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Interpolation is the technique of estimating the unknown value of dependent variable (y) for a given value of independant variable (x) which is within the limits or range of the independent variable.

It is a positive correlation.

\(\int_{0}^{\infty}\frac{\displaystyle{dx}}{\displaystyle{(1+x)^3}} = \left.\frac{\displaystyle{-1}}{\displaystyle{2(1+x)^2}}\right|_0^\infty =0.5 \\ Remember \:that\\ \int \frac{1}{x^n}dx =\int x^{-n}dx = \frac{\displaystyle{x^{-n+1}}}{\displaystyle{-n+1}} + C \)

Here D₅ = 50, which divides into two equal parts. 

∴ P₅₀ also divides into two equal parts 

i.e. P₅₀ = 50.

\(I_n= \int\limits^\infty_0 e^{-x} sin^nx\,dx\)

\(= \int\limits^\infty_0\underset{I}{\underline{e^{-x}sin^{n-1}x}}\,\,\underset{II}{\underline {sinx\, dx}}\)

\(= \left[e^{-x}sin^{n -1}x\int sinx\,dx - \int\frac{e^{-x}(n - 1)sin^{n -2}x cos x - sin^{n -1}x}{x - cos x \,dx}\right]^\infty_0\)

\(= \left[-e^{-x}sin^{n -1}x cos\,x\right]^\infty _0 + \int\limits ^\infty_0(n - 1)e^{-x}sin^{n -2}x . cos^2x \, dx \\\,\,\,\,\,\,\,\,\, -\int\limits ^\infty_0e^{-x} sin^{n -1} x\, dx\)

\(= (n - 1)\int\limits^\infty_0e^{-x}sin^{n -2}x\,dx - \int\limits^\infty_0(n - 1)e^{-x}sin^nx\,dx -\int\limits ^\infty_0e^{-x} sin^{n -1} x\, cosx\,dx\)

(As \(e^{-\infty} = 0\) and \(cos^2x = 1 - sin^2x\))

\(= (n - 1)I_{n -2}- (n -1) I_n - \left[e^{-x}\frac{sin^nx}{n}\right]^\infty_0 +\int\limits^\infty_0- e^{-x} \frac{sin^nx}{n}dx\)

⇒ \(I_n = (n - 1)I_{n - 2} - (n - 1)I_n - \frac1nI_n\)     \((\because e^{-0} = 0)\)

⇒ \(I_n + (n - 1)I_n + \frac 1nI_n = (n - 1)I_{n -2}\)

⇒ \(I_n = \frac{n - 1}{1 + n - 1 + \frac1n}\) \(I_{n - 2} = \frac{n(n - 1)}{n^2 + 1}I_{n -2}\)

⇒ \(I_n = \frac{n(n - 1)}{n^2 + 1}I_{n - 2}\)

which is required relation.

We know that GM = √(a x b) = √(2 x 8) = √16 = 4.

The number of observations corresponding to a particular class is known as the class frequency.

Classification reduces the size of the data. 

Midpoint: x/m = Lower limit + Upper limit/2.

Ogives (less than Ogive) are used to locate the value of median.

Ogives are used to locate median.

F(x) = \(\frac{1}{B(p,q)}\) \(\frac{X^{p-1}}{(1+X)^{p+q}}\)

Harmonic mean = \(\frac{\displaystyle\sum_{n=0}^{\infty} F(X_n)}{\displaystyle\sum_{n=0}^{\infty} \frac{F(X_n)}{X}}\)

and we know sum change into integration in continuous series.

Harmonic mean of F(x) = \(\frac{\frac{1}{B(p,q)}\int_0^\infty \frac{X^{p-1}}{(1+X)^{p+q}}}{\frac{1}{B(p,q)}\int_0^\infty \frac{X^{p-1}}{X(1+X)^{p+q}}}\)

= \(\frac{\int_0^\infty\frac{X^{p-1}}{(1+X)^{p+q}}}{\int_0^\infty\frac{X^{(p-1)-1}}{(1+X)^{p+q+1}}}\) = \(\frac{B(p,q)}{B(p-1,q+1)}\)    \(\Big(\because \int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} = B(m,n)\Big)\)

= \(\frac{p!q!}{(p+q)!}\) x \(\frac{(p+q)!}{(p-1)!(q+1)!}\)

= \(\frac{p(p-1)!q!}{(p-1)!(q+1)q!}\) = \(\frac{p}{q+1}\)

G.M = √(3 x 27) = √81 = √9² = 9.

Kurtosis means peakedness or steepness of a frequency curve as compared to a normal curve.

γ = ± 1 or – 1 ≤ γ ≤ + 1.

Interpolation is the procedure of estimating the unknown value of the dependent variable for a given value of the independent variable which is within the limits of independent variable.

We know that ; S.D.(X) = √var(x)

Squaring both the sides,

[S.D (X)]² = E(X² ) – [E(X)]²

By substituting ; 122 = E (X² ) – (10)²

∴ E(X² )= 144 – 100 = 44. 

Coefficient of correlation measure the degree of relation between variables, where associates of attributes measures the degree of relationship between attributes such as sex and literacy. Success in examination and marriage.

If A and B are independent, then P(A∩B)= P(A) x P(B)

Here 1/6 = 1/2 x 1/3 = 1/6

Yes, A and B are independent.

The following are the limitations of statistics: 

• It does not deal with qualitative data. It deals only with quantitative data. 
• It does not deal with individual facts, but deals with aggregates of facts. 
• Statistical results are not exact. 
• It can be misused.
• Common people cannot handle statistics properly.

The important methods are:

• Direct personal interview method. 
• Indirect investigation method. 
• Local corresponding method. 
• Questionnaire method.

Direct personal Interview method: In this method the investigator personally meets the informants. He puts a number of questions relating to the enquiry and collects necessary and desired information from the informants. 

Merits:

• The information obtained is more accurate. 
• Confusion, wrong interpretation of the questions by the respondants can be avoided.

Demerits : 

• Time required is more. 
• There is ar chance of personal influence on the outcome of the enquiry.

The different methods of collection of primary data. 

• Direct personal interview/observation. 
• Indirect oral interview. 
• Information through agencies. 
• Mailed questionnaire.
• Schedules sent through enumerator.

Schedules sent through enumerator: In this method, a trained enumerator meets informants and collects the necessary information. Here schedule is a list of questions where the answers for the questions will be supplied by the informants and recorded by the enumerator.

1. Statistics does not deal with qualitative data. It deals with only quantitative data. 

2. Statistical results are riot exact.

Primary data are those which are directly collected by the investigator himself initialls for his own purpose. Method of collection of primary date is direct personal interview method.

The parts of table are

Range is the difference between the highest and the lowest value in the data. 

The lowest and the highest values which are taken to define the boundaries of a class are called class limits.

Explanation:

Types of sections used in India:

1. Rail section in use on BG

• 60 kg/m (UIC)
• 52 kg/m (IRC)
• 90R
• Length = 13 m

2. Rail section in use on MG

• 90R (90 lb/yd)
• 75R (75 lb/yd)
• 60R (60 lb/yd)
• Length = 12 m (13 m in case of 90R rails)

3. Rail section in use on NG

• 50R (50 lb/yd)
• length = 12 m

Temperature = Constant

By Boyle's law

PV = Constant

P₁V₁= P₂V₂

At NTP

1 atm x V = P₂ x V/4

P₂ = 4 atm

In case of a breach of warranty, the injured party is liable to be compensated. The injured party can refuse to accept the goods as well as claim damages in case of breach of condition. A condition can be treated as a warranty on the wish of the buyer. A warranty cannot be treated as a condition.

1. They cannot be used for further statistical analysis. 

2. They can be easily misled and can create wrong impression about the data.

P.I. = \(\frac{1}{D^3+1}\) cos(2x - 1)

= \(\frac{1}{(D+1)(D^2-D+1)}\) cos(2x - 1)

= \(\frac{1}{(D+1)(-4-D+1)}\) cos(2x - 1)

=   \(\frac{1}{(D+1)(-3-D)}\) cos(2x - 1)

= \(\frac{-1}{D^2+4D+3}\) cos(2x - 1)

= \(\frac{-1}{-4+4D+3}\) cos(2x - 1)

= \(\frac{-(4D+1)}{(4D-1)(4D+1)}\) cos(2x - 1)

= \(\frac{-(4D+1)}{16D^2-1}\)  cos(2x - 1)

= \(\frac{-(4D\,cos(2x - 1))+cos(2x-1)}{16\times-4-1}\)

= \(\frac{-(-8\,sin(2x-1)+cos(2x-1)}{-65}\)

= \(\frac{1}{65}\) cos(2x-1)-8 sin (2x-1)

(x + 1) \(\frac{dy}{dx} - y\) = e^3x (x + 1)

\(\Rightarrow\) \(\frac{dy}{dx} - \frac{y}{x + 1} = e^{3x}\)

∴ \(P = \frac{-1}{x + 1},\) Q = e^3x

^{I.F. \(= e^{\int\,p\,dx} = e^{\int\,\frac{-1}{x + 1}}dx\) \(= e^{- \log (x + 1)}\)}

^{\(= e^{\log \left(\frac{1}{x + 1}\right)} = \frac{1}{x + 1}\)}

∴ y × I.F. \(= \int Q \times (I.F.)dx\)

\(\Rightarrow\) \(\frac{y}{x + 1} = \int \frac{e^{3x}}{x + 1}dx\)

which is complete solution of given linear differential equation.

dy/dx = \(\frac{(x^2+y^2)}{xy}\); y(1) = 2

dy/dx = \(\frac{1+(y/x)^2}{y/x}\)  _________(1)

Let y = vx.

Then dy/dx = v + x dv/dx

Then from (1), we get

v + x dv/dx = \(\frac{1+v^2}v\)

⇒ x dv/dx = \(\frac{1+v^2}v\) - v = \(\frac{1+v^2-v^2}v\) = 1/v

⇒ v dv = 1/x dx

⇒ ∫ v dv = ∫ 1/x dx

⇒ v²/2 = log x + log c (where log c is an integral constant)

⇒ v² = 2 log cx

⇒ (y/x)² = 2 log cx

⇒ y² = 2 x² log cx   _________(2)

∵ y(1) = 2

∴ 4 = 2 log c

⇒ log c = 4/2 = 2

⇒ c = c²

∴ From (2), y² = 2 x² log (e²x), which represent the solution of given differential equation.

dy + dx = (x+2y) (dx - dy)

⇒ dy + (x+2y) dy = (x+2y) dx - dx

⇒ (x+2y+1) dy = (x+2y-1) dx

⇒ dy/dx = \(\frac{x+2y-1}{x+2y+1}\)

Let x+2y = z

⇒ 1 + 2 dy/dx = dz/dx

∴ dz/dx = 1 + 2 (z-1/z+1)

⇒ dz/dx = \(\frac{z+1+2z-2}{z+1}\) = 3z-1/z+1

⇒ z+1/3z-1 dz = dx

⇒ 1/3 ∫ 3z+3/3z-1 dz = dx

⇒ ∫\(\frac{3z-1+4}{3z-1}\) dz = ∫3 dx

⇒ ∫ dz + ∫ 4/3z-1 dz = ∫3 dx

⇒ z + 4/3 log |3z-1| = 3x + c

⇒ (x+2y) + 4/3 log |3(x+2y-1)| = 3x + c.

(y+x) dx + x dy = 0

⇒ (y dx + x dy) + x dx = 0

⇒ ∫(y dx + x dy) + ∫x dx = c   (By integrating both sides)

⇒ ∫d(xy) + x²/2 = c

⇒ xy + x²/2 = c which is solution of given differential equation.

\(\frac{dy}{dx} - e^{x-y} = e^{x-y}\)

⇒ \(\frac{dy}{dx} = 2e^{x-y}\)----(i)

Let x - y = t

Then 1 - \(\frac{dy}{dx}=\frac{dt}{dx}\) 

\(\therefore\) \(\frac{dt}{dx} = 1-2e^t\)  (From (i))

⇒ \(\frac{dt}{1-2e^t}=dx\) 

⇒ \(\frac{e^{-t}dt}{e^{-t}-2}=dx\) 

⇒ \(\int\frac{e^{-t}}{e^{-t}-2}dt = \int dx\) 

⇒ -log|e^-t - 2| = x + c

\(\therefore\) -log|e^{-(x - y)} - 2| = x + c is a solution of given differential equation.

Given differential equation is

\(\frac{d^2y}{dx^2}+\frac{dy}{dx}=\frac{1}{1+e^x}\)\(=\frac{e^{-x}}{e^{-x}(1+e^x)}=\frac{e^{-x}}{e^{-x}+1}\) 

⇒ \(\int\frac{d^2y}{dx^2}dx+\int\frac{dy}{dx}dx\) = \(\int\frac{e^{-x}}{e^{-x}+1}dx\)(By taking integration w.r.t. x)

⇒ \(\frac{dy}{dx}+y = -log(e^{-x}+1)\) 

which is solution of given differential equation.

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

Let x + y = t

Then 1 + \(\frac{dy}{dx} = \frac{dt}{dx}\)

\(\therefore\frac{dt}{dx}=1+t^2\) 

(\(\because\frac{dt}{dx}=(x+y)^2=t^2\)) 

⇒ \(\frac{dt}{1+t^2}=dx\) 

⇒ \(\int\frac{dt}{1+t^2}=\int dx\)

⇒ tan^-1 t = x + c

⇒ tan^-1(x + y) = x + c-----(1)

(\(\because t=x+c\))

\(\because\) y(0) = 0

\(\therefore \) tan^-1(0 + 0) = 0

⇒ C = tan^-1(x + y) = x is (From (1))

⇒ x + y = tan x

⇒ y = tan x - x

which is solution of given DE.

\(\phi\)(x) = ||x - 1| - |x + 1||

\(=\begin{cases}1-(x-1)-(-(x+1))&;x\leq-1\\1-(x-1)-(x+1)&;-1\leq x\leq1\\|x-1-(x+1)|&;x\geq1\end{cases}\)

\(=\begin{cases}2&;x\leq-1\\2|x|&;-1\leq x\leq 1\\2&;x\geq 1\end{cases}\) 

Now,  f(x)  = min\(\phi\)(f) ; -3 \(\leq t\leq x\) 

\(=\begin{cases}2&;-3\leq x\leq-1\,and\,x\geq 1\\0&;-1\leq x\leq1\end{cases}\)

\(\left(\because Min |t| = 0\\for\,-1\leq t\leq x \leq 1\right)\)

\(\therefore\) f(x) is not differentiable at x = -1 and x = 1

(C) f(100) = 2 (\(\because 100>1\))

(D)

\(\int\limits_{-3}^{10}f(x)dx=\int\limits_{-3}^{-1}2dx+\int\limits_{-1}^10dx+\int\limits_1^{10}2dx\) 

\(=2(x)^{-1}_{-3}+0+2(x)_1^{10}\) 

= 2(-1-(-3)) + 2(10 - 1)

 = 2(-1 + 3) + 2 x 9

 = 4 + 18 = 22

(a) Given 65°F convert into °C

 °C = 5/9 (°F - 32)

= 5/9 (65 - 32)

= 5/9 x 33

°C = 18.33

(b) 37°C into °F

°F = 9/5 °C + 32

°F  = 9/5 x 37 + 32

°F = 66.6 + 32

°F = 98.6

(c) 273 k into °C

°C = k - 273

°C = 273 - 273

°C = 0

(d) 96°F into k

°k - 273 = 10/18 [°F - 32]

°k - 273 = 5/9 [96 - 32]

°k - 273 = 5/9 x 64

°k = 35.5 + 273

⇒ 308.5

Given P(1)=1,P(2)=2,P(3)=3,P(4)=5

Let f(x)=(P(x)−x)

f(1)=P(1)−1=1−1=0

f(2)=P(2)−2=2−2=0

f(3)=P(3)−3=3−3=0

∴f(x)=0,x=1,2,3

⇒f(x)=a(x−1)(x−2)(x−3)

P(x)=a(x−1)(x−2)(x−3)+x

Put x=4

5=a(3)(2)(1)+4

⇒a= 1/6 ​ 

∴P(x)= 1/6(x−1)(x−2)(x−3)+x 

∴P(6)= 1/6 ​ (5)(4)(3)+6=16

(6^-1+8^-1)×(5/2)^-1

={(1/6)+(1/8)}×(2/5)

={(4+3)/24}×(2/5)     (taking the LCM of 8 and 6 which is 24)

=(7/24)×(2/5)

=14/120

=7/60.         (we get this by dividing 14 and 120 by 2)

Homework 22:- 

(1) 

Sol:- Dimensions of cuboid = 10 × 0.8 × 0.16 (1dm = 0.1m)

Volume of cuboid = lbh

Volume of cuboid = 10 × 0.8 × 0.16m³ = 1.28m³

Hence, the volume of the cuboid is 1.28m³ 

(2) 

Sol:- Dimensions of cuboid = 15 × 10 × 8 

Volume of the cuboid = lbh

Volume of the cuboid = 15 × 10 × 8cm³ = 1200cm³

(3) Edge of cube = 8 cm. 

Volume of the Cube = (Edge)³

Volume of the Cube = (8)³ = 512cm³

Hence, the volume of the cube is 512cm³ 

(4) 

Sol:- Volume of the Box = 40 × (Edge)³

Edge of the cube = 3cm 

Volume of the box = 40 × 3³ = 1080cm³ 

(5) 

Sol:- Volume of 1 brick = 23 × 11 × 8cm³ 

Volume of Wall = 920 × 600 × 22cm³ 

No of bricks = Volume of Wall / Volume of 1 brick 

No of bricks = (920 × 600 × 22) / (23 × 11 × 8) = 600 × 2 × 5 = 6000

Hence, the no of bricks required is 6000

Homework 21:- 

(1) Area of the square = (Side)²

Area of the square = (25)² = 625cm²

Therefore, the area of the square is 625cm²

(2) Area of the square = (Side)² 

Area of the square = (32)² = 1024m²

Therefore, the area of the square is, 1024m²

(3)  Area of the square = (Side)² 

Area of the square = (17)² = 289m² 

Therefore, the area of the square is, 289m² 

(4)  Area of the square = (Side)² 

Area of the square = (35)² = 1225dm² 

Therefore, the area of the square is, 1225dm² 

(5) Area of the rectangle = l × b

Area of the rectangle = 22 × 18 = 396cm²

(6) Area of the rectangle = 34 × 20 = 680cm²

(7) Area of the rectangular park = 15 × 9 = 135m²

(D² + 3D + 2)y = e^2x-sin x

m² + 3m + 2 = 0 is its auxiliary equation

⇒ (m+ 1)(m + 2) = 0

⇒ m = -1, -2

∴ C.F. = C₁e^-x + C₂e^-2x

P.I. = \(\frac1{D^2+3D+2}e^{2x}-\frac1{D^2+3D+2}sin x\)

 = \(\frac{e^{2x}}{12}-\frac{sin x}{-1+3D+2}\)

 = \(\frac{e^{2x}}{12}-\frac{sin x}{3D+1}\) 

 = \(\frac{e^{2x}}{12}-\frac{(3D-1)}{9D^2-1}sin x\) 

 = \(\frac{e^{2x}}{12}-\frac{(3D-1)sin x}{-9-1}\) 

 = \(\frac{e^{2x}}{12}-\frac{3cos x-sin x}{-10}\) 

 = \(\frac{e^{2x}}{12}+\frac3{10}cos x-\frac3{10} sin x\)

Correct option is A) bridge