text:
If the difference between the mode and median is 2, then find the difference between the median and mean(in the given order).

1. 2
2. 1
3. 3
4. 4

Option 2 : 1

Detailed Solution

Concept:

Relation between mode, median and mean is given by:

Mode = 3 × median – 2 × mean

Calculation:

Given:

Mode – median = 2

As we know

Mode = 3 × median – 2 × mean

Now, Mode = median + 2

⇒ (2 + median) = 3median – 2mean

⇒ 2Median - 2Mean = 2

⇒ Median - Mean = 1

∴ The difference between the median and mean is 1.

Consider the below data:$$\begin{array}{*{20}{c}} x&:&0&1&2\\ {f\left( x \right)}&:&4&3&{12} \end{array}$$The value of$$\mathop \smallint \nolimits_0^2 f\left( x \right)dx$$ by Trapezoidal rule will be:

1. 11
2. 12
3. 15
4. 9

Option 1 : 11

Detailed Solution

Concept:

Trapezoidal rule states that for a function y = f(x)

 x x0 x1 x2 x3 …… xn y y0 y1 y2 y3 …… yn

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

$$\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]$$   ---(1)

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.

Calculation:

$$\begin{array}{*{20}{c}} x&:&0&1&2\\ {f\left( x \right)}&:&4&3&{12} \end{array}$$

Here: x0 = 4, x1 = 3, x2 = 12, h = 1

From equation (1);

$$\mathop \smallint \limits_0^2 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{x_0} + {x_2}} \right) + 2\left( {{x_1}} \right)} \right]$$

$$= \frac{1}{2}\left[ {\left( {{4} + {12}} \right) + 2\left( {{3}} \right)} \right]={22\over2}=11$$

Key Points:

Apart from the trapezoidal rule, other numerical integration methods are:

Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

$$\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]$$     ..2)

Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

$$\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]$$

For what value of λ, do the simultaneous equation 2x + 3y = 1, 4x + 6y = λ have infinite solutions?

1. λ = 0
2. λ = 1
3. λ ≠ 2
4. λ = 2

Option 4 : λ = 2

Detailed Solution

Concept:

Non-homogeneous equation of type AX = B has infinite solutions;

if ρ(A | B) = ρ(A) < Number of unknowns

Calculation:

Given:

2x + 3y = 1

4x + 6y = λ

The augmented matrix is given by:

$$\left( {A{\rm \ {|}} \ B} \right) = \left[ {\begin{array}{*{20}{c}} 2&3&{1\;}\\ 4&6&\lambda \end{array}} \right]$$

Applying R2 → R2 – 2R1

$$\left( {A{\rm \ {|}} \ B} \right) = \left[ {\begin{array}{*{20}{c}} 2&3&{1\;}\\ 0&0&{\lambda - 2} \end{array}} \right]$$

For the system to have infinite solutions, the last row must be a fully zero row.

So if λ = 2 then the system of equations has infinitely many solutions.

Key Points:

Remember the system of equations

AX = B have

1) Unique solution, if ρ(A : B) = ρ(A) = Number of unknowns.

2) Infinite many solutions, if ρ(A : B) = ρ(A) < Number of solutions

3) No solution, if ρ(A : B) ≠ ρ(A).

The value of $$\mathop {\lim }\limits_{x \to 0} \left( {x\sin \frac{1}{x}} \right)$$ is:

1. - 1
2. 0
3. 1

Option 3 : 0

Detailed Solution

Concept:

sin θ can take from - to +  but sin θ gives - 1 to + 1, i.e. for values of θ from  to ∞, sin θ always lies between -1 and 1.

$$- 1 \le \sin θ \le 1$$

sin gives finite value.

Analysis:

Let,

$$f(x)=\mathop {\lim }\limits_{x \to 0} \left( {x\sin \frac{1}{x}} \right)$$

When x = 0

f(x) = 0 × sin

f(x) = 0 × Finite value

$$f(x)=\mathop {\lim }\limits_{x \to 0} \left( {x\sin \frac{1}{x}} \right)=0$$

A bag contains 3 white, 2 blue and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is not red?

1. 3/10
2. 1/5
3. 1/2
4. 4/5

Option 3 : 1/2

Detailed Solution

Calculation:

A bag contains 3 white, 2 blue and 5 red balls.

Total number of balls = 3 + 2 + 5 = 10

Number of balls that are not red = 10 - 5 = 5

Probability of balls drawn is not red = (number of balls which are not red)/(total number of balls) = 5/10 = 1/2

The differential equation 2y dx – (3y – 2x) dy = 0 is

1. exact and homogenous but not linear
2. exact, homogenous and linear
3. exact and linear but not homogenous
4. homogenous and linear but not exact

Option 2 : exact, homogenous and linear

Detailed Solution

Concept:

Homogenous equation: If the degree of all the terms in the equation is the same then the equation is termed as a homogeneous equation.

Exact equation: The necessary and sufficient condition of the differential equation M dx + N dy = 0 to be exact is:

$$\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}$$

Linear equation: A differential equation is said to be linear if the dependent variable and its differential coefficient only in the degree and not multiplied together.

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation is:

$$\frac{{dy}}{{dx}}+Py=Q$$

where, P, Q is a function of x.

or, $$\frac{{dx}}{{dy}}+Px=Q$$

where, P, Q is a function of x.

Condition 1:

2y dx + (2x - 3y) dy = 0   ---.(1)

(It is Homogeneous)

Condition 2:

Equation (1) can be written as  $$\frac{{dy}}{{dx}}=\frac{{2y}}{{2x\;-\;3y}}$$ .

It is not a linear form.

or $$\frac{{dx}}{{dy}}=\frac{{2x-3y}}{{2y}}$$

$$\frac{{dx}}{{dy}}+\frac{{x}}{{y}}=\frac{{3}}{{2}}$$

It is in linear form

Condition 3:

M dx + N dy = 0

2y dx – (3y – 2x) dy = 0

hence, M = 2y and N = 2x - 3y

$$\frac{{\partial M}}{{\partial y}} =\frac{{\partial (2y)}}{{\partial y}}= 2$$ and $$\frac{{\partial N}}{{\partial x}}= \frac{{\partial (2x+3y)}}{{\partial x}}=2$$

As $$\frac{{\partial M}}{{\partial y}}=\frac{{\partial N}}{{\partial y}}$$

so, it is an exact equation.

The standard ordered basis of R3 is {e1, e2, e3} Let T : R3 → R3 be the linear transformation such that T(e1) = 7e1 - 5e3, T (e2) = -2e2 + 9e3, T(e3) = e1 + e2 + e3. The standard matrix of T is:

1. $$\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)$$
2. $$\left( {\begin{array}{*{20}{c}} 7&-2&1\\ -5&{ 9}&1\\ { 0}&0&1 \end{array}} \right)$$
3. $$\left( {\begin{array}{*{20}{c}} 7&0&-5\\ 0&{ - 2}&9\\ 1&1&1 \end{array}} \right)$$
4. $$\left( {\begin{array}{*{20}{c}} 7&-5&0\\ -2&{ 9}&1\\ { 1}&1&1 \end{array}} \right)$$

Option 1 : $$\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)$$

Detailed Solution

Concept:

Matrix transformations:

Theorem: Suppose L: Rn → Rm is a linear map. Then there exists an m×n matrix A such that L(x) = Ax for all x ∈ Rn. Columns of A are vectors L(e1), L(e2), . . . , L(en), where e1, e2, . . . , en is the standard basis for Rn.

Calculation:

Given linear transformation are:

T(e1) = 7e1 - 5e3,

T(e2) = -2e2 + 9e3,

T(e3) = e1 + e2 + e3

Let the standard matrix be A with respect to the basis e1, e2, e3

Now T(e1) = 7e1 + 0e2 - 5e3,

T(e2) = 0e1 -2e2 + 9e3,

T(e3) = e1 + e2 + e3.

The standard matrix will be (transpose of linear combinations)

$$\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)$$

The value of $$\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2}\sin \theta dr\;d\theta \;d\phi$$ will be:

1. $$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 + \sqrt 3 } \right)$$
2. $$\frac{{{2\pi } }}{3}\left( {\sqrt 3 -1 } \right)$$
3. $$\frac{{ {2\pi } }}{3}\left( {\sqrt 2 - \sqrt 3 } \right)$$
4. $$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)$$

Option 4 : $$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)$$

Detailed Solution

$$\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2} \sin \theta drd\theta d\phi$$

$$\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \mathop \smallint \limits_0^1 \left( {{r^2} \times dr} \right) \times \sin \theta d\theta d\phi$$

$$\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \left[ {\frac{{{r^3}}}{3}} \right]_0^1 \times \sin \theta d\theta d\phi$$

$$\frac{1}{3}\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \sin \theta \times d\theta \times d\phi$$

$$\frac{1}{3}\mathop \smallint \limits_0^{2\pi } - \left[ {\cos \theta } \right]_0^{\frac{\pi }{4}} \times d\phi$$

$$\frac{1}{3}\mathop \smallint \limits_0^{2\pi } - \left[ {\frac{1}{{\sqrt 2 }} - 1} \right] \times d\phi$$

$$\frac{1}{{3\sqrt 2 }}\left[ {\sqrt 2 \phi - \phi } \right]_0^{2\pi }$$

$$\frac{{2\pi }}{{3\sqrt 2 }}\left[ {\sqrt 2 - 1} \right]$$

$$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)$$

A complete solution of partial differential equation $$\frac {\partial z}{\partial x} - 3x^2 = \left(\frac {\partial z}{\partial y}\right)^2 - y$$ will be ________, where a and b are arbitrary constants.

1. z2 = ax2 + by2 + 1
2. $$z = ax + x^3 + \left(\frac 2 3\right) (a + y)^{\frac 3 2} + b$$
3. z1 / 2 = (x + a)1 / 2 + (y + b)1 / 2
4. z = (ax2 + by2)3 / 2 + 2

Option 2 : $$z = ax + x^3 + \left(\frac 2 3\right) (a + y)^{\frac 3 2} + b$$

Detailed Solution

Concept:

Separable equation of the form f(x, p) = g(y, q)

Let f(x, p) = g(y, q) = a (constant)

Solve f(x, p) = a and g(y, q) = a for p and q, we get

p = ϕ(x, a) and q = ψ(y, a)

We have dz = p dx + q dy

Bu integration, we get the required solution as follows,

z = ∫ ϕ(x, a) dx + ∫ ψ(y, a) dy + b

Where a, b are arbitrary constants.

Calculation:

Given Partial differentiation equation is $$\frac {\partial z}{\partial x} - 3x^2 = \left(\frac {\partial z}{\partial y}\right)^2 - y$$

Writing in terms of p,q ⇒ p - 3x2 = q2 - y

Let p - 3x2 = q2 - y = a (constant)

Now p - 3x2 = a ⇒ p = 3x2 + a

q2 - y = a ⇒ $$q = \sqrt {y+a}$$

Substituting the values of p and q in dz = p dx + q dy, we get

$$dz = (a+3x^2)dx + (\sqrt {y+a})dy$$

Integrating on both sides, we get the general solution as

$$z = ax + x^3 + \left(\frac 2 3\right) (a + y)^{\frac 3 2} + b$$

For the curve xy3 - yx3 = 6, the slope of the tangent line at the point (1, -1) is:

1. 1 / 2
2. -1
3. 2
4. 1

Option 2 : -1

Detailed Solution

Concept:

Let y = f(x) be the equation of a curve, then the slope of the tangent at any point say (x1, y1) is given by:

$$m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}$$

Calculation:

Given curve is xy3 - yx3 = 6

Now by partially differentiating the equation of curve with respect to x we get;

$$3xy^2\frac{dy}{dx}+y^3-3x^2y-x^3\frac{dy}{dx}=0$$

$$\frac{dy}{dx}({3xy^2-x^3})={3x^2y-y^3}$$

$$\frac{dy}{dx}=\frac{3x^2y-y^3}{3xy^2-x^3}$$

$$\frac{dy}{dx}=\frac{y(3x^2-y^2)}{x{(3y^2-x^2)}}$$

The slope(m) i.e. dy/dx of the tangent at (1, -1) is:

m =  -1

A partial differential equation derived from the equation z = aeby sinbx will be:

1. $$\frac{{\partial z}}{{\partial y}} = 2y{\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}$$
2. $$\left( {1 + \frac{{\partial z}}{{\partial y}}} \right)\frac{{\partial z}}{{\partial x}} = z$$
3. $$\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$$
4. $$2z = x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}}$$

Option 3 : $$\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$$

Detailed Solution

z = aeby sinbx

Now by differentiating w.r.t. x we get:

$$\frac{{\partial z}}{{\partial x}} = a \times {e^{by}} \times \cos bx \times b$$

Again differentiating:

$$\frac{{{\partial ^2}z}}{{\partial {x^2}}} = - a \times {b^2} \times {e^{by}} \times \sin bx$$  ----(1)

z = aeby sinbx

Now by differentiating w.r.t. y we get:

$$\frac{{\partial z}}{{\partial y}} = a \times \sin bx \times {e^{by}} \times b$$

Again differentiating:

$$\frac{{{\partial ^2}z}}{{\partial {y^2}}} = a \times {b^2} \times {e^{by}} \times \sin bx$$  ----(2)

Now by adding (1) and (2) we get:

$$\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$$

Hence, Option 3 is correct.

The value of x for which of the following series converges is$$x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \frac{{{x^5}}}{5} - \ldots \infty ,$$

1. The series converges f0r -1 < x < 1
2. The series converges for -1 < x < 1
3. The series diverges for -1 < x < 1
4. The series converges for x > 1

Option 2 : The series converges for -1 < x < 1

Detailed Solution

Explanation:

Given series is,

$$x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \frac{{{x^5}}}{5} - \ldots \infty ,$$

$${u_n} = {\left( { - 1} \right)^{n - 1}}\frac{{{x^n}}}{n}$$

$${u_{n + 1}} = {\left( { - 1} \right)^n}\frac{{{x^{n + 1}}}}{{n + 1}}$$

$$\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{ - n}}{{n + 1}}x$$

$$\mathop {{\rm{lt}}}\limits_{n \to \infty } \left| {\left. {\frac{{{u_n} + 1}}{{{u_{n\;\;}}}}} \right| = \mathop {{\rm{lt}}}\limits_{n \to \infty } \left( {\frac{n}{{n + 1}}} \right)\left( {\left| x \right|} \right)} \right.$$

$$= \left| x \right|$$

By ratio test, the given series converges for |x| < 1 and diverges for |x| > 1

Let us examine the series for x = ± 1

For x = 1, the series reduces to

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \ldots$$

This is an alternating series and is convergent.

For x = -1 the series becomes

$$- \left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots } \right)$$

This is a divergent series as can be seen by comparison with P-series with P = 1

Hence the given series is converges for -1 < x < 1

The value of $$\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}$$ is:

1. $$\frac1 3$$
2. $$\frac 4 3$$
3. 1
4. $$\frac 2 3$$

Option 2 : $$\frac 4 3$$

Detailed Solution

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}$$

II. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }$$

Then we can apply L-Hospital Rule ⇔ $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$

Calculation:

Given:

$$\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}$$

As we can see, $$\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}=\frac{0}{0}$$

So, Apply the L-Hospital rule here,

$$\mathop {\lim }\limits_{x \to 2} \frac{{{2x} }}{{3}}$$

After putting the limit we'll get:

$$\mathop {\lim }\limits_{x \to 2} \frac{{{2x} }}{{3}}=\frac{4}{3}$$

Hence the required value of the limit is $$\frac4 3$$.

$$\mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1}}{{\sin x - x}}$$ is equal to

1. Undefined
2. 1
3. 0

Option 2 :

Detailed Solution

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}$$

II. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }$$

Then we can apply L-Hospital Rule as:

$$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$

Calculation:

Given:

$$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x\; -\;1}}{{sin\;x\; - \;x}} = \left( {\frac{0}{0}} \right)$$ form

Applying L’ Hospital  rule:

$$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x - 1}}{{sin\;x-\;x}} =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({cos\;x\; -\;1})}{{\frac{d}{dx}(sin\;x\;-\;x})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\;x {\rm{}}}}{{\cos {\rm{x}\;-\;1}}}$$

$$\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\; {\rm{x}}}}{{\cos {\rm{x}\,-1\;}}} = \left( {\frac{0}{0}} \right){\rm{form}}$$

Once again by L’ Hospital rule,

$${\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-cos\; {\rm{x}}}}{{{\rm{-sin~x}}}} = \frac{-1}{0} = \infty{\rm{}}$$

The general solution of partial differential equation $$x\left( {{y^2} + z} \right)\frac{{\partial z}}{{\partial x}} - y\left( {{x^2} + z} \right)\frac{{\partial z}}{{\partial y}}=\left( {{x^2} - {y^2}} \right)z$$ will be:

1. φ(xyz, x2 + y2 - 2z) = 0, where φ is an arbitrary function.
2. φ(x + y + z,x2 + y2) = 0, where φ is an arbitrary function.
3. φ(x2 + y2 + z2, x + y + z) = 0 where φ is an arbitrary function
4. φ(x + y - z, xy + yz + zx) = 0, where φ is an arbitrary function

Option 1 : φ(xyz, x2 + y2 - 2z) = 0, where φ is an arbitrary function.

Detailed Solution

Concept:

Linear Partial Differential Equation of First Order: A linear partial differential equation of the first order, commonly known as Lagrange's Linear equation, is of the form Pp + Qq = R where P, Q, and R are functions of x, y, z. This equation is called a quasi-linear equation.

Thus, to solve the equation of the form Pp + Qq = R, we have to follow this solution procedure:

1) Form the subsidiary equations as:

$$\frac {dx}{P} = \frac {dy}{Q} = \frac {dz}{R}$$

2) Solve any two simultaneous equations by any method giving u = a and v = b as its solutions.

3) Write the complete solution as φ (u, v) = 0 or u = f (v).

Lagrange's method of multipliers:

Consider the partial differentiation equation Pp + Qq = R

then auxiliary equations are given by:

$$\frac {dx}{P} = \frac {dy}{Q} = \frac {dz}{R}$$

Now choose multipliers P1, Q1, R1 in such a way that in ratio

$$\frac {{{P_1}{dx}}+{{Q_1}{dy}}+{{R_1}{dz}}}{{{P_1}{P}}+{{Q_1}{Q}}+{{R_1}{R}}}$$, denominators vanish.

⇒ P1 dx + Q1 dy + R1 dz = 0

By integration, the solution is given by u (x, y, z) = c1

Similarly, by choosing another set of multipliers, function v(x, y, z) is determined.

Then complete solution is given by f (u, v) = 0.

Calculation:

Given:

$$x\left( {{y^2} + z} \right)\frac{{\partial z}}{{\partial x}} - y\left( {{x^2} + z} \right)\frac{{\partial z}}{{\partial y}}=\left( {{x^2} - {y^2}} \right)z$$

Comparing with Pp + Qq = R,

P = x (y2 + z); Q = -y (x2 + z); R = (x2 - y2) z;

∴ The auxiliary equations are:

$$\frac{{{\rm{dx}}}}{{{\rm{x\;}}\left( {{\rm{y^2}} + {\rm{z}}} \right){\rm{\;}}}} = \frac{{{\rm{dy}}}}{{{\rm{\;-y\;}}\left( {{\rm{x^2}} + {\rm{z}}} \right){\rm{\;}}}} = \frac{{{\rm{dz}}}}{{{\rm{z\;}}\left( {{\rm{x^2}} - {\rm{y^2}}} \right)}}$$

1st function:

Using multipliers x, y and -1 ⇒

$${\rm{\;}}\frac{{{\rm{xdx}} + {\rm{ydy}} - {\rm{dz}}}}{{{\rm{x^2\;}}\left( {{\rm{y^2}} + {\rm{z}}} \right) - {\rm{\;y^2\;}}\left( {{\rm{x^2}} + {\rm{z}}} \right) - {\rm{z\;}}\left( {{\rm{x^2}} - {\rm{y^2}}} \right)}} = \frac{{{\rm{xdx}} + {\rm{ydy}} - {\rm{dz}}}}{0}$$

∴ x dx + y dy - dz = 0 ⇒ x2 + y2 - 2z = constant;

2nd function:

Using multipliers 1/x, 1/y and 1/z

⇒ $${\rm{\;}}\frac{{{\rm{{\frac{1}{x}}dx}} + {\rm{{\frac{1}{y}}dy}} + {\rm{{\frac{1}{z}}dz}}}}{{\left( {{\rm{y^2}} + {\rm{z}}} \right) - \left( {{\rm{x^2}} + {\rm{z}}} \right) + \left( {{\rm{x^2}} - {\rm{y^2}}} \right)}}$$

$$= \frac{{{\rm{{\frac{1}{x}}dx}} + {\rm{{\frac{1}{y}}dy}} + {\rm{{\frac{1}{z}}dz}}}}{0}$$

∴ $${{\rm{{\frac{1}{x}}dx}} + {\rm{{\frac{1}{y}}dy}} + {\rm{{\frac{1}{z}}dz}}} = 0$$

xyz = constant

So both the equations are:

x2 + y2 - 2z = constant; xyz = constant;

Complete solution is φ(xyz, x2 + y2 - 2z) = 0, where φ is an arbitrary function.

A matrix X has a dimension of 2 × 2. If the eigenvalues of this matrix is 5 and 6, what would be the eigen values of X2?

1. 2.5 and 3
2. 5 and 6
3. 10 and 12
4. 25 and 36

Option 4 : 25 and 36

Detailed Solution

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.

Properties of Eigenvalues:

(1) If λ is an eigenvalue of a matrix A, then λn will be an eigenvalue of a matrix An.

(2) If λ is an eigenvalue of a matrix A, then kλ will be an eigenvalue of a matrix kA where k is a scalar.

(3) Sum of eigenvalues is equal to the trace of that matrix.

(4) The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A.

(5)  If λ is an Eigenvalue of matrix A, then λ2 will be an Eigenvalue of matrix A2.

(6) If λ1 is an Eigenvalue of matrix A, then (λ1 + 1) will be an Eigenvalue of the matrix (A + I).

(7) Eigenvalues of a matrix and its transpose are the same because the transpose matrix will also have the same characteristic equation.

Calculation:

If  λ1, λ2, λ3 …. λ4 are the given values of A, then

eigenvalues of Awill be $$\lambda _1^m,\lambda _2^m,\lambda _3^m, \ldots \ldots ..$$

‘S’ matrix has eigenvalues 5 and 6.

S2 matrix has eigenvalues 52 and 362

i.e. 25 and 36.

The value of $$\mathop \oint \nolimits_c \frac{1}{{{z^2} + 4}}dz$$ Where C |z - 2i| = 1 is will be:

1. 0
2. 1/5
3. π/2
4. π /3

Option 3 : π/2

Detailed Solution

Concept:

Cauchy's Integral Formula

If f(z) is analytic within and on a closed curve and if is any point within C, then:

$$\mathop \smallint \limits_C^{} \frac{{f\left( z \right)}}{{z - a}}dz = 2\pi if\left( a \right)$$

Analysis:

Given:

$$f\left( z \right) = \smallint \frac{{{1}}}{{{z^2+4}}}dz$$

|z - 2i| = 1

|x + iy - 2i| = 1

|x + i(y - 2)| = 1

Taking the magnitude of the above, we get:

x2 +(y - 2)= 1

This is the equation of a circle with:

Center = (0, 2)

Poles of f(z) = z2 + 4

z = (+2i, -2i)

2i lies Inside the circle

-2i lies outside the circle

∴ We can write:

$$For\;pole\;\left( {z\; = \;-2i} \right) = \smallint \frac{{{1}}}{{\left( {z-2i} \right)\left( { z+2i} \right)}}dz = 0$$

Now, for pole z = 2i (lies inside the circle)

$$\smallint \frac{{{1}}}{{\left( {z+2i} \right)\left( {z-2i} \right)}}dz = 2\pi i{\left[ {\frac{{{1}}}{{2i+2i}}} \right]_{z = 2i}}$$

$$2\pi i\left[ {\frac{1}{4i}} \right] ={\frac{\pi}{2}}$$

If Rank (A) = 2 and Rank (B) = 3 then Rank (AB) is:

1. 6
2. 5
3. 3

Detailed Solution

Concept:

Rank:

The rank of a matrix is a number equal to the order of the highest order non-vanishing minor, that can be formed from the matrix.

For matrix A, it is denoted by ρ(A).

The rank of a matrix is said to be r if,

• There is at least one non-zero minor of order r.
• Every minor of matrix having order higher than is zero.

Property of Rank of Matrix:

ρ(AB) ≤ min [ρ(A), ρ(B)]

Calculation:

Given:

ρ(A) = 2, ρ(B) = 3

Using properties

ρ(AB) ≤ min [ρ(A), ρ(B)]

ρ(AB) ≤ min (2,3)

ρ(AB) ≤  2

Alternate Method

Let order of matrix A be 2 × m and order of matrix B be m × 3 (∵ for multiplication we need the column of A and row of B to be same)

∴ Order of matrix AB will be 2 × 3

Using properties

ρ(AB) ≤ min (Row, Column)

⇒ ρ(AB) ≤ min (2, 3)  [only when the column of A and row of B is the same]

⇒ ρ(AB) ≤ 2.

we don't know the dimension of A and B, we cannot predict the exact rank of AB but its maximum rank will be 2.

Important Points

Other properties of rank of a matrix are:

• The rank of a matrix does not change by elementary transformation, we can calculate the rank by changing the matrix into Echelon form. In the Echelon form, the rank of a matrix is the number of non-zero rows of the matrix.
• The rank of a matrix is zero if the matrix is null.
• ρ(A) ≤ min (Row, Column)
• ρ(AB) ≤ min [ρ(A), ρ(B)]
• ρ(AA) = ρ(A AT) = ρ(A) = ρ(AT)
• If A and B are matrices of the same order, then ρ(A + B) ≤ ρ(A) + ρ(B) and ρ(A - B) ≥ ρ(A) - ρ(B).
• If Aθ is the conjugate transpose of A, then ρ(Aθ) = ρ(A) and ρ(A Aθ) = ρ(A).
• The rank of a skew-symmetric matrix cannot be one.

The approximate solution of the system of simultaneous equations2x - 5y + 3z = 7x + 4y - 2z = 32x + 3y + z = 2by applying Gauss-Seidel method one time (using initial approximation as x - 0, y - 0, z - 0) will be:

1. x = 2.32, y = 1.245, z = -3.157
2. x = 1.25, y = -2.573, z = -3.135
3. x = 2.45, y = -1.725, z = -3.565
4. x = 3.5, y = -0.125, z = -4.625

Option 4 : x = 3.5, y = -0.125, z = -4.625

Detailed Solution

Concept:

Gauss Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variable as 0.

2x - 5y + 3z = 7

Putting y = 0, z = 0 ⇒ x = 3.5

x + 4y - 2z = 3

Putting x = 3.5, z = 0y = - 0.125

2x + 3y + z = 2

Putting x = 3.5, y = - 0.125 ⇒ z = 2 – 3(-0.125) – 2(3.5)

z = - 4.625

The value of $$\mathop \smallint \nolimits_0^1 \frac{{{{\sin }^{ - 1}}x}}{x}dx$$ is:

1. π log 2
2. $$\frac{2 \pi }{3}$$
3. $$\frac{\pi }{4}$$
4. $$\frac{\pi }{2}log2$$

Option 4 : $$\frac{\pi }{2}log2$$

Detailed Solution

Let $$I = \mathop \smallint \nolimits_0^1 \frac{{{{\sin }^{ - 1}}x}}{x}dx$$

Assume sin-1 x = θ

x = sin θ

dx = cos θ dθ

When x = 0: θ = 0

$$x = 1$$

$$\theta = \frac{\pi }{2}$$

$$I = \mathop \smallint \nolimits_0^{\pi /2} \frac{\theta }{{\sin \theta }} \cdot \cos \theta \;d\theta = \mathop \smallint \nolimits_0^{\pi /2} \theta \cdot \cot \theta \cdot d\theta$$

Let u = θ ; v = cot θ

We have:

$$\smallint uv = u\smallint v - \smallint u'\smallint v$$

$$I = \theta \mathop \smallint \nolimits_0^{\pi /2} \cot \theta \cdot d\theta - \mathop \smallint \nolimits_0^{\pi /2} \left[ {\frac{{d\theta }}{{d\theta }} \cdot \smallint \cot \theta \cdot d\theta } \right]d\theta$$

$$= \theta \cdot \log \left| {\sin \theta } \right|_0^{\pi /2} - \mathop \smallint \nolimits_0^{\pi /2} \log \left| {\sin \theta } \right|d\theta$$

In the range of $$\left[ {0,\frac{\pi }{2}} \right]\;;\sin \theta$$ is positive ⇒ |sin θ| = sin θ:

$$I = \theta \cdot \left| {\log \sin \theta } \right|_0^{\frac{\pi }{2}} - \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta d\theta$$      ---(1)

Let $$\mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \;d\theta = {I_1}$$      ---(2)

Using $$\mathop \smallint \nolimits_0^a f\left( x \right)dx = \mathop \smallint \nolimits_0^a f\left( {a - x} \right)dx$$

$${I_1} = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \log \cos \theta \;d\theta$$      ---(3)

From (2) and (3), we get:

$$2{I_1} = \mathop \smallint \nolimits_0^{\pi /2} \left( {\log \sin \theta + \log \cos \theta } \right)d\theta$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \log (\sin \theta \cdot \cos \theta ) \cdot d\theta$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \log \left( {\frac{{\sin 2\theta }}{2}} \right) \cdot d\theta$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \left( {\log \sin 2\theta - \log 2} \right)d\theta$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \log \sin 2\theta \;d\theta - \mathop \smallint \nolimits_0^{\pi /2} \log 2 \cdot d\theta$$

$$= \frac{1}{2}\mathop \smallint \nolimits_0^\pi \log \sin 2\theta \;d\left( {2\theta } \right) - \log 2\;\mathop \smallint \nolimits_0^{\pi /2} d\theta$$

Property of integral:

$$\mathop \smallint \nolimits_a^b \;F\left( x \right)dx = \mathop \smallint \nolimits_a^b f\left( t \right)dt$$

$$2{I_1} = \frac{1}{2}\mathop \smallint \nolimits_0^\pi \log \sin \theta \;d\theta - \log 2\;\left| \theta \right|_0^{\pi /2}$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \;d\theta - \log 2 \cdot \frac{\pi }{2}$$

$$= \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \cdot d\theta - \frac{\pi }{2} \cdot \log 2$$

Using (2), we get:

$$2{I_1} = {I_1} - \frac{\pi }{2}\log 2$$

$${I_1} = - \frac{\pi }{2}\log 2$$

Substituting valve of I1 in equation (1), we get:

$$I = \frac{\pi }{2} \cdot \log \sin \frac{\pi }{2} - 0 \cdot \log \sin \theta - \left( {\frac{{ - \pi }}{2} \cdot \log 2} \right)\;$$

$$= 0 + \frac{\pi }{2} \cdot \log 2$$

$$I = \frac{\pi }{2} \cdot \log 2$$

For $$0 \cdot \log \sin \theta \;;\;\;P = \mathop {\lim }\limits_{\theta \to 0} \theta \cdot \log \sin \theta = 0$$


text:
## 效果

1. 目标的护甲有效值会根据实际使用的武器和单位的伤害类型，护甲类型和生命类型而变动。
2. 所受的伤害会根据护甲类型得到一定程度的减免，具体的数值则是通过一个伤害减免方程得到的。

### 装甲类型

Grineer 铁制装甲 合金装甲
Corpus 合金装甲
Infested 感染肌腱

Tenno Tenno装甲

+50%
-15%
-25%
+75%

+15%
-50%
+25%
-50%
-50%
+75%

+25%
+25%
-50%
+50%

Tenno装甲
*Tenno装甲没有抗性或易伤。

• 能无视护甲，但不能无视护盾。

### 护甲类型倍率

• 护甲是在将伤害类型纳入考虑之前的护甲值。
• 护甲类型倍率可参照上下文中出现的图表。

• 补充说明，对于护甲，伤害类型克制会带来两个效果，百分比增伤效果（对于生命值和护盾，也有该效果）和百分比无视护甲效果（仅针对护甲）。
• 举个例子，对带有铁质装甲的目标，计算伤害时，既考虑百分比增伤效果（$\text{实际伤害}=\text{伤害}*(1+0.75)$），也考虑百分比无视护甲效果（$\text{总护甲}=\text{护甲}*(1-0.75)$）。

### 伤害削减公式

 伤害削减 = 1 − 300 300 + 总护甲

## 有效生命值

 ( 护甲 ) 有效生命值 = 面板生命值 × 1 + 300
• 面板生命值指的是游戏中实际显示出来的生命值，也就是你游戏中画面右上方的生命值。
• 有效生命值指的是实际有效的生命值，它能作为衡量战甲伤害抵抗能力的标准。

## 提升护甲

### MOD

• MOD护甲加成倍率指的是MOD上显示出的加成值。满级的钢铁纤维是1.1，满级的灵活装甲则是0.45，两者同时装备时则是1.55。
• 如果能够确保触发生命转换或的完整效果，事实上它们提供的护甲加成（在大多数战甲上）要远远高于使用其他的、直接按百分比增加护甲的Mod。

 有效生命提升值 = MOD护甲加成倍率 × (1 − 1 ) 1 + 护甲值 300

### 能力

 能量75快捷键3
 石化凝视Atlas 强硬的凝视将石化敌人、治疗石头人，并创造出石化壁垒。当石化的敌人碎裂后，将掉出可以治疗 Atlas 的碎石。
 强度:100% (石者治疗量) 持续时间:5 / 10 / 15 / 20 秒 范围:10 / 11 / 12 / 14 米 (锥形区域半径) 其他:60° (视角)+50% (生命类伤害系数提升)1 (每个敌人死亡后掉落的碎石)
 能量100快捷键4
 漫步石者召唤两个元素石头近战打手。召唤动作将石化临近 Atlas 的敌人。当它们的生命结束时，石头人将碎成一堆用于治疗的碎石。
 强度:500 / 750 / 1000 / 1200 (生命)50 / 150 / 250 / 500 (护甲) 500 / 750 / 1000 / 1250 (爆炸伤害)800 / 1000 / 1400 / 2000 (近战伤害)200 / 250 / 350 / 500 (投石伤害) 持续时间:15 / 20 / 30 / 45 秒 范围:3 / 4 / 5 / 6 米 (石化半径)3 / 4 / 5 / 6 米 (爆炸范围)1 (速度倍率) 其他:2 (石者数量)10 秒 (石化时间)1 (每个石者掉落的碎石数)≤ 50 (碎石产生的生命/护甲值)
When killed, petrified enemies and ' Rumblers drop a resource called Rubble that grants bonus armor if he is at full health. Bonus armor caps at 1,500 points. This bonus armor decays over time at a rate of 5 armor points per second.

 能量50快捷键2
 元素之护根据 Chroma 的元素属性，制造攻击性的范围效果。而 Chroma 和附近的盟友会被这股防御能量保护。
 强度: 25 / 50 / 75 / 100 (伤害) 50% / 75% / 100% / 200% (生命加成) 2% / 5% / 7% / 10% (异常触发几率) 2.5x / 3x / 5x / 10x (伤害倍率) 50% / 65% / 75% / 100% (护盾加成) 10% / 15% / 20% / 25% (异常触发) 15% / 20% / 30% / 35% (武器切换速率) 15% / 20% / 30% / 35% (装填速率) 25% / 35% / 40% / 50% (每秒伤害几率) 1.5x / 2x / 2.5x / 3x (伤害倍率) 25% / 75% / 100% / 150% (护甲加成) 10% / 15% / 20% / 25% (异常触发几率) 持续时间:10 / 15 / 20 / 25 秒 范围:6 / 8 / 10 / 12 米
 能量75快捷键3
 怨怒护甲当护盾被击中时，Chroma和附近友军的护甲变得更强，当生命值被击中时，武器伤害增加。
 强度:200% / 250% / 300% / 350%（最大护甲倍率）200% / 225% / 250% / 275%（最大伤害倍率） 持续时间:10 / 15 / 20 / 25 秒 范围:8 / 10 / 15 / 18 米
As shields are depleted while is active, 's armor is increased by 350%. This is a total of 1250 before considering the effect of mods.

 能量25快捷键1
 大块朵颐将一个敌人囫囵吞入Grendel的腹内。没吃饱？那就接着吃吧。但请注意，Grendel肚子里的每个敌人都会缓慢消耗能量。按住技能键即可将腹内的敌人吐出，而这些敌人身上会覆盖着毒性消化液。 能量 / （敌人 * 秒） 1.5 / 1.5 / 1.5 / 1.5 吐出能量消耗: 0
 强度:2 / 3 / 4 / 5 % （每秒护甲移除）250 / 300 / 400 / 500 （吐出时的 伤害） 持续时间:N/A 范围:10 / 15 / 20 / 25 m （吞噬范围） 其他:1.5 s （大口张开时长）1 s （每个敌人的额外时长）100% (喷吐时触发率)8 m（锥形喷吐范围）

Grendel gains additional armor for every enemy swallowed via , but each enemy in his stomach drains 1.5 points of energy per second, which ramps up the longer they remain trapped. Once all his energy is drained, or 's ability key is held, Grendel vomits out all stored enemies and loses his armor bonus.

While there is technically no limit to how many enemies Grendel can devour, and by extension how much bonus armor he can get, the amount of energy drained eventually becomes too much to manage even if Grendel is modded specifically for 技能效率 in mind.

 能量25快捷键4
 圣甲虫群充能来将生命值转化成坚硬的圣甲虫护甲。释放充能来使圣甲虫群卷向敌人；活下来的敌人生命值将被消耗及转移到同伴身上。
 强度:100 / 150 / 175 / 200 持续时间:6 / 10 / 12 / 15 秒 范围:20 / 22 / 25 / 30 米 （施放范围）6 / 10 / 12 / 15 米 (治疗半径)5 米 （传播范围）
Indefinitely increases ' armor by up to 100%. This bonus is additive to armor mods and not affected by 技能强度.

 能量25快捷键3
 疗愈脉动Oberon将源自自身的能量治愈波导向友军，使他们随时间持续恢复生命。能量消耗: 2 点/秒每个目标的能量消耗： 3 点/秒
 强度:50 / 75 / 100 / 125 (首次治愈)15 / 20 / 25 / 40 (持续治愈)125 / 150 / 175 / 200 (钢铁再生护甲加成） 持续时间:20% / 25% / 35% / 45% (濒死延缓)20 秒 (钢铁再生)2 / 3 / 3 / 4 秒 (回复效果持续时间) 范围:10 / 15 / 20 / 25 米

If Oberon or any allies affected by Renewal stands on or moves onto his , they will receive the Iron Renewal buff, which grants 125 / 150 / 175 / 200 bonus armor for as long as Renewal is active, which can be increased by 技能强度 to 598 bonus armor. Once Renewal is deactivated, the armor buff will remain on all affected units for 20 more seconds. Bonus armor is added after other armor increases. An Oberon with a maxed and under the effect of the Iron Renewal buff has armor equaling:

$\text{Total Armor}=\text{Base Armor}(1+\text{Mod Multiplier})+(\text{Iron Renewal}\times \text{Power Strength})$ $\text{Total Armor}=150(2.1)+(300\times1.3)=575$

Oberon's (4) can be modified with the ability augment（翻译提示：战甲强化MOD）, which increases armor by a fixed value of 250 at max rank. Because this is a fixed value and not a percentage, it provides a greater protection to frames with minimal armor.

 能量75快捷键3
 谐韵节拍给予随着 Octavia 的音乐做出相应动作的人各类增益效果。合拍的跳跃提供轻挑快板速度加成。跟上节拍的下蹲给予夜曲的隐匿增益。有节奏地射击可获得歌剧戏曲的多重射击加成。合拍的近战攻击给以强音奏鸣曲伤害加成。
 强度:10 / 15 / 20 / 35 % (护甲加成) 10 / 15 / 20 / 30 % (速度加成) 12 / 20 / 25 / 30 % (多重射击加成) 20 / 25 / 25 / 30 % (近战伤害加成) 持续时间:8 / 12 / 16 / 20 s (技能持续时间) 5 / 8 / 12 / 15 s (加成持续时间) 范围:6 / 8 / 10 / 12 米

Rhino's (1) can be modified with the ability augment（翻译提示：战甲强化MOD）, which increases his armor by 50%, or 142% with maximized Ability Strength, for each enemy hit. This bonus is calculated multiplicatively with other armor modifiers, such as Steel Fiber. A Rhino with a max rank Steel Fiber and unmodified Ability Strength that has charged through 5 enemies increases the armor according to the following equation:

$\text{Total Armor}=\text{Base Armor}(1+\text{Mod Multiplier})(1 + \text{Ironclad Modifier} \times \text{Enemies Struck})$ $\text{Total Armor}=190(1 + 1.1)(1 + 0.5 \times 5 )=1,397$

In the above example, using Ironclad Charge against 5 enemies increases Rhino's damage reduction from 57.08% at 399 armor to 82.32% at 1396.5 armor.

 能量75快捷键2
 咆哮Valkyr发出振奋的战吼来增强她伙伴的近战攻击速度并且减缓附近敌人的速度。
 强度:15% / 20% / 25% / 50% (加速效果)25% / 35% / 45% / 50% (护甲加成)15% / 20% / 25% / 30% (缓慢效果) 持续时间:7 / 10 / 12 / 15 秒 范围:15 / 20 / 22 / 25 米

Valkyr's (2) increases the armor of all friendly units within range by 50%, or up to 142% with maximized 技能强度. This bonus is calculated additively with other base armor multipliers, such as Steel Fiber. A Valkyr with a max rank Steel Fiber and unmodified Ability Strength increases her armor according to the following equation:

$\text{Total Armor}=\text{Base Armor}(1+\text{Mod Multiplier}+\text{Warcry Multiplier})$

$\text{Total Armor}=600(1+1.1+0.5)=1,560$

 能量50快捷键3
 死命顽抗Wukong 与他的分身进入无敌状态，同时嘲讽敌人，吸引火力。在此期间受到的所有伤害都会被积攒起来，而后 Wukong 会以一记狂怒的棍击将其释放。根据积攒的伤害量，Wukong 还会得到护甲加成。
 强度:4 / 5 / 6 / 7.5 x（伤害倍率）0.8 / 1 / 1.2 / 1.5 x （护甲倍率） 持续时间:2 / 2 / 2 / 2 s （无敌持续时间）15 / 18 / 20 / 25 s （护甲强化持续时间） 范围:7 / 8 / 10 / 12 m （长棍挥舞范围）
gains an armor bonus based on how much damage he absorbs. Minimum armor bonus is 50 while the maximum is 1,500.

## 受护甲影响的技能

 能量50快捷键2
 钢化皮肤Rhino硬化他的皮肤，隔绝一切伤害。
 强度:400 / 600 / 800 / 1200 (基础生命) 持续时间:1.5 / 2 / 3 / 3 秒 (无敌时间) 范围:N/A
• Base health and armor multiplier are affected by 技能强度.
• Iron Skin's health uses the following expression when accounting for Ability Strength:
Modified Health = (Base Health + (Armor Multiplier × Base Armor × (1 + Armor Bonus))) × (1 + Ability Strength) + Absorbed Damage.
• Note that the base armor value is different between Rhino and Rhino Prime; while Rhino has 190 Base Armor, Rhino Prime has 275 Base Armor.
• As an example for Rhino:
• With a maxed and , a Rank 3 Iron Skin cast by Rhino will have an initial health of (1200 + (2.5 × 190 × 2.1)) × 1.3 = 2,856.75 before absorbing damage.

 能量50快捷键3
 冰雪护罩冻结周围所有水汽和水分，制造出能用来防御的球形冰罩；冰罩生成时有短暂的无敌时间，此时敌人对冰罩造成的伤害能强化冰罩的耐久度；在相同位置重复释放冰罩可强化其耐久度。
 强度:3000/ 4000/ 4500/ 5000 (基础生命)5x (护甲倍率)40% / 50% / 60% / 67% (缓慢效果)? / ? / ? / 150 (爆炸伤害) 持续时间:1 / 2 / 3 / 4 秒 (无敌时间)? / ? / ? / 3 s (冻结时间) 范围:5 米(半径)
• Base health and armor multiplier are affected by 技能强度.
• Snow Globe's health uses the following expression when accounting for Ability Strength:
Modified Health = (Base Health + 5 × Frost's Base Armor × Base Armor Bonus) × (1 + Ability Strength) + Absorbed Damage.
• For example, with a maxed and , rank-3 Snow Globe will have an initial health of
(5000 + 5 × 300 × 2.1) × (1 + 0.3) = 10,595 before converting absorbed damage.

 能量75快捷键3
 混天火绫制造出守护火环，对靠近的敌人造成伤害并晕眩。
 强度:500 / 650 / 800 / 1000 （基础生命）50 / 75 / 100 / 125 （每秒伤害）1.15x / 1.25x / 1.75x / 2.50x （护甲倍率）2.5x （伤害吸收倍率） 持续时间:N/A 范围:1.25 / 1.5 / 1.75 / 2 米 其他:3 秒（发动前无敌时间）1 秒（失效后无敌时间）90%（伤害减免）
• Base health, armor multiplier, absorption multiplier and damage per second, are affected by 技能强度.
• Warding Halo's health uses the following expression when accounting for Ability Strength:
Modified Health = (Base Health + Armor Multiplier × Nezha's Base Armor × (1 + Base Armor Bonus)) × (1 + Ability Strength) + Absorbed Damage.(More testing and confirmation needed)
• As an example, with a maxed and , a rank-3 Warding Halo will have an initial health of (900 + 2.5 × 175 × 2.1) × 1.3 = 2,364.375 before absorbing damage.

 能量50快捷键2
 大地构筑召唤一堵石墙，再次使用可使巨石滚动压倒前方敌人。
 强度:5x（护甲倍率）2,850 / 3,150 / 3,450 / 3,750 (基础生命)150 / 300 / 450 / 600 (碰撞伤害)250 / 300 / 350 / 500 (爆炸伤害) 持续时间:N/A 范围:3 / 3 / 4 / 5 米 (爆炸范围) 其他:1 / 2 / 3 / 4 秒 (无敌时间)15 米 (滚动距离)30 (石化后滚动距离)
• The bulwark's health uses the following expression when accounting for 技能强度:
Modified Health = (Base Health + 5 × (Atlas' Base Armor × (1 + Armor Mods)) × (1 + Strength Mods) + Additive Armor) + Absorbed Damage.
• For example, with a maxed and , rank-3 Tectonics will produce a bulwark with an initial health of
(3750 + 5 × 450 × 2.1) × (1 + 0.3) = 11,017.5 before converting absorbed damage.
 能量100快捷键4
 漫步石者召唤两个元素石头近战打手。召唤动作将石化临近 Atlas 的敌人。当它们的生命结束时，石头人将碎成一堆用于治疗的碎石。
 强度:500 / 750 / 1000 / 1200 (生命)50 / 150 / 250 / 500 (护甲) 500 / 750 / 1000 / 1250 (爆炸伤害)800 / 1000 / 1400 / 2000 (近战伤害)200 / 250 / 350 / 500 (投石伤害) 持续时间:15 / 20 / 30 / 45 秒 范围:3 / 4 / 5 / 6 米 (石化半径)3 / 4 / 5 / 6 米 (爆炸范围)1 (速度倍率) 其他:2 (石者数量)10 秒 (石化时间)1 (每个石者掉落的碎石数)≤ 50 (碎石产生的生命/护甲值)
• Rumbler armor uses the following expression when accounting for 技能强度:
Modified Armor = Armor × (1 + Armor Mods + Ability Strength).
• For example, with a maxed and , rank-3 Rumblers will have a modified armor value of
500 × (1 + 1.1 + 0.3) = 1,200.

 能量75快捷键4
 广域玻璃化创造一个逐渐扩张的熔融玻璃环，进入环内的敌人将逐渐结晶化。当扩张完毕时，玻璃环将硬化并能阻挡外界火力。环可以从结晶化敌人的护盾和生命值中获取额外力量。使用「玻璃碎击」可以击碎环，并将尖锐的玻璃碎片向外击飞 能量消耗: 5 / 5 / 5 / 3 点/秒
 强度:1.2 / 1.3 / 1.4 / 1.5 (伤害倍率)1,125 / 1,375 / 1,625 / 2,225 (每一扇面玻璃的基础生命值) 持续时间:3 秒 (扩张时间)10 / 12 / 14 / 16 秒 (晶体化持续时间)10 / 14 / 18 / 22 秒 (圆环持续时间) 范围:2 米 (圆环初始半径)8 / 8.75 / 10.25 / 11 米 (圆环最大半径)5 / 6 / 7 / 8 米 (爆炸范围) 其他:3 米 (圆环初始高度)12 米 (圆环最大高度)3 秒 (晶化时间))12 (玻璃环的扇数)50% (吸收伤害)
• Barrier segment base health and segment explosion damage are affected by 技能强度, while crystallize converted health is affected by both Ability Strength and 护甲 mods such as equipped on Gara.
• Barrier segment health uses the following expression:
Modified Health = (Base Health + (5 × Base Armor × (1 + Armor Mods))) × (1 + Strength Mods)
• With a maxed and a wall segment would have
[2225 + (5 × 125 × (1 + 1.1))] × (1 + 0.3) = 4,598.75 Health before accounting for converted health from crystallized enemies (values resulting from intermediate calculations are rounded down to the nearest whole number).

## 战甲护甲和有效生命值

#### 增加生命与抵消伤害

• 更高的伤害抗性可以提升恢复定量的生命点数的治疗方式的效率。生命球、使用时的能量球、或的生命恢复效果都属于这种治疗方式。
• 更高的面板生命值可以提高效果将损失的生命值转化为能量的总量。

## 敌人护甲

$f_1(x) = 1 + 0.005 × ( x - 基础等级 )^{1.75}$ (Original scaling 更新 27.2之前)

$f_2(x) = 1 + 0.4 × ( x - 基础等级 )^{0.75}$

\begin{aligned}f_3(x) = min( 1,\frac{max( x, 60 + 基础等级 ) - ( 60 + 基础等级 )}{20})\end{aligned}

\begin{aligned}护甲系数 = 1 + (f_1(当前等级) - 1) × (1 - f_3(当前等级)) + (f_2(当前等级) - 1) × (f_3(当前等级))\end{aligned}

（译者注：令x为当前等级，用分段函数来表示该公式，即为：

$护甲系数= \begin{cases} f_1(x) & & {x - 基础等级 \leq 60},\\ 1 + (f_1(x) - 1) × (1 - f_3(x)) + (f_2(x) - 1) × (f_3(x)) & & {60 < x - 基础等级 < 80},\\ f_2(x) & & {x - 基础等级 \geq 80}. \end{cases}$

。）

\begin{aligned}当前等级的护甲 = 基础护甲 × 护甲系数\end{aligned}

\begin{aligned}\text{所受伤害} = \frac{300}{\text{基础护甲}\times\text{护甲系数} + 300}\end{aligned}

### 消除敌人护甲

#### 技能

This ability augment makes find weaknesses in the target's armor, temporarily reducing its armor values by 70% for 8 seconds at max rank. The armor reduction is multiplicative on the target's total armor. 技能强度 increases this armor reduction percentage; at 143%, a Seeking Shuriken removes all armor from a foe.

$\text{Net Armor}=\text{Total Armor}(1-0.70)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Overlapping successive casts will stack its armor reduction effect multiplicatively. Casting 2 Shurikens will reduce an enemy's total armor to 9% of its original value, and Corrosive Projection will remove 30% of an enemy's total armor before Shuriken's effect, leaving 21% of the target's armor still functioning.

This ability augment makes strain the targets' armor, temporarily reducing their armor values by 70% for 8 seconds.The armor reduction is multiplicative on the target's total armor. 技能强度 can increase this armor reduction percentage and this augment can strip all the armor off of an affected enemy if the Ability Strength is increased above 143%.

$\text{Net Armor}=\text{Total Armor}(1-0.70)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects (although with the ability to remove all the armor there is no need.) Overlapping successive casts will stack its armor reduction effect multiplicatively. Corrosive Projection will remove 30% of an enemy's total armor before Sonic Boom's effect, leaving 21% of the target's armor still functioning.

 能量100快捷键4
 雪崩唤来变化莫测的雪崩，瞬间冻结范围内的所有敌人；被雪崩冻结的敌人，其护甲会变得更为脆弱。
 强度:25 / 30 / 35 / 40% (护甲削减)800 / 1000 / 1200 / 1500 (初始伤害)100 / 200 / 300 / 400 (爆炸伤害) 持续时间:5 / 6 / 7 / 8 秒 范围:8 / 10 / 12 / 15 米 (冰冻半径)3 / 3.5 / 4 / 4.5 米 (爆炸半径)

This ability freezes enemies and reduces their armor by 40% for the duration of the effect. The armor reduction is multiplicative on the target's total armor. 技能强度 increases this armor reduction percentage; at 250% Ability Strength, Avalanche can remove all armor from affected enemies.

$\text{Net Armor}=\text{Total Armor}(1-0.40)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Overlapping successive casts do not stack its armor reduction effect, but do refresh its duration. Corrosive Projection will remove 30% of an enemy's total armor before Avalanche's effect, leaving 42% of the target's armor still functioning.

 能量25快捷键1
 大块朵颐将一个敌人囫囵吞入Grendel的腹内。没吃饱？那就接着吃吧。但请注意，Grendel肚子里的每个敌人都会缓慢消耗能量。按住技能键即可将腹内的敌人吐出，而这些敌人身上会覆盖着毒性消化液。 能量 / （敌人 * 秒） 1.5 / 1.5 / 1.5 / 1.5 吐出能量消耗: 0
 强度:2 / 3 / 4 / 5 % （每秒护甲移除）250 / 300 / 400 / 500 （吐出时的 伤害） 持续时间:N/A 范围:10 / 15 / 20 / 25 m （吞噬范围） 其他:1.5 s （大口张开时长）1 s （每个敌人的额外时长）100% (喷吐时触发率)8 m（锥形喷吐范围）
Enemies swallowed by this ability are stored in Grendel's stomach and lose 5% of their armor every second they remain swallowed, but each enemy in his stomach drains 1.5 points of Grendel's energy per second, ramping up the longer they remain ingested. 技能强度 increases armor reduction percentage.

 能量50快捷键3
 热能抽离在附近区域吸取动能，为电池蓄电并为附近的敌人带来 状态。长按按键来逆转这一过程，通过消耗的电池来使附近的敌人处于 伤害的状态中。
 强度:75 - 375 / 100-500 / 125 - 625 / 150 - 750 ( 或 伤害)150 - 750 / 200-1000 / 250 - 1250 / 300 - 1500 ( 伤害) 持续时间:4 / 6 / 8 / 10 秒 (持续时间)2.5 - 5 / 3-6 / 3.5 - 7 / 4 - 8 s (触发时间) 范围:6 / 8 / 10 / 12 米 (初始半径)? / ? / ? / ? m (最终半径) 其他:10%（单击施放获得电池充能）10%（长按施放消耗电池电量）100%（状态在期间冻结几率）100% (状态在期间爆燃几率)50% (伤害在期间护甲削减率)
This ability creates a field of when ability tapped or a field of when ability held. When used while is active and the battery gauge is above 80%, stacking the two fields creates a zone that instantly and permanently strips armor on cast, equal to the battery gauge up to 100% at full battery.

 能量0快捷键2
 掠夺掠夺邻近敌人一定比例的护盾与装甲，用以补充自身的护盾与过载护盾。使用技能消耗护盾：150
 强度:10 / 15 / 20 / 25 % (每个敌人 护盾/装甲) 持续时间:2 / 2 / 2 / 2 秒 (脉冲扩散时间) 范围:5 / 6 / 7 / 8 米 (脉冲基础距离) 其他:100% 清除负面状态
This ability releases a pulse that removes shields and armor by 25% to all enemies hit by the pulse. 技能强度 increases debuff percentage. If the affected enemy has both shield and armor, shield is removed first.

At maximum rank this ability augment causes the explosions of projectiles to apply a guaranteed Corrosive status effect.

While it may only remove 25% of an enemy's current armor with each hit, it can stack quickly, especially with multiple casts on one location, and this form of armor removal is permanent, unlike most of the other abilities/augments listed here.

The ability can also be used to apply the effect to large groups, at any location within line-of-sight, up to a considerable range.

This ability augment makes strain the targets' armor, temporarily reducing their armor values by 50% for 7 seconds. The armor reduction is multiplicative on the target's total armor. 技能强度 does not increase this armor reduction percentage.

$\text{Net Armor}=\text{Total Armor}(1-0.50)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Overlapping successive casts will stack its armor reduction effect multiplicatively. Corrosive Projection will remove 30% of an enemy's total armor before Crush's effect, leaving 35% of the target's armor still functioning.

 能量75快捷键3
 护盾极化释放一道能撤销敌人护盾与护甲的能量脉冲、制造出碎片，与磁吸立场搭配时将更为致命。碰触到此能量脉波的盟友将会回复护盾。
 强度:250 / 300 / 350 / 400（伤害）1x / 1.5x / 2x / 2.5x（伤害倍率） 持续时间:2 / 3 / 4 / 5 秒（脉冲扩散时间） 范围:5 / 6 / 7 / 8 米（爆炸半径）

 能量75快捷键2
 惊骇将恐惧感深植周遭敌人心中，使他们窜逃于恐惧之中。
 强度:7 / 12 / 15 / 20 (受影响的敌人数量)5% / 10% / 15% / 20% (护甲削减) 持续时间:10 / 15 / 20 / 25 秒 范围:5 / 10 / 12 / 15 米

This ability causes affected enemies to flee and reduces their armor by 20% for the duration of the effect. The armor reduction is multiplicative on the target's total armor. 技能强度 increases this armor reduction percentage; at maximized（翻译提示：最大化） strength, Terrify can remove 56.8% of armor from foes.

$\text{Net Armor}=\text{Total Armor}(1-0.20)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Overlapping successive casts will stack its armor reduction effect multiplicatively. Corrosive Projection will remove 30% of an enemy's total armor before Terrify's effect, leaving 56% of the target's armor still functioning.

 能量50快捷键2
 精神飞刃Nyx朝敌人投出数把精神飞刃，并使用精神力量来控制他们追击目标和附近的敌人。可命中一些敌人的防御。
 强度:20 / 40 / 60 / 80 %(削减量)N/A (范围伤害)3 / 4 / 5 / 6 (飞弹数量) 持续时间:5 / 7 / 9 / 11 秒 范围:60 米 (索敌范围)

This ability releases 6 telekinetic bolts that removes shield and armor by 80% from up to 6 enemies in a single cast. 技能强度 increases debuff percentage; at 125%, Psychic Bolts can remove all shield and armor.

Psychic Bolts will be active for its duration until all affected enemies die. It can be recasted at any time but doing so will cancel debuff on currently affected enemies.

 能量100快捷键4
 惩戒清算将附近的敌人迅速举到天空，再以坚定的信念将他们猛地摔向地面。遭此力量击杀的敌人有几率生成一颗生命球。
 强度:500 / 750 / 1000 / 1250 (伤害)100 / 225 / 400 / 625 (额外伤害)25% / 35% / 45% / 50% (生命球产生几率)30% (护甲削减) 持续时间:1 / 2 / 3 / 4 秒 (目盲持续时间) 范围:8 / 10 / 12 / 15 米 (施放半径)2.5 / 3 / 3.5 / 4 米 (致盲半径)

This ability knockdowns affected enemies and if used along with , will permanently reduce their armor by 30%. The armor reduction is multiplicative on the target's total armor. 技能强度 increases this armor reduction percentage; at Maximized Ability Strength, Reckoning can remove 111.9% armor from affected enemies.

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Overlapping successive casts will stack its armor reduction effect multiplicatively. Corrosive Projection will remove 30% of an enemy's total armor before Reckoning's effect, leaving 49% of the target's armor still functioning.

This ability augment makes drain the targets' armor, temporarily reducing their armor values by 45% for the ability's duration. The armor reduction is multiplicative on the target's total armor. 技能强度 increases this armor reduction percentage; at 223%, Abating Link can remove all armor from a foe.

$\text{Net Armor}=\text{Total Armor}(1-0.45)$

As a multiplicative modifier, it does not stack additively with other armor reduction effects. Corrosive Projection will remove 30% of an enemy's total armor before Link's effect, leaving 38.5% of the target's armor still functioning.

 能量100快捷键4
 力场牢狱建立一个包围力场来捕获敌人，使他们陷入停滞状态并剥离其护甲。长按技能键来使所有力场牢狱坍塌成为一个破坏性的漩涡。 坍缩立场所需能量： 0
 强度:15 （力场产生的伤害）6 / 8 / 10 / 12 （最大滞空敌人数量）2.5 / 5 / 7.5 / 10 % （每秒剥离护甲）50 (漩涡中每秒伤害) 持续时间:8 / 10 / 12 / 15 秒 （力场持续时间）10 / 12 / 13 / 15 s （漩涡持续时间）3 s （自动坍缩为漩涡持续时间） 范围:5 / 7 / 8 / 10 米 （捕获范围） 其他:?% （ 触发几率）∞ （力场内护甲加成持续时间）10 s （力场外护甲加成持续时间）1,000 （护甲加成最大值）?% （触发几率）
Enemies suspended by will be stripped of their armor at a rate of 10% per second.

#### 专精

 撕裂冲刺使用虚空冲刺穿过敌人，降低其10% / 20% / 30% / 40% / 50% / 75%的护甲。 解锁花费：80,000升级花费：120,000 / 162,426 / 240,000 / 340,454 / 459,411容量占用： 1 / 4 / 7 / 10 / 13 / 16

## 彻底消除护甲

• 注意，上述的多种百分比削甲效果之间可以累乘。
• 假设一个敌人带有10层腐蚀异常和1层火焰异常，则$(1-0.8)*(1-0.5)=0.1$，即敌人的护甲将会减少至原来的10%。
• 显然，当一种百分比削甲效果的数值大于等于100%，必定能完全移除敌人的护甲。
• 一般来说的，异常状态的削甲效果仅在持续时间内生效，技能的削甲效果也仅在技能持续时间内生效，腐蚀投射的削甲效果仅在玩家存活时生效。

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Tenno的装甲类型是铁质装甲

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# What do computers see?

Tricks to visualize and understand how neural networks see.

August 15, 2018
Eye (source: Pixabay)

In the last decade or so, computer vision has made tremendous strides forward, mostly because of developments in deep learning. It is not just that we have new and better algorithms—digital cameras and the web have provided us with a near infinite set of training data. And maybe even more importantly, graphics cards developed for computer gaming turn out to have super computer powers when it comes to training deep neural networks.

This is all good news for anybody wanting to experiment with deep learning and image recognition. All it takes to build a cat versus dog classifier these days is Keras and a Python notebook with 100 lines of code. But doing this doesn’t tell us what computers see.

## Learn faster. Dig deeper. See farther.

Join the O'Reilly online learning platform. Get a free trial today and find answers on the fly, or master something new and useful.

If we wanted to understand how humans see, we could open their skulls and try to figure out how information flows from the eye’s photoreceptor cells through the visual cortex to the rest of the brain. This would be rather hard, though. It’s much easier to poke that opened-up brain with an electrode until the subject sees the color blue. So, how do we prod a neural net with an electrode?

Modern networks typically consist of a large number of layers stacked on top of each other. The image to be recognized gets fed into the lowest layer, and as the information travels through the network, the image representation becomes more abstract until at the other end, a label comes out and the network says, I see a cat!

Poking a neural network with an electrode boils down to running this process in reverse; rather than showing the network a picture and asking it what it sees, we’re going to give the network some noise and ask it to make changes such that a particular neuron has a maximum activation. The image that does that represents what this particular neuron sees, what a human would report seeing if we prodded that neuron.

model = vgg16.VGG16(weights='imagenet', include_top=False)
model.summary()

That last statement shows us the structure of the network.

We define a loss function that optimizes the output of the given neuron, then create an iterate Keras function that optimizes toward that by changing the input image. We then start with an image filled with noise and run the iteration 16 times. (All code referred to in this post is available on GitHub as both a standalone script and a Python notebook. See the references at the end of the article.)

loss = K.mean(layer_output[:, :, :, neuron_idx])
img_data = np.random.uniform(size=(1, 256, 256, 3)) + 128.
for i in range(16):

We can try this on some random neuron and we see some artifacts appear that seem to tickle this specific neuron:

That’s cute! Not quite what I’d imagine brain-surgery-induced hallucinations look like, but it is a start! This neuron was picked from a random layer. Remember as we go through the stacked layers, the abstraction level is supposed to go up. What we can do now is sample neurons from different layers and order them by layer; that way we get an idea of the sort of features that each layer is looking out for:

This aligns nicely with our intuition that abstraction goes up as we traverse the layers. The lowest layers are looking for colors and simple patterns; if you go higher, the patterns become more complex.

Unfortunately, there’s a limit to how far this trick will get us; if you find the bit in the highest layer that represents cat-ness, you can optimize all you want, but no obvious cats will roll out.

We can get more detail, though, by zooming in on the image as we run the optimization function. The idea here is if you optimize an image for neuron activation, it tends to get stuck in a local minimum since there is no good way for it to influence the overall structure of the image. So instead of starting with a full image, we start with a small 64×64 image that we optimize. We then scale the image up a bit and optimize again. Doing this for 20 steps gives us a nice and full result that has a certain plasticity to it.

The zooming itself is somehow pleasing, like something organic unfolding.

We can make this into a movie of arbitrary length if we keep zooming, but once we reach a certain size, we also need to start center cropping so the image always remains the same size. This movie has a nice fractal-like mesmerizing effect. But why stop there? We can loop through a set of random neurons while zooming, making for a wonderful acid-like movie:

In this post, we’ve seen some tricks to visualize what a neural network sees. The visualizations are interesting and somehow compelling by themselves, but they also give us an understanding into how computer vision works. As information flows through, the level of abstraction increases, and we can to some extent show those abstractions.

The notebook version of the code can be found on GitHub. A script-based version can be found on GitHub as well.

Post topics: Artificial Intelligence
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Students can Download 2nd PUC Electronics Previous Year Question Paper June 2018, Karnataka 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Electronics Previous Year Question Paper June 2018

### 2nd PUC Electronics Previous Year Question Paper June 2018

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
Expand JFET.
Junction Field Effect Transistor.

Question 2.
Define CMRR.
CMRR is the ratio of differential mode gain to common mode gain.

Question 3.
How many frequency components are present in amplitude modulated wave?
3

Question 4.
Mention any one application of digital communication system.
TV transmission or mobile signal transmission.

Question 5.
Draw the circuit symbol of TRIAC.

Question 6.
Write Excess-3 code of (26)10.
(26)10 = (010110O1)2

Question 7.
It is circuit that performs arithmetic sum of 3 binary bits.

Question 8.
Give the meaning of CLR A.
Clear the content of Accumulator.

Question 9.
Write the C-equivalent expression for the mathematical expression
(X * Y) / Z

Question 10.
What is Transponder?
Transponder is transmitter receiver combination in a satellite.

Part – B

Answer any five questions.  5 x 2 = 10

Question 11.
Define amplification factor. Write its relation with rd and gm.
Amplification factor is the ratio of small change in drain to source voltage to the corresponding change in Gate to source voltage for a constant drain current.
ICBO and ICEO

Question 12.
Write the steps involved in drawing AC equivalent circuit of an amplifier.

• Reduce all dc sources to zero.
• Short all capacitors.

Question 13.
The open loop gain and closed loop gain of an amplifier are 100 and 50 respectively. Calculate the feedback fraction.
Aβ = 100
Af = 50
$$\mathrm{Af}=\frac{\mathrm{A}}{1+\mathrm{A} \beta}$$
β = 0.01

Question 14.
Write any two advantages of negative feedback.

• Stabilisation of Gain
• Increase in the bandwidth

Question 15.
Draw the neat circuit diagram of Wein-bridge oscillator.

Voltage gain A = $$1+\frac{\mathrm{R}_{3}}{\mathrm{R}_{4}}=3$$

Question 16.
Explain double-injection meahanism in power diode.
At higher forward bias, the holes from p+ region reach n~ n+junction and attract electrons from n+ cathode. This leads to electron injection into n~ drift region from n+ cathode layer. This mechanism is called double injection method.

Question 17.
Realize XOR gate using only NAND gates.

Question 18.
Mention any two applications of bluetooth.

• Wireless communication between PC or laptops or cell phones.
• Used in wireless audio head sets.

Part – C

Answer any five questions.  5 x 3 = 15

Question 19.
What is meant by leakage current? Name two types of leakage currents.
The flow of current through the device due to the motion of minority charge carriers under reverse bias condition is called leakage current.
ICBO and ICEO

Question 20.
What is Ionosphere? Explain F-layer.
Ultraviolet radiation from the sun cause the upper atmosphere to ionize i.e. to become electrically charged. Hence a thick layer of ions formed at heights of 50 km to 400km. The ionosphere supports MF and HF wave propagation.
Its three layers are:

D layer: It is at a height of 70 km and has an average thickness of 10 km. It disappears in the night time. It reflects VLF and LF waves and absorbs MF and HF waves.

E layer: It is at a height of 100 km with a thickness of 25 km. This layer disappears at night. It helps in MF and HF wave propagation.

F layer: It is at a height of 150 km and extends upto 400 km. This layer exists during both day and night time. During day time, it splits into two layers F, and F2, and combines to form a single F layer during night time.

 F2 region. 250-400 km. F1 region. 160-250 km. E region. 95-130 km D region. 50-95 km. Troposphere Earth.

Question 21.
With a neat circuit diagram explain linear diode detector.

LC tank circuit is used as a parallel resonant circuit. The resonant frequency of the circuit can be varied by varying the value of capacitance C and hence RF signal of any desired frequency can be tuned in.

When a selected modulated signal is applied to the diode, the diode conducts only during the positive half of the modulated wave. Thus the diode removes the entire negative half cycles. Across R1C1 only positive half cycles of carrier modulated wave appears. Thus the low pass filter made up of R1 C1 removes the RF.

During positive half cycle, diode conducts the capacitor C1 gets charged. During negative half cycle of the carrier and diode doesnot conduct but C1 discharges through R1. Hence output is obtained. The spikes can be reduced by proper choice of R1 C1 and depth of modulation. The capacitor CB removes Dc component produced by the detector.

Question 22.
Draw the circuit diagram, gate pluses and load voltage waveform of power inverter.

Question 23.
Determine Vdc and Idc of SCR full wave frectifier. Given firing angle is 60°. The rms voltage of ac input to the rectifier is 150V, when load resistance of 15Ω is connected.

Question 24.
Name the addressing modes of the following instructions:
(i) MOV R2, A
(ii) MOV A, 25H
(iii) MOV A, @ R0.

Question 25.
What is Debugging in C-program? Write the syntax of the if else statement.
Debugging is detecting and correcting errors.
if (condition)
{
statement – 1 ;
}
else
{
statement -2;
}

Question 26.
Draw the block diagram of RADAR system and mention the function of duplexer.

• Transmitter: The radar transmitter produces high-frequency RF pulses and is transmitted into space by the antenna.
• Duplexer: It switches the antenna between transmitter and receiver alternately. This is required because high power pulses of the transmitter will destroy the receiver if energy were allowed to enter the receiver.
• Receiver: It amplifies and demodulates the received RF signals. It provides video signals on the output.
• Radar antenna: The antenna transfers the transmitting signal to space with the required distribution and efficiency.

Part – D

Answer any Three questions:  3 x 5 = 15

Question 27.
For the CE amplifier circuit shown below, find:
(i) Voltage drop across 8.2 Ω
(ii) IZ
(iii) re
(iv) Zin(base)
(v) Z out

Question 28.
Find the output voltage ‘V0‘ for the following circuit: Given: V1, = IV, V2 = 2V, V3 = -1 V

Question 29.
Calculate the frequency and feedback fraction of the circuit shown below:

Question 30.
A 100 KW carrier power is amplitude modulated at 50% depth of modulation by a sinusoidal modulating signal. Calculate:
(i) Total power
(ii) Sideband power
(iii) Power in each sideband.

Question 31.
Simplify the following Boolean expression by using K-MAP. If(A, B, C, D) = Σm(0, 1, 2, 4, 5, 8, 9, 10, 12, 13) Draw the logic circuit for the simplified expression using only NAND gates.

Part – E

Answer any Four questions.   4 x 5 = 20

Question 32.
Explain with circuit diagram the working of class-B Push-Pull power amplifier.

Class B push-pull amplifier is as shown in the diagram. T1 and T2 are two centre tapped transformers and Q1, Q2 are two identical transistors. Transformer T1 produces signal voltages V1 and V2 which are 180° out of phase with each other.

These two signals are applied to the Polarities reverse during negative half cycle of input voltage. Q2then is ON and Q1 is OFF. Q2amplifies the signal and the alternate half cycle appears across the loud speaker.

two transistors. Transformer T2 couples AC output signal from collector to loud speaker. The two emitters are connected to centre tap of transformer T1 secondary and Vcc to the centre tap of T2 secondary.

During positive half cycle of input voltage, secondary winding of T1 has voltages V, and V2. Transistor Q1, conducts and Q2 is cut off. The collector current through Q1 produces an amplified and inverted voltage which applied to loud speaker through a transformer.

Question 33.
What is a differentiator? With the help of circuit diagram obtain an expression for
output voltage of OP-AMP differentiator.

Question 34.
Along with waveforms, derive an expression for instantaneous voltage of frequency modulated wave.
Frequency modulation is the process of varying the frequency of high frequency carrier wave in accordance with instantaneous amplitude of the modulating signal.
Let the modulating signal be given by vm=vm coswt.
Where w is the angular frequency and vm is the peak amplitude. Let the carrier voltage be given by the equation vc= vcsin (Wct +Ø) where wc is the angular frequency of the carrier and Vc is the peak amplitude of the carrier phase angle θ = Wct + θ.
This is instantaneous phase angle of carrier voltage.
Then carrier voltage Vc = Vc sin Ø
The angular frequency $$\omega _{ c }=\frac { d\emptyset }{ dt }$$
or Ø = ∫ ωdt
The angular frequency of carrier after modulation is
ω = ω + kvm
ω = ω+ kVmcos ωt.
Where K is a constant of proportionality called as the modulation deviation constant. On integrating,

Question 35.
Explain the working of four bit synchronous up counter with the help of logic diagram. Write its truth table.

Counter is a logic circuit used for counting the pulses. It is a set of flip flops whose state changes in response to pulses applied to their inputs.

Counters are of two types:

• Synchronous counters and
• Asynchronous counters.

In a synchronous circuit, the clock signal is applied to all flip flops simultaneously.

Question 36.
Write an assembly language program to multiply 08 H and OB H. What are the contents of register A and register B after the execution of the program?
MOV A, # 08H → Load the number 08H into the accumulator.
MOV B, # OBH → Load the number OBH into register B.
MOV AB → 08H  OBH
08H= 1000(2)              0BH=1011(2)
1000 x 1011 =01011000(2) = 58H
∴ A = 58H    B = OOH

Question 32.
Write a C-program to find the sum of first n
# include <stdio.h>
# include < conico.h>
main ( )
{
int n, i, sum = 0;
printf (” Enter the value of n”);
scanf(” % d”, & n);
i= i;
while (i< = n)
{
sum = sum + i; i + +;
}
printf(“The sum = %d”, sum); getch  ( );
}


text:
Can this expression involving pseudoinverse be simplified?

I have this expression embedded in a larger equation:

$$A^T (A D A^T)^+ A$$

Where $A \in \mathbf{R}^{m \times n}$ is an arbitrary matrix, $D \in \mathbf{R}^{n \times n}$ is diagonal (with some elements possibly $0$), and $X^+$ is the Moore-Penrose Pseudoinverse of $X$.

Can the expression be simplified? In particular, I'd like to "distribute" the pseudoinverse in terms of the pseudoinverse of $A$ and $D$, if that's possible.

If $D = I$, for instance, it's easy to show that $A^T (A D A^T)^+ A = V V^T$, where $A = U \Sigma V^T$, $V^T \in \mathbf{R}^{r \times n}$ is the "thin" SVD of $A$ with rank $r$.

Well obviously if you replace $B = A \sqrt{D}$ you get

$A^T (ADA^T)^+ A^T = D^{-1/2} W W^T D^{-1/2}$ with $W$ from the thin svd from of $B$ (in place of the $V$ in your question).

That is the only thing I would see. Of course that does not strictly work for you, since you have possibly zero elements in $D$.

• Ah, I hadn't thought about turning the problem inside out like that! So simple in hindsight. If there are zero terms in $D$, is there a way to show that the form above still holds or not? Since they're going to be zeroed out in $B$ anyway when you multiply, I'm wondering if it doesn't matter. – Jay Lemmon Jan 25 '14 at 5:53
• Good question, I dont know. I would say it depends what the limit of the $\infty \cdot 0$ expression inside of $D^{-1/2} W$ would be. If it is always zero, then you are right. If it is some number, than we are out of luck. But before starting an mathematical investigation, you could just try it out numerically. – Andreas H. Jan 25 '14 at 17:54
• Matlab experiments seem to indicate you can't just do something like sqrt(pinv(D)) when D has a 0 element :( Still, thanks for the help. – Jay Lemmon Jan 26 '14 at 22:29
• too bad, thanks for the notice – Andreas H. Jan 26 '14 at 23:29


text:
# MD5 pre-fixing with an unkown postfix

Is the MD5 prefixing attack still valuable to an attacker if he doesn't know the postfix of the string?

For instance, what if the following sanity check was being used:

if attacker_controlled_sanity_code == md5(attacker_controlled + secret_key)

Clearly not an RFC compliant HMAC. Despite the fact that the attacker doesn't know the entire md5'ed message mean that the prefix attack in md5 is useless? Is there another problem with using a broken hash function in this way?

• What's the objective of the construct; message integrity, like HMAC? If so, the usual setup is not a cryptographic_nonce, suggesting non-reuse by legitimate parties, but not secrecy. Rather, usual setup is a secret_key, assumed reused and shared. – fgrieu Aug 13 '12 at 11:09
• @fgrieu yes, you are correct. – Rook Aug 13 '12 at 14:45

If the attacker does not know the value of $secretkey$, he cannot predict in advance the value of $MD5( attackercontrolled + secretkey)$. However, what he can to is create two different messages $M$, $M'$ such that:
$MD5( M + secretkey) = MD5( M' + secretkey)$
He cannot predict this common value in advance; however, if he is able to submit $M$ and observe the generated tag, he immediately knows the tag for $M'$


text:
Expectation Maximization Machine Learning Tools¶

This package is part of the signal-processing and machine learning toolbox Bob. It contains routines for learning probabilistic models via Expectation Maximization (EM).

The EM algorithm is an iterative method that estimates parameters for statistical models, where the model depends on unobserved latent variables. The EM iteration alternates between performing an expectation (E) step, which creates a function for the expectation of the log-likelihood evaluated using the current estimate for the parameters, and a maximization (M) step, which computes parameters maximizing the expected log-likelihood found on the E step. These parameter-estimates are then used to determine the distribution of the latent variables in the next E step.

The package includes the machine definition per se and a selection of different trainers for specialized purposes:

• Maximum Likelihood (ML)
• Maximum a Posteriori (MAP)
• K-Means
• Inter Session Variability Modelling (ISV)
• Joint Factor Analysis (JFA)
• Total Variability Modeling (iVectors)
• Probabilistic Linear Discriminant Analysis (PLDA)
• EM Principal Component Analysis (EM-PCA)

Installation¶

Complete Bob’s installation instructions. Then, to install this package, run:

\$ conda install bob.learn.em

Contact¶

For questions or reporting issues to this software package, contact our development mailing list.


text:
# Proof of an admissible Airy stress function

## Homework Statement

Show that a stress function of the form $\phi(r,\theta) = f(r)\cos(2\theta)$, where f(r) is a function of r only, satisfies the biharmonic equation and hence an admissible Airy stress function.

## Homework Equations

$g \equiv \nabla^2 \phi = \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta}$

$\nabla^4 \phi = \nabla^2 g = \frac{\partial^2 g}{\partial r^2} + \frac{1}{r}\frac{\partial g}{\partial r} + \frac{1}{r^2}\frac{\partial^2 g}{\partial \theta}$

## The Attempt at a Solution

Evaluate the harmonic equation:
$g = \frac{\partial^2f(r)}{\partial r^2}\cos(2\theta) + \frac{1}{r}\frac{\partial f(r)}{\partial r}\cos(2\theta) -\frac{4}{r^2}f(r)\cos(2\theta)$

Evaluate the biharmonic equation:
I will split the equation into three parts to better show the different derivatives

$\frac{\partial^2 g}{\partial r^2} = \left[\frac{\partial^4 f(r)}{\partial r^4} + \frac{\partial}{\partial r}\left(-\frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r}\frac{\partial^2f(r)}{\partial r^2} + \frac{8}{r^3}f(r) - \frac{4}{r^2}\frac{\partial f(r)}{\partial r}\right)\right]\cos(2\theta)$

$=\left[\frac{\partial^4 f(r)}{\partial r^4} + \frac{1}{r^2}\frac{\partial f(r)}{\partial r} - \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} - \frac{1}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{1}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{24}{r^4}f(r) + \frac{8}{r^3}\frac{\partial f(r)}{\partial r} + \frac{8}{r^3}\frac{\partial f(r)}{\partial r} - \frac{4}{r^2}\frac{^2 f(r)}{\partial r^2}\right]\cos(2\theta)$

$\frac{1}{r}\frac{\partial g}{\partial r} = \left[\frac{1}{r}\left(\frac{\partial^3f(r)}{\partial r^3} - \frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^3}f(r) - \frac{4}{r^2}\frac{\partial f(r)}{\partial r}\right)\right]\cos(2\theta)$

$=\left[\frac{1}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^4}f(r) - \frac{4}{r^3}\frac{\partial f(r)}{\partial r}\right]\cos(2\theta)$

$\frac{1}{r^2}\frac{\partial^2g}{\partial \theta^2} = \left[\frac{1}{r^2}\left(-4\frac{\partial^2f(r)}{\partial r^2} - \frac{4}{r}\frac{\partial f(r)}{\partial r} + \frac{16}{r^2}f(r)\right)\right]\cos(2\theta)$

$= \left[\frac{-4}{r^2}\frac{\partial^2f(r)}{\partial r^2} - \frac{4}{r^3}\frac{\partial f(r)}{\partial r} + \frac{16}{r^4}f(r)\right]\cos(2\theta)$

Combining these terms:

$\nabla^4 \phi = \left[\frac{\partial^4 f(r)}{\partial r^4} - \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} + \frac{2}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{8}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^3}\frac{\partial f(r)}{\partial r}\right]\cos(2\theta)$

The final form of the biharmonic equation above does not equal 0 (and for this problem it is supposed to). Is there anywhere I went wrong?

Related Calculus and Beyond Homework Help News on Phys.org
pasmith
Homework Helper

## Homework Statement

Show that a stress function of the form $\phi(r,\theta) = f(r)\cos(2\theta)$, where f(r) is a function of r only, satisfies the biharmonic equation and hence an admissible Airy stress function.

[snip]

The final form of the biharmonic equation above does not equal 0
Why would you expect it to for arbitrary $f$? Clearly you must impose some additional condition.

Since $\phi$ is separable we have
$$\nabla^2 \phi = \nabla^2(f(r) \cos 2\theta) = \left(\frac{1}r\frac{\partial}{\partial r} \left(r\frac{\partial f}{\partial r}\right) - 4\frac{f}{r^2}\right)\cos 2\theta = D_r^2(f)\cos 2 \theta$$
which is separable. Hence
$$\nabla^4 \phi = \nabla^2(\nabla^2 \phi) = \nabla^2(D_r^2(f) \cos 2\theta) = D_r^2(D_r^2 f) \cos 2\theta = D_r^4 (f) \cos 2\theta$$
which again is separable, and it is this separability which allows us to impose a condition on $f(r)$ so that $\nabla^4 \phi = 0$ for all $r$ and $\theta$: we must have $D_r^4 f = 0$.

(I suspect your expression for $D_r^4(f)$ is wrong, because it is dimensionally inconsistent: I expect terms of the form $\frac 1{r^n} \frac{\partial^m}{\partial r^m}$ where $n + m = 4$.)

Last edited:
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text:
# Simulating a simulation study In rsimsum: Analysis of Simulation Studies Including Monte Carlo Error

options(width = 150)
knitr::opts_chunkset( collapse = TRUE, comment = "#>", fig.align = "center", fig.height = 6, fig.width = 6, out.width = "66.66%" )  # Introduction In this vignette, we show how the simulated data included as an example dataset in simsum has been generated. # Motivation Say we want to run a simulation study in which we want to compare the sensitivity of parametric and semiparametric survival models on relative risk estimates. # Data generating mechanisms We simulate an hypotetical trial with a binary treatment. We fix the log-treatment effect to-0.50$, and we generate a treatment indicator variable for each simulated individual via a$Binom(1, 0.5)$random variable. We simulate two different sample sizes (50 and 250 individuals) and we assume two different baseline hazard functions: exponential with scale parameter$\lambda = 0.5$, and Weibull with scale parameter$\lambda = 0.5$and shape parameter$\gamma = 1.5$. Finally, we apply administrative censoring at time$t = 5$. exp_basehaz <- function(t, lambda = 0.5) lambda * 1 * t^0 exp_weibull <- function(t, lambda = 0.5, gamma = 1.5) lambda * gamma * t^(gamma - 1) curve(exp_basehaz, from = 0, to = 5, lty = 1, ylim = c(0, 2), ylab = expression(h[0](t)), xlab = "Follow-up time t") curve(exp_weibull, from = 0, to = 5, lty = 2, add = TRUE) legend(x = "topleft", lty = 1:2, legend = c("Exponential baseline hazard", "Weibull baseline hazard"), bty = "n")  The survival times are estimated using the approach of Bender et al. (2005), based on drawing from a$U(0, 1)$random variable and applying the following transformations: 1. for an exponential baseline hazard, the survival time$t$is simulated as: $$t = -\frac{log(U)}{\lambda \exp(\beta ^ T X)}$$ 2. for a Weibull baseline hazard, the survival time$t$is simulated as: $$t = \left(-\frac{log(U)}{\lambda \exp(\beta ^ T X)}\right) ^ {1 / \gamma}$$ The R function to simulate a dataset for our simulation study is defined as follows: simulate_data <- function(dataset, n, baseline, params = list(), coveff = -0.50) { # Simulate treatment indicator variable x <- rbinom(n = n, size = 1, prob = 0.5) # Draw from a U(0,1) random variable u <- runif(n) # Simulate survival times depending on the baseline hazard if (baseline == "Exponential") { t <- -log(u) / (params$lambda * exp(x * coveff))
} else {
t <- (-log(u) / (params$lambda * exp(x * coveff)))^(1 / params$gamma)
}
# Winsorising tiny values for t (smaller than one day on a yearly-scale, e.g. 1 / 365.242), and adding a tiny amount of white noise not to have too many concurrent values
t <- ifelse(t < 1 / 365.242, 1 / 365.242, t)
t[t == 1 / 365.242] <- t[t == 1 / 365.242] + rnorm(length(t[t == 1 / 365.242]), mean = 0, sd = 1e-4)
# ...and make sure that the resulting value is positive
t <- abs(t)

# Make event indicator variable applying administrative censoring at t = 5
d <- as.numeric(t < 5)
t <- pmin(t, 5)
# Return a data.frame
data.frame(dataset = dataset, x = x, t = t, d = d, n = n, baseline = baseline, stringsAsFactors = FALSE)
}

# Methods

We compare the Cox model (Cox, 1972) with a fully parametric survival model assuming an exponential baseline hazard and a flexible parametric model with 2 degrees of freedom for modelling the baseline hazard (Royston and Parmar, 2002). The Cox model can be fit via the coxph function from the survival package, the exponential model can be fit via the phreg function from the eha package, and the Royston-Parmar model can be fixed via the stpm2 function from the rstpm2 package.

# Performance measures

Say we are interested in the following performance measures:

• Bias in the estimated log-treatment effect, and corresponding $95\%$ Monte Carlo confidence intervals
• Coverage of confidence intervals for the log-treatment effect, defined as the proportion of simulated data sets for which the true log-treatment effect of $-0.50$ lies within the $95\%$ confidence intervals obtained from the model

# Sample size

We are primarily interested in bias, and assume that the variance of the estimated log-treatment effect is $0.1$. The Monte Carlo standard error for the bias is:

$$\text{MCSE} = \sqrt{\frac{\text{variance}}{# \text{simulations}}}$$

Aiming for a Monte Carlo standard error of 0.01 on the estimated bias, we would require $1,000$ replications.

The Monte Carlo standard error for converage is:

$$\text{MCSE} = \sqrt{\frac{\text{coverage} \times (1 - \text{coverage})}{# \text{simulations}}}$$

This Monte Carlo standard error is maximised for a coverage = $0.5$. In that setting, the Monte Carlo standard error with $1,000$ replications would be $0.01581139$, which is deemed to be acceptable.

Therefore, we will run $1,000$ replications of this simulation study.

# Running the simulation study

## Generate data

We generate $1,000$ datasets for each data-generating mechanism.

First, we set a random seed for reproducibility:

set.seed(755353002)

Then, we simulate the data:

data <- list()
data[["n = 50, baseline = Exp"]] <- lapply(
X = 1:1000,
FUN = simulate_data,
n = 50,
baseline = "Exponential",
params = list(lambda = 0.5)
)
data[["n = 250, baseline = Exp"]] <- lapply(
X = 1:1000,
FUN = simulate_data,
n = 250,
baseline = "Exponential",
params = list(lambda = 0.5)
)
data[["n = 50, baseline = Wei"]] <- lapply(
X = 1:1000,
FUN = simulate_data,
n = 50,
baseline = "Weibull",
params = list(lambda = 0.5, gamma = 1.5)
)
data[["n = 250, baseline = Wei"]] <- lapply(
X = 1:1000,
FUN = simulate_data,
n = 250,
baseline = "Weibull",
params = list(lambda = 0.5, gamma = 1.5)
)

## Run models

We define a function to fit the models of interest:

library(survival)
library(rstpm2)
library(eha)

fit_models <- function(data, model) {
# Fit model
if (model == "Cox") {
fit <- survival::coxph(Surv(t, d) ~ x, data = data)
} else if (model == "RP(2)") {
fit <- rstpm2::stpm2(Surv(t, d) ~ x, data = data, df = 2)
} else {
fit <- eha::phreg(Surv(t, d) ~ x, data = data, dist = "weibull", shape = 1)
}
# Return relevant coefficients
data.frame(
dataset = unique(data$dataset), n = unique(data$n),
baseline = unique(data$baseline), theta = coef(fit)["x"], se = sqrt(ifelse(model == "Exp", fit$var["x", "x"], vcov(fit)["x", "x"])),
model = model,
stringsAsFactors = FALSE,
row.names = NULL
)
}

We now run the models for each simulated dataset:

results <- list()
results[["n = 50, baseline = Exp, model = Cox"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Exp"]],
FUN = fit_models,
model = "Cox"
)
)
results[["n = 250, baseline = Exp, model = Cox"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Exp"]],
FUN = fit_models,
model = "Cox"
)
)
results[["n = 50, baseline = Wei, model = Cox"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Wei"]],
FUN = fit_models,
model = "Cox"
)
)
results[["n = 250, baseline = Wei, model = Cox"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Wei"]],
FUN = fit_models,
model = "Cox"
)
)

results[["n = 50, baseline = Exp, model = Exp"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Exp"]],
FUN = fit_models,
model = "Exp"
)
)
results[["n = 250, baseline = Exp, model = Exp"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Exp"]],
FUN = fit_models,
model = "Exp"
)
)
results[["n = 50, baseline = Wei, model = Exp"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Wei"]],
FUN = fit_models,
model = "Exp"
)
)
results[["n = 250, baseline = Wei, model = Exp"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Wei"]],
FUN = fit_models,
model = "Exp"
)
)

results[["n = 50, baseline = Exp, model = RP(2)"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Exp"]],
FUN = fit_models,
model = "RP(2)"
)
)
results[["n = 250, baseline = Exp, model = RP(2)"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Exp"]],
FUN = fit_models,
model = "RP(2)"
)
)
results[["n = 50, baseline = Wei, model = RP(2)"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 50, baseline = Wei"]],
FUN = fit_models,
model = "RP(2)"
)
)
results[["n = 250, baseline = Wei, model = RP(2)"]] <- do.call(
rbind.data.frame,
lapply(
X = data[["n = 250, baseline = Wei"]],
FUN = fit_models,
model = "RP(2)"
)
)

## Aggregating results

relhaz <- do.call(
rbind.data.frame,
results
)
row.names(relhaz) <- NULL

We save the final results, that will be included as an example in the R package rsimsum.

library(devtools)
devtools::use_data(relhaz, overwrite = TRUE)

## Summarising results

Finally, we obtain summary statistics by calling the simsum function:

library(rsimsum)
s <- rsimsum::simsum(data = relhaz, estvarname = "theta", se = "se", true = -0.50, methodvar = "model", ref = "Cox", mcse = TRUE, by = c("n", "baseline"))
s

summary(s)

# Conclusions

With this vignette we showed how to simulate survival data and run a small, simple simulation study.

# References

• Cox D.R. Regression models and life-tables. Journal of the Royal Statistical Society, Series B (Methodological), 1972, 34(2):187-220

• Royston P. and Parmar M.K. Flexible parametric proportional-hazards and proportional-odds models for censored survival data, with application to prognostic modelling and estimation of treatment effects. Statistics in Medicine, 2002, 21(15):2175-2197

• Bender R., Augustin T., and Blettner M. Generating survival times to simulate Cox proportional hazards models. Statistics in Medicine, 2005, 24(11):1713-1723

## Try the rsimsum package in your browser

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rsimsum documentation built on June 21, 2018, 5:04 p.m.


text:
# Percent By Weight Formula

Solutions can be described in other concentration besides molarity, normality or molality. Solutions are sometimes represented in terms of relative per cent concentration of solute in a solution. To determine the weight per cent of a solution, divide the mass of solute by mass of the solution (solute and solvent together) and multiply by 100 to obtain per cent.

The per cent by weight formula can be expressed as

$\dpi{120}&space;\large&space;Percent\;&space;by\;&space;weight=\frac{gram\;&space;of\;solute}{&space;100g\;&space;of\;&space;solution}$

Example 1

Calculate how many grams of NaOH are required to make a 30.0% solution by using De-ionized water as the solvent.

Solution

Substitute the numbers that are known in the basic formula for per cent weight

X g of solute in 100g of solution = 30.0%

30.0 g of NaOH = X

Example 2

Determine the % of a solution which has 25.0g of NaCl dissolved in100mL of De-ionized water.

Solution

Substitute the known values in the per cent weight formula.

25.0g NaCl 100mL of solution = X%

Therefore, X = 25%


text:
Force Out The Inequality!

Algebra Level 4

$\large \dfrac{x^2+2-\sqrt{x^4+4}}{x}$ For real $$x$$, find the closed form of the maximum value of the expression above.