The entries in column (4) and (6) are identical  ~(p↔️q) ≡ p↔️~q

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

Observe that the given distribution is continuous
By definition \(\mu =E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -2x } \right) dx } +\int _{ 0 }^{ \infty }{ x\left( { \lambda e }^{ -\lambda x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ x\left( { e }^{ -\lambda x } \right) dx } \) 
=  \(0+\lambda \left( \frac { 1 }{ { \lambda }^{ 2 } } \right) \) (using Gamma integral for positive integer n,\(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\cfrac { n }{ { a }^{ n+1 } } \))
= \(\frac { 1 }{ \lambda } \) 
Variance :
By definition,\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -\lambda x } \right) } dx+\int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \lambda { e }^{ -2x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( { e }^{ -2x } \right) dx } \)
 (using Gamma integral for positive integer)
Therefore Var(X ) = E(X² )- E(X )²
= \(\frac { 2 }{ { \lambda }^{ 2 } } -\left( \frac { 1 }{ \lambda } \right) ^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } } \) 
Hence the mean and variance are respectively \(\frac { 1 }{ \lambda } \)  and \(\frac { 1 }{ { \lambda }^{ 2 } } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

Let V be the velocity and the retardation (negative acceleration) be -\(\frac{dv}{dt}\)
Given \(\frac{dv}{dt}\) = -V
Separating the variables,
\(\frac{dv}{v}=-dt\)
\(\Rightarrow \int { \frac { dv }{ v } } =-\int { dt } \)
\(\Rightarrow log\quad v=-t+logC\)
\(\Rightarrow logv-logC=-t\)
\(\Rightarrow log\left( \frac { v }{ { C }_{ v } } \right) =-t\)
\(\Rightarrow ={ e }^{ -t }\)
\(\Rightarrow \frac { v }{ { C }_{ v } } ={ Ce }^{ -t }...(1)\)
Given when t = 0, v = m/sec
\(\therefore\) (1) become 10 = C_e0 \(\Rightarrow\) C = 10
\(\therefore\) (1)  v = 10e^-t
When t = 2, v = 10e^-2
\(\Rightarrow v=\frac { 10 }{ { e }^{ 2 } } \)

\(x\frac { dy }{ dx } +2y=x^2logx\)
Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 2 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 2 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 2 }{ x } } dx=2logx=logx^2\)
\(\therefore I.F={ e }^{ \int { pdx } }={ e }^{ log\ x ^2}=x ^2\)
\(\therefore\) The solution is \({ e }^{ \int { u\ dv} }=uv-\int { vdu}\)
\(u=log\ x;dv=x^3\)
\(du=\frac { 1 }{ x }dx;v=\frac { { x }^{ 4} }{ 4} \)
\({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { xlogx.({ x }^{ 2 })dx } \)
\(\Rightarrow { x }^{ 2 }y=\int { { x }^{ 3 } } log\quad xdx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\int { \frac { { x }^{ 4 } }{ 4 } .\frac { 1 }{ x } } dx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { 1 }{ 4 } \int { { x }^{ 3 }dx } \)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { { x }^{ 4 } }{ 16 } +c\)

Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 1 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 1 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 1 }{ x } } dx=logx\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ log\quad x }=x\)
\(\therefore\) The solution is \({ e }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }=dx+c } \)
\(u=cos\quad x;dv=x\)
\(du=\frac { 1 }{ x } ,v=\frac { { x }^{ 2 } }{ x } \)
\(\int { udv } =uv-\int { vdu } \)
\(yx=\int { xlogxdx+c } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\int { \frac { { x }^{ 2 } }{ 2 } } .\frac { 1 }{ x } dx\)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } \int { xdx } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } .\frac { { x }^{ 2 } }{ 2 } +c\)
\(\Rightarrow xy=\frac { { 2x }^{ 2 }logx-{ x }^{ 2 }+4c }{ 4 } \)
\(\Rightarrow 4xy=2{ x }^{ 2 }logx-{ x }^{ 2 }+4c\)

\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)
\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)dx
put 25 - x² = t²
⇒ -2x dx = 2t dt
x dx = -dt
Putting in (1),
⇒ x dx = \(\frac{-dt}{2}\)
dy = t \(\times\)(-t) dt
Integrating on both sides,
\( \int d y =-\int t^2 d t \)
\(y =\frac{-t^3}{3}+C_1 \Rightarrow 3 y+t^3=3 C_1 \)
\(3 y+\left(25-x^2\right)^{\frac{3}{2}} =C\)

x⁴+ 3x³- 3x - 1 = 0
Sum of the co-efficients = 1 + 3 - 3.- 1 = 0 
⇒x = 1 is a root ⇒(x - 1) is a factor

[Using synthetic division]
∴ x = 1, 1 are the roots and the remaining factor
\(\Rightarrow x=\frac { -3\pm \sqrt { 9-4(1)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -3\pm \sqrt { 5 } }{ 2 } \)
∴ The roots are 1, -1, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

The sample space S = {H,T} \(\times\) {H,T}
That is S = {TT, TH, HT, HH}
Let X be the random variable denoting the number of heads.
Therefore
X (TT ) = 0 , X (TH ) = 1,
X (HT) = 1, and X (HH) = 2 .
Then the random variable X takes on the values 0, 1 and 2

The probabilities are given by

\(f(0)=P(X=0)=\cfrac { 1 }{ 4 } \) 
\(f(1)=P(X=1)=\cfrac { 1 }{ 2 } \) 
and \(f(2)=P(X=2)=\cfrac { 1 }{ 4 } \) 
The function f (x) satisfies the conditions
(i) f (x) ≥ 0 , for x = 0, 1, 2
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=2 }{ f(x) } =f(0)+f(1)+f(2)\) 
= \(\cfrac { 1 }{ 4 } +{ \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } =1 }\) 
Therefore f (x) is a probability mass function.
The probability mass function is given by

(or)
\(f(x)\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & forx=0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } & forx=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 4 } & forx=2 \end{matrix} \end{cases}\)

The possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(i) The sample space
S = {1, 2, 3, 4, 5, 6}\(\times\){1, 2, 3, 4, 5, 6}
consists of 36 ordered pairs (α, β) where α and β can take any integer value between 1 and 6 as shown. X is assigned to each point (α, β) the sum of the numbers on the dice .
That is X (α, β) = α + β
Therefore
X (1,1) = 1+1 = 2
X (1, 2) = X (2,1) = 3
X (1,3) = X (2,2) = X (3,1)= 4
X (1, 4) = X (2,3) = X (3, 2)  X (4,1) = 5
X (1,5) = X (2,4) = X (3,3) = X (4, 2) = X (5,1) = 6
X (1,6) = X (2,5) = X (3, 4) = X (4,3 = X (5, 2) X (6,1) = 7
X (2,6) = X (3,5) = X (4,4) = X (5,3) = X (6,2) = 8
X (3,6) = X (4,5) = X (5,4)  X (6,3) = 9
X (4,6) = X (5,5) X (6,4) = 10
X (5,6) = (6,5) = 11
X (6,6) = 12
(ii) Then the random variable X takes on the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
(iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}.
(iv) The number of inverse images are given below

Taking u = x² and v = cos nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ \pi }{ { x }^{ 2 }cos\quad nxdx } ={ \left[ ({ x }^{ 2 })\left( \frac { sin\quad nx }{ n } \right) -(2x)\left( -\frac { cos\ nx }{ { n }^{ 2 } } \right) +(2)\left( -\frac { sin\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ \pi }\)
\(=\frac { 2\pi { (-1) }^{ n } }{ { n }^{ 2 } } \), since cos n\(\pi\) = (-1)ⁿ and sin n\(\pi\) = 0

\(\int _{ 0 }^{ 1 }{ \frac { 2x+7 }{ { 5x }^{ 2 }+9 } dx } =\int _{ 0 }^{ 1 }{ \frac { 2x }{ { 5x }^{ 2 }+9 } } +7\int _{ 0 }^{ 1 }{ \frac { dx }{ (5{ x }^{ 2 })+{ 3 }^{ 2 } } =\frac { 1 }{ 5 } } log{ [{ 5x }^{ 2 }+9] }_{ 0 }^{ 1 }+\frac { 7 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }{ \left( \frac { 3 }{ \sqrt { 5 } } \right) }^{ 2 } } } \)
\(=\frac { 1 }{ 5 } [log14-log9]+\frac { 7 }{ 5 } \times \frac { \sqrt { 5 } }{ 3 } { \left[ { tan }^{ -1 }\frac { x }{ \left[ \frac { 3 }{ \sqrt { 5 } } \right] } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 5 } log\frac { 14 }{ 9 } +\frac { 7 }{ 3\sqrt { 5 } } { tan }^{ -1 }\frac { \sqrt { 5 } }{ 3 } \)

\(sin\left( \frac { dy }{ dx } \right) =a\)
\(\Rightarrow \frac { dy }{ dx } ={ sin }^{ -1 }(a)\)
\(\Rightarrow dy={ sin }^{ -1 }(a)dx\)
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
Taking Integration on both sides, we get
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
\(\Rightarrow y={ sin }^{ -1 }(a)x+c ...(1)\)
Initial condition:
Since y(0) = 1 we get,
1 = sin^-1(a)(0) +C
0 + C ⇒ C = 1
equation (1) ⇒ y = sin^-1(a) x + 1
y^-1 = sin^-1(a) + x
\(\Rightarrow \frac { y-1 }{ x } =sin(a)\Rightarrow sin\left( \frac { y-1 }{ x } \right) =a\)

Equation of family of parabolas with axis as y axis is given by,
(x-0) = 4a(y-k) .... (1)

Given: Vertex at (0, - 1).
Putting k = -1 in (1), we get
⇒ x² = \(\pm\)4a(y + 1) ....(2)
Differentiating with respect to 'x'  
2x = \(\pm\)4a\(\left( \frac { dy }{ dx } \right) \) ....(3)
⇒ 4a = \(\frac { 2x }{ \frac { dy }{ dx } } \) 
\(\frac{x^2}{2x} = \frac{y+1}{\frac{dt}{dx}}\)
ie) \(x \frac{dy}{dx}-2(y+1) =0\)
This is the required differential equation.

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \wedge \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) =\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

p ∨ ¬ t ( p ∨ ¬s) contains 3 variables p, s, and t. Hence the corresponding truth table will contain 2³ = 8 rows

Given \(p=\frac { 3 }{ 4 } \) 
n = 5
P(X = x) = nC_xp^x (1-p)^n-x
\(P(X=3)={ 5C }_{ 3 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( 1-\frac { 3 }{ 4 } \right) ^{ 2 }\) 
\(P(X=3)={ 5C }_{ 2 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) [∵nCr = nC_n-r]

= \(\frac { 135 }{ 512 } \)

\(f(x)= \begin{cases}2(x-1) & 1
\(Mean=E(X)=\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ 2((x-1)dx } } \) 
\( =2\left[\frac{8}{3}-\frac{4}{2}-\frac{1}{3}+\frac{1}{2}\right] \)
\( =2\left(\frac{7}{3}-\frac{3}{2}\right) \)
\( =2 \times \frac{5}{6} \)
\( =\frac{5}{3} \)
\(E({ x }^{ 2 })=\int _{ 1 }^{ 2 }{ { x }^{ 2 }f(x)dx } \) 
= \(\int _{ 1 }^{ 2 }{ { x }^{ 2 }.2\left( x-1 \right) } dx\) 
= \(2\int _{ 1 }^{ 2 }{ ({ x }^{ 3 }-{ x }^{ 2 })dx } \) 
= \(2\left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\) 
= \(2\left[ \left( 4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right) \right] \) 
= \(2\left[ \frac { 4 }{ 3 } +\frac { 1 }{ 12 } \right] =2\left[ \frac { 16+1 }{ 12 } \right] \) 
= \(\frac { 17 }{ 6 } \) 
ஃ Var(X) = E(X²) - [E(X)]²
= \(\frac { 17 }{ 6 } -(\frac{5}{ 3 }^{ 2 })=\frac { 17 }{ 6 } -\frac { 25 }{ 9 } \)
= \(\frac{51-50}{18}\)
= \(\frac{1}{18}\)

We know that \(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x) } dx\quad if(2a-x)=f(x)\)
Take 2a = \(\pi\) and f(x) = g(sinx)
Then, f(2a-x) = g(sin(\(\pi\)-x)) = g(sinx) = f(x).
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)
\(\int _{ 0 }^{ \pi }{ g(sinx)dx } =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ g(sinx)dx } \)

 P(0.2≤X < 0.6)
= \(\int _{ 0.2 }^{ 0.6 }{ f(x)dx } =\int _{ 0.2 }^{ 0.6 }{ x.dx } =\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0.2 }^{ 0.6 }\)  
= 0.18 - 0.02
= 0.16

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

Given equation of curve is
xy = ae^x + be^−x + x² ...(1)
where a &b are aribitrary constant. differentiate equation (1) twice successively, because we have two arbitray constant.
\(x \frac{d y}{d x}+y(1)=a \mathrm{e}^{x}-\mathrm{be}^{-x}+2 x\) ...(2)
\( x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)+\frac{d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \)
\( x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \) ...(3)
From (1), we get \(x y-x^{2}=\mathrm{ae}^{x}+\mathrm{be}^{-x}\) ...(4)
Substituting equation (a) in (3), we get
\(\therefore x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}-x y+x^{2}-2=0\) which is the required differential equaiton.