Engineering Mathematics Test 2

Concept:

If x is Eigen vector matrix corresponding to the matrix A & λ is the Eigen value then AX = λX

Calculation:

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

\(\lambda X = \lambda \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 2 \end{array}} \right]\)

\(AX = \lambda X \Rightarrow 2\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 2 \end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 2 \end{array}} \right] \Rightarrow \lambda = 2\)

According to question, λ = x

Hence [λ = 2 or λ = x = 2]

Concept: 

A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

Calculation:

For the function to be continuous,

\(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = f\left( 3 \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)\)

\(\mathop {\lim }\limits_{x \to {3^ - }} \left( {-4 + x^2} \right) = \mathop {\lim }\limits_{x \to 3} \left( {-4 +x^2} \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {-2x +a} \right)\)

⇒ - 4 + 3² = - 2 × 3 + a ⇒ 5 = - 6 + a

⇒ a = 11

Concept:

\(\begin{array}{l} \int\limits_0^3 {f(x)dx} = \frac{h}{3}\left[ {\left( {\mathop y\nolimits_0 + \mathop y\nolimits_n } \right) + 4(\mathop y\nolimits_1 + \mathop y\nolimits_3 + ......) + 2(\mathop y\nolimits_2 + \mathop y\nolimits_4 + ......)} \right]\\ \end{array}\)

Calculation:

\(\begin{array}{l} \int\limits_0^3 {f(x)dx} = \frac{h}{3}\left[ {\left( {\mathop y\nolimits_0 + \mathop y\nolimits_3 } \right) + 4(\mathop y\nolimits_1 ) + 2(\mathop y\nolimits_2 )} \right] = \frac{1}{3}\left[ {\left( {44 + 2} \right) + 4(5) + 2(16)} \right] = 32.67\\ \end{array}\)

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 2\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

If f(x)  is odd function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

Calculation:

Let I = \(\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \rm \dfrac{sin^3 \ x \ e^{-x^2}}{x^4}dx\)

Let f(x) = \( \dfrac{sin^3 \ x \ e^{-x^2}}{x^4}\)

Replaced x by -x, 

⇒ f(-x) = \(\rm \dfrac{sin^3 \ (-x) \ e^{-{(-x)^2}}}{(-x)^4}\)

As we know sin (-θ) = - sin θ

= \( \dfrac{-sin^3 \ x \ e^{-x^2}}{x^4}\)

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

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 accruing in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

\({\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^{\frac{1}{2}}} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^{\frac{1}{3}}}\)

Highest derivative in the given differential equation is 4

Hence order is 4.

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\({\left[ {{{\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)}^{\frac{1}{2}}}} \right]^6} = {\left[ {{{\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]}^{\frac{1}{3}}}} \right]^6}\)

\(\Rightarrow {\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^3} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^2}\)

Now the differential equation is free from radicals

Degree of highest derivative = 3

Order and degree of the given differential equation = 4 and 3

Concept:

If f(z) is analytic within & on a closed curve C and “a” is a point inside the curve C then,

\(2π if\left( a \right) = \mathop \oint \nolimits_C \frac{{f\left( z \right)dz}}{{(z\;-\;a)}}\); we consider f(z)/(z-a) to be analytic at all points within C except at z = a.

If multiple singularities occur then we use this;

\(2π i{\{f'\}^{n}}\left( a \right) = \mathop \oint \nolimits_C \frac{{f\left( z \right)\;dz}}{{{{\left( {z\;-\;a} \right)}^{n+1}}}}\)        ……(1)

Calculation:

Given:

\(\mathop \smallint \limits_c \frac{{{z^2}\;-\;z\;+\;1}}{{(2z\;-\;1)}{(z\;-\;2)}}dz\) and |z| = 1

The point of singularities can be calculated by putting the denominator equal to zero.

(2z - 1)(z - 2) = 0

∴ z = 1/2 and 2

|z| = 1 represents circle with unit radius, ∴ z = 1/2 lies inside and z = 2 lies outside the circle. 

Rearranging the integrand;

\(\mathop \smallint \limits_c \frac{{{z^2}\;-\;z\;+\;1}}{{(2z\;-\;1)}{(z\;-\;2)}}dz=\int_{c}^{}\frac{\frac{z^2\;-\;z\;+\;1}{2(z\;-\;2)}}{{(z\;-\;\frac{1}{2})}}dz\)

\(\therefore f(z)=\frac{z^2\;-\;z\;+\;1}{2(z\;-\;2)}\)

Integral ⇒ 2πi f(1/2) = - 0.5πi

Concept:

Laplace transform of f(t):

L {f(t)} = F (s)

Then, Multiplication by tⁿ:           

L {tⁿ f(t)} = (-1)^{n\(\frac{{{d^n}}}{{d{s^n}}}\)} [ F(s)]

Now,

L (cos at) = \(\frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}\)

L {t cos at} = (-1) \(\frac{{\rm{d}}}{{{\rm{ds}}}}\left( {\frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}} \right)\)

= \(\frac{{{\rm{s\;}}\left( {2{\rm{s}}} \right) - \left( 1 \right)\left( {{{\rm{s}}^2} + {{\rm{a}}^2}} \right)}}{{{{\left( {{{\rm{s}}^2} + {{\rm{a}}^2}} \right)}^2}}}\)

f(x) = 3x3 – 40.5x2 + 180x + 7

f’(x) = 9x2 – 81x + 180 = 0

put f’(x) = 0 to find the point at which maximum and minimum value exists

9x2 – 81x + 180 = 0

x2 – 9x + 20 = 0

(x – 5)(x – 4 ) = 0

∴ x = 5 or x =4

Interval [4, 5]

f'’(x) = 2x – 9

put x = 4

f’’(x) = -1 < 0 (maximum value might exist)

put x = 5

f’’(x) = 1 > 0 (minimum value might exist)

Also check border values:

The minimum value is 269.5

Concept:

\(S.D = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \)

Calculation:

\(E\left( X \right) = 2\left( {\frac{1}{3}} \right) + \left( { - 1} \right)\left( {\frac{2}{3}} \right)\)

∴ E(X) = 0

Now,

\(E\left( {{X^2}} \right) = \sum x_i^2P\left( {{x_i}} \right)\)

\(E\left( {{X^2}} \right) = {2^2}\left( {\frac{1}{3}} \right) + {\left( { - 1} \right)^2}\left( {\frac{2}{3}} \right)\)

∴ E(X²) = 2

We know that,

\(S.D\;\left( \sigma \right) = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \) 

\(\therefore S.D\;\left( \sigma \right) = \sqrt {{2^2} - 2} \)

∴ σ = √2 = 1.414

Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.

\({\rm{P}}\left( {\frac{{{{\rm{E}}_1}}}{{{{\rm{E}}_2}}}} \right) = {\rm{Probability\;of\;getting\;the\;event\;}}{{\rm{E}}_1}{\rm{\;when\;}}{{\rm{E}}_2}{\rm{is\;already\;occured}}.\)

\(P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}\)

Calculation:

Given:

P (E₁) = Probability of stopping at the gas station and ask for tyre checked = 0.12

P (E₂) = Probability of stopping at the gas station and ask for oil checked = 0.29

P (E₁∩ E₂) = Probability of both checked = 0.07

\({\rm{P}}\left( {\frac{{{{\rm{E}}_2}}}{{{{\rm{E}}_1}}}} \right) = {\rm{Probability\;of\;person\;who\;has\;his\;tyre\;checked\;will\;also\;have\;oil\;checked}}\)   

∵ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}}\)

∴ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{0.07}}{{0.12}} = 0.58\)

y = \(A{e^{3x}} + B{e^{2x}}\)

, \(\frac{{dy}}{{dx}} = 3A{e^{3x}} + 2B{e^{2x}}\)

, \(\frac{{{d^2}y}}{{d{x^2}}} = 9A{e^{3x}} + 4B{e^{2x}}\)

Now, according to option (a)

\(\frac{{{d^2}y}}{{d{x^2}}} + 5\frac{{dy}}{{dx}} - 6y = \;9A{e^{3x}} + 4B{e^{2x}} + 15A{e^{3x}} + 10B{e^{2x}} - {e^{3x}} + 6B{e^{2x}} \ne 0\)

Hence, option (a) is not correct.

Now, according to option (b)

\(\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} + 6y = \;9A{e^{3x}} + 4B{e^{2x}} - 15A{e^{3x}} - 10B{e^{2x}} + {e^{3x}} + 6B{e^{2x}} = 0\)

So, option (b) is the correct option.

Concept:

According to Newton-Raphson Method

\({X_{n + 1}} = {X_n} - \frac{{f\left( {{X_n}} \right)}}{{f'\left( {{X_n}} \right)}}\)

Calculation:

f(x) = e^x – 1

f’(x) = e^x

x₀ = 1

Iteration 1:

At x₀ = 1, f(x) = 1.718

f'(x) = 2.718

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

\( = 1 - \frac{{1.718}}{{2.718}} = 0.3678\)

Iteration 1:

At x₁ = 0.3678, f(x) = 0.4445

f'(x) = 1.4445

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

\( = 0.3678 - \frac{{0.4447}}{{1.4447}} = 0.06005\)

By inspection, the actual solution for the given equation is, x = 0

Therefore, the error = 0.06005

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

Rolle’s Theorem:

Suppose f(x) is continuous on [a, b] differentiable on (a, b) and f(a) = f(b). Then there exists some point c ϵ [a, b] such that f'(c) = 0.

Calculation:

The given function is continuous and differentiable at x = 0

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ - }} \end{array}f\left( x \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ + }} \end{array}f\left( x \right)\)

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}f\left( {1 + x} \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}\left( {1 - x} \right)\left( {px + q} \right)\)

⇒ 1 = q

The function to be differentiable at x = 0

f'(x = 0) = f'(x = 0⁺)

⇒ 1 = (-1) (px + q) + (1 – x) p

At, x = 0

⇒ 1 = -q + p ⇒ p = 1 + q = 2

Concept:

Taylor series Method: Consider the first-order differential equation as:

\({y}'=\frac{dy}{dx}=f\left( x,y \right)~and~y\left( {{x}_{0}} \right)={{y}_{0}}\)

Taylor’s series state that:

\(y\left( x \right)=y\left( {{x}_{0}} \right)+\frac{\left( x-{{x}_{0}} \right)}{1!}{y}'\left( {{x}_{0}} \right)+\frac{{{\left( x-{{x}_{0}} \right)}^{2}}}{2!}{y}''\left( {{x}_{0}} \right)+\ldots \)

Where, \({y}'\left( {{x}_{0}} \right)={{\left( \frac{dy}{dx} \right)}_{at~{{x}_{0}}}}\)

\(and~{y}''\left( {{x}_{0}} \right)={{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{at~{{x}_{0}}}}\)

Application:

Given,

\({y}'=\frac{dy}{dx}={{x}^{2}}y-1~and~y\left( 0 \right)=1\)

As we know, y(x₀) = y₀

∴ x₀ = 0 and y₀ = 1

Now,

y’ = x² y - 1

\(y_{0}^{'}={{\left( {{x}_{0}} \right)}^{2}}y+0-1\)

\(\Rightarrow y_{0}^{'}=-1\)

Now, y’’ = x² y’ + y.2x

\(i.e.~y_{0}^{''}=0\)

y’’’ = x²y’’ + y’ 2x + 2y + 2xy

\(y_{0}^{'''}~=~2\)

Putting these values and x = 0.1 in equation 1, we get:

\(y\left( x \right)=1+\frac{\left( x-0 \right)}{1!}\left( -1 \right)+\frac{{{\left( x-0 \right)}^{2}}}{2!}\times 0+\frac{{{\left( x-0 \right)}^{3}}}{3!}\times 2+\ldots ~\)

\(y\left( x \right)=1-x+0+\frac{{{x}^{3}}}{3}\)

∴ Neglecting the higher-order teams, we get:

\({\rm{I\;}} = {\rm{\;}}\smallint \frac{1}{{\left( {x + 1} \right)\sqrt {1 - 2x - {x^2}} }}dx\)

Put \(x + 1\; = \;\frac{1}{t}\; \Rightarrow \;dx\; = \;\frac{{ - 1}}{{{t^2}}}dt\)

\(\begin{array}{l} \Rightarrow \;I\; = \;\smallint \frac{{\frac{{ - 1}}{{{t^2}}}dt}}{{\frac{1}{t}\sqrt {1 - 2\left( {\frac{1}{t} - 1} \right) - {{\left( {\frac{1}{t} - 1} \right)}^2}} }}\\ = \; - \smallint \frac{{dt}}{{\sqrt {2{t^2} - 1} }}\end{array}\)

\(\begin{array}{l}\; = \; - \frac{1}{{\sqrt 2 }}\smallint \frac{{dt}}{{\sqrt {{t^2} - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} }}\\ = \; - \frac{1}{{\sqrt 2 }}\cos {h^{ - 1}}\left( {\frac{t}{{y\sqrt 2 }}} \right)\;\\ = \; - \frac{1}{{\sqrt 2 }}\cos {h^{ - 1}}\left( {\frac{{\sqrt 2 }}{{x + 1}}} \right)\end{array}\)

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  

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:

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

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

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

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

Explanation:

Mean \( = \mu = np = \left( {180} \right) \cdot \frac{1}{6} = 30\) 

Standard deviation \( = \sigma = \sqrt {npq} = 5\) 

Let x = number of times the face 4 will turn up

\(z = \frac{{x - \mu }}{\sigma }\)

\(\therefore z = \frac{{x - 30}}{5}\)

Now,

When x = 35, z = 1

Required probability = P(z ≥ 1)

Required probability = 1 – P(z < 1)

Required probability = 1 – (0.5 + P(0 < z < 1))

Required probability = 1 – (0.5 + 0.3413)

∴ The required probability = 0.1587

Concept:

By stokes theorem,

\(\mathop \oint \limits_C^\; F.dr = \mathop \int\!\!\!\int \limits_S^\; Curl\;F.Nds\)

Calculation:

\(\vec F\left( {x,\;y,\;z} \right) = x\hat i + y\hat j + 3\left( {{x^2} + {y^2}} \right)\hat k\)

\(\nabla \times \vec F = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ x&y&{3\left( {{x^2} + {y^2}} \right)} \end{array}} \right|\)

= i (6y) – j(6x) = 6y î - 6x ĵ

S: x² + y² + z² – 64 = 0

\(ds = + 2x\hat i + 2y\hat j + 2z\hat k\)

\(\left( {\nabla \times \vec F} \right) \cdot ds = \left( {6y\hat i - 6x\hat j} \right) \cdot \left( {2x\hat i + 2y\hat j + 2z\hat k} \right)\)

= 12xy – 12xy + 0 = 0

\( \Rightarrow \mathop \smallint \nolimits_C \vec F \cdot \overrightarrow {dr} = 0\)

Concept

Note

i) Gradient (∇) converts scalar function into vector function

ii) Divergence (∇.\({\rm{\bar F}}\)) Converts vector function to scalar (Dot product)

iii) Curl (∇ × \({\rm{\bar F}}\)) converts vector function to vector function (Cross product)

Calculation

Given, \(f = {x^2} - {y^2} + 2{z^2}\) 

Point P is (1, 2, 3), point Q is (5, 0, 4)

Line \(\overline {PQ} = \left( {5 - 1} \right)\bar i + \left( {0 - 2} \right)\bar j + \left( {4 - 3} \right)\bar k\)

\(\overline {PQ} = 4\bar i - 2\bar j + 1\bar k\)

Gradient of f (x, y, z) is given by ∇f

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

\(= \;\frac{{\partial \left( {{x^2} - {y^2} + 2{z^2}} \right)}}{{\partial x}}\;\bar i + \;\frac{{\partial \left( {{x^2} - {y^2} + 2{z^2}} \right)}}{{\partial y}}\;\bar j + \frac{{\partial \left( {{x^2} - {y^2} + 2{z^2}} \right)}}{{\partial z}}\overline {\;k} \)

\(= 2x\;\overline {\;i} - 2y\overline {\;j} + 4z\overline {\;k}\)

Directional derivative at point P (1, 2, 3) is

\({\left( {\nabla f} \right)_p} = 2 \times \left( 1 \right)\overline {\;i} - 2 \times \left( 2 \right)\bar j + 4 \times \left( 3 \right)\;\bar k\)

\({\left( {\nabla f} \right)_p} = 2\overline {\;i} - 4\;\bar j + 12\;\bar k\;\;\)

i) Maximum value of the directional derivative is

\( = \left| {\nabla {f_p}} \right| = \sqrt {{2^2} + {{\left( { - 4} \right)}^2} + {{\left( {12} \right)}^2}} \)

= 12.80

ii) Directional derivative of f (x, y, z) at P along the line \(\overline {PQ} \) is

\(D.D = {\left( {\nabla f} \right)_p}.\frac{{\overline {PQ} }}{{\left| {\overline {PQ} } \right|}}\)

\( = \left( {2\;\bar i - 4\;\bar j + 12\;\bar k} \right)\;.\frac{{4i - 2j + 1k}}{{\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {1^2}} }}\)

As it is dot product

\(= \frac{{8 + 8 + 12}}{{\sqrt {21} }}\)

\(= \frac{{28}}{{\sqrt {21} }}\)

Let \(y = \mathop {\lim }\limits_{x \to 0} {\left[ {\tan \left( {\frac{\pi }{4} + x} \right)} \right]^{1/x}}\) 

It is an indeterminate form.

Thus using log-concept,

log y \( = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{x}} \right]\) 

Applying L-H Rule, 

\(\log y = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\frac{1}{{\tan \left( {\frac{\pi }{4} + x} \right)}} \times {{\sec }^2}\left( {\frac{\pi }{4} + x} \right)}}{1}} \right]\) = 2

Concept:

Fourier series of f(x)

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos \frac{{n\pi x}}{L} + \mathop \sum \limits_{n = 1}^\infty {n_n}\sin \frac{{n\pi x}}{l}\)

\({a_0} = \frac{1}{l}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx,\;{a_n} = \frac{1}{l}\mathop \smallint \limits_{ - l}^l f\left( x \right)\cos \frac{{n\pi x}}{L}\;dx,\;{b_n} = \frac{1}{l}\;\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

Calculation:

\({a_0} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)dx = \frac{1}{\pi }\;\mathop \smallint \limits_0^\pi \sin x\;dx = \frac{2}{\pi }\)

\({a_n} = \frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin x\cos nx\;dx = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi [\sin \left( {n + 1} \right)x - \sin \left( {n - 1} \right)x]\;dx\)

\({a_n} = \frac{1}{{2\pi }}\left[ { - \frac{{\cos \left( {n + 1} \right)x}}{{n + 1}} + \frac{{\cos \left( {n - 1} \right)x}}{{n - 1}}} \right]_0^\pi \)

= 0, n is odd and \(\frac{{ - 2}}{{\pi \left( {{n^2} - 1} \right)}}\), n is even

Now,

\({b_n} = \frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin x\sin n\;x\;dx\)

\({b_n} = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi \left[ {\cos \left( {n - 1} \right)x - \cos \left( {n + 1} \right)x} \right]dx\)

\({b_n} = \frac{1}{{2\pi }}\left[ {\frac{{\sin \left( {n - 1} \right)x}}{{n - 1}} - \frac{{\sin \left( {n + 1} \right)x}}{{n + 1}}} \right]_0^\pi = 0\left( {n \ne 1} \right)\)

When n = 1

\({b_1} = \frac{1}{2}\)

\(\therefore f\left( x \right) = \frac{1}{\pi } - \frac{2}{\pi }\left[ {\frac{{\cos 2x}}{{{2^2} - 1}} + \frac{{\cos 4x}}{{{4^2} - 1}} + \ldots } \right] + \frac{1}{2}\sin x\)

\({\rm{Put\;}}x = \frac{\pi }{2}\)

\( \Rightarrow 1 = \frac{1}{\pi } - \frac{2}{\pi }\left( { - \frac{1}{{1.3}} + \frac{1}{{3.5}} - \frac{1}{{5.7}} + \ldots \infty } \right) + \frac{1}{2}\)

\(\therefore \frac{1}{{1.3}} - \frac{1}{{3.5}} + \frac{1}{{5.7}} - \ldots \infty = \frac{1}{4}\left( {\pi - 2} \right) = 0.285\)

Concept:

Binomial Probability Distribution:

Let X is the discrete Random variable such that its p.m.f is defined as

\(P\left( {X = r\;success} \right) = {\;^n}{C_r}{p^r}{q^{n - r}} \approx {\left( {q + p} \right)^n}\)

Where,

p = probability of success & q = probability of failure & p + q = 1

then X is said to follow Binomial distribution.

1) Parameter/statistical Attributes: Those unknowns which are necessary to evaluate complete distribution are called parameters.

2) Mean : E(x) = ∑ x_ip_i = np

3) Variance: E(x²) – (E(x))² = npq

4) Standard Deviation: \(\sqrt {npq}\)

Calculation:

Comparing \({\left( {q + p} \right)^n}\;with\;{\left( {\frac{1}{3} + \frac{2}{3}} \right)^{10}}\)

\(q = \frac{1}{3},\;p = \frac{2}{3},n = 10\)

∴ Standard Deviation \(= + \sqrt {npq}\) \(= \sqrt {\frac{1}{3} \times \frac{2}{3} \times 10} \approx 1.49\)

Variance \(= npq = \frac{1}{3} \times \frac{2}{3} \times 10 = \frac{{20}}{9}\)

∴ Required sum \(= 1.49 + \frac{{20}}{9} = 3.712\;\)