\frac{1}{{2x}}\frac{{dt}}{{dx}} = 1 - \frac{{2t - 1}}{{t + 1}}\) 

\(= \frac{{2 - t}}{{t + 1}}\) 

\(2xdx = \frac{{t + 1}}{{2 - t}}dt\) 

\(2xdx + \left( {1 + \frac{3}{{t - 2}}} \right)dt = 0\) 

Integrating, we have

x² + t + 3 log (t – 2) = C

2x² + y² + 3 log (x² + y² – 2) = C

Which is the required solution.

The given integration can be written as

\(I = \mathop \smallint \limits_{x = 0}^{x = 2} \mathop \smallint \limits_{y = 0}^{y = \lambda \left( x \right)} \mathop \smallint \limits_{z = 0}^{z = \mu\left( {x,y} \right)} dz\;dy\;dx\)

where the limit of z varies from

z = 0 to z = 4 – (2x + y)

∵ μ(x,y) is a function of x,y therefore μ(x,y) is a limit of z.

∴ μ(x,y) = 4 – (2x + y)         

After eliminating z value λ(x) is a function of x, hence λ(x) is a limit of y.

For obtaining limit of y, we have to put z = 0 in the equation of plane,

hence

y = 4 – (2x + z)

y = 4 – 2x (∵ z = 0)

Hence, the limit of y will vary from y = 0 to y = 4 – 2x

∴ λ(x) = 4 – 2x

∴ λ(x) – μ(x,y) = 4 – 2x – (4 – (2x+y))

= 4 – 2x – 4 + 2x + y

= y

Hence, λ(x) – μ(x,y) = y

Concept:

A square matrix ‘A’ is said to be orthogonal if it follows the following condition.

AAT = ATA = I

Calculation:

\(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\)

\({M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right]\)

\(M{M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right]\)

For orthogonal matrix, MM^T = I

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

By comparing both the sides, we get

\(\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }} = 0 \Rightarrow a + b = 0\)

a² + b² = 1

\(\Rightarrow b = \pm \sqrt {1 - {a^2}}\)

Therefore, \(b = \sqrt {1 - {a^2}}\) is not always true.

⇒ (a + b)² – 2ab = 1

\(\Rightarrow ab = - \frac{1}{2}\)

By using the relation, a + b = 0 i.e. b = -a, we get

\( \Rightarrow a\left( { - a} \right) = - \frac{1}{2} \Rightarrow a = \pm \frac{1}{{\sqrt 2 }}\)

For \(a = \frac{1}{{\sqrt 2 }},b = - \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)

\({M^2} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {I_2}\)

For \(a = \frac{{ - 1}}{{\sqrt 2 }},b = \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\({M^2} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right] \ne {I_2}\)

Therefore, M² = I₂ is not always true.