~(p V q) V (~p ∧ q) and ~p

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and ∗ be the matrix multiplication.
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)and 
B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)∈M
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & xy \\ 2xy & 2xy \end{matrix} \right) \in M\\ \)
[∵ 2xy∈R-{0}]
∴ M is closed under M
Identity:
Since identity of 2\(\times\)2 matrices is I =\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)∉M 
∴ M has no identity under *.
Inverse:
Since it has no identity, it won't have inverse also.

∴ ((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
The given statement is a tautology.

The table for the operation x₁₁ is as follows.

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 5 }{ kdx } +\int _{ 5 }^{ \infty }{ 0dx } =1\) 
\(0+k\left( x \right) _{ 1 }^{ 5 }+0=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 0 & Otherwise \end{matrix} \end{cases}\)
(i) Distribution function
The distribution function 
\(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1, \(F(x)=\int _{ -\infty }^{ x }{ f(u)du } =\int _{ -\infty }^{ x }{ oldu } =0\) 
When 1 ≤ x ≤ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u)du=\int _{ -\infty }^{ x }{ 0du } +\int _{ 1 }^{ x }{ odu } +\int _{ 1 }^{ x }{ \frac { 1 }{ 4 } du } =\frac { 1 }{ 4 } (x-1) } \) 
When x ≥ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du=\int _{ -\infty }^{ x }{ odu } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 5 }^{ 5 }{ odu } =1\) 
Thus \(F(x)=\begin{cases} \begin{matrix} 0 & x<1 \end{matrix} \\ \begin{matrix} \frac { x-1 }{ 1 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 1 & x>5 \end{matrix} \end{cases}\) 
(ii) P(X < 3) = P(X ≤ 3) = F(3) = \(\frac { 3-1 }{ 2 } =\frac { 1 }{ 2 } \) (Since F(x) is continuous)
(iii) P(2 < X < 4) = P(2 ≤ X ≤ 4) F(4) - F(2) = \(\frac { 3 }{ 4 } -\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \) 
(iv) P(3 ≤ X ) = P(X ≥ 3) = 1− P(X < 3) = 1 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Since X is a discrete random variable, from the given data, X takes on the values
−2, −1, 0, and 1.
For discrete random variable X, by definition, we have f (x) = P(X = x)
Therefore left hand limit of f(x) at x = -2 is F(− 2⁻ ) 
f (−2) = P(X =-2 ) = F(-2 ) - F(- 2^- )= 0.25-0 = 0.25
Similarly for other jump points, we have
f (−1) = P(X = -1) = F(-1) - F(-2) = 0.60 - 0.25 = 0.35.
f (0) = P(X ) 0) = F(0) - F(-1) = 0.90 - 0.60 = 0.30 ,
f (1) = P(X =1) = F(1) - F(0) 1- 0.90 = 0.10 .
Therefore the probability mass function is

The distribution function F(x) has jumps at x = -2, -1, 0, and 1. The jumps are respectively 0.25, 0.35, 0.30, and 0.1 is shown in the figure given below.
These jumps determine the probability mass function

(i) \(P(X<0)=\sum _{ -\infty }^{ -1 }{ P(X=x)=P(X=-1)=0.25+0.35 } =0.60\)
(ii) \(P(X\ge -1)=\sum _{ -1 }^{ 1 }{ P(X=x)=P(x=-1) } +P(X=0)+P(X=1)=0.35+030+0.10=0.75\)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

Let T be the temperature of the coffee at time t
and T_m' the temperature of the kitchen.
By Newton's law of cooling
\(\frac { dT }{ dt } =K(T-{ T }_{ m })\)
\(\Rightarrow \frac { dT }{ dt } =K(T-70)\)
\(\Rightarrow \int { \frac { dT }{ T-70 } =K\int { dt } } \)
\(\Rightarrow log(T-70)=kt+logC\)
\(\Rightarrow log(T-70)-logC=Kt\)
\(\Rightarrow log\left( \frac { T-70 }{ C } \right) =Kt\)
\(\Rightarrow \frac { T-70 }{ C } ={ e }^{ Kt }\)
\(\Rightarrow T-70={ Ce }^{ Kt }...(1)\)
\(\\ When\ t=0,\ T={ 180 }^{ o }F\)
\(\therefore { 180 }^{ o }-{ 70 }^{ o }={ Ce }^{ 0 }\)
\(\Rightarrow C={ 11 }0^{ 0 }\)
\(\\ \therefore (1)\Rightarrow T-70=110{ e }^{ Kt } ..(2)\)
\(When\ t=0,T=160\)
\(\therefore 160-70=110{ e }^{ 10K }\)
\(90=110{ e }^{ 10K }\)
\(\Rightarrow { e }^{ 10K }=\frac { 9 }{ 11 } \)
\(\Rightarrow { e }^{ K }={ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } }...(3)\)
(i) when t = 15, (2) becomes,
\(\Rightarrow T-70=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } \times 15 }\)
\(=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 3 }{ 2 } }\)
\(=110\times { \left( \frac { 9 }{ 11 } \right) }\left( \sqrt { \frac { 9 }{ 11 } } \right) \)
\(=110\times \frac { 9 }{ 11 } \times \frac { 3 }{ \sqrt { 11 } } \)
\(=\frac { 270 }{ \sqrt { 11 } } =\frac { 270 }{ 3.32 } =81.33\)
\(\Rightarrow\) T=81.33+70=151.3F
\(\therefore\) T = 151.3F
\(\therefore\) The temperature of the coffee at 10.15 am is 151.3F
(ii) when T = 130F, (2) becomes
T-70 = 110e^kt  ...(2)
\(\Rightarrow\) 130-70 = 110e^kt 
60 = 110e^kt 
e^kt  = \(\frac{6}{11}\)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 6 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 6 }{ 11 } \right) }{ log\left( \frac { 9 }{ 11 } \right) } \)
\(=\frac { log(0.545) }{ log(0.818) } =\frac { -0.264 }{ -0.087 } \)
= 3.34
t = 30.34min
T = 140F (2)becomes
140-70 = 110e^kt ...(2)
\(\Rightarrow 70={ 110e }^{ kt }\)
\({ e }^{ kt }=\frac { 7 }{ 11 } \)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 7 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 7 }{ 11 } \right) }{ log\left( \frac { 7 }{ 11 } \right) } =\frac { -0.197 }{ -0.087 } \)
= 2.26
t = 22.6min
\(\therefore\) Between 10.22 min to 10.30 min, the woman should have drunk the coffee.

Let P be denote the population of a city
Given that \(\frac{dP}{dt}\infty\)
\(\Rightarrow P=\frac{dP}{dt}kP\)
\(\Rightarrow \frac{dP}{P}=kdt\)
\(\Rightarrow \int { \frac { dP }{ P } = } k\int { dt } \)
\(\Rightarrow logP=kt+logc\)
\(\Rightarrow log\left( \frac { P }{ c } \right) =kt\)
\(\Rightarrow \frac { P }{ c } ={ e }^{ kt }\)
\(_{ }^{ c }{ P }={ c.e }^{ kt } ..(1)\)
Given when t = 0, P = 3,00,000
\(\therefore(1)\rightarrow\) = ce⁰ \(\Rightarrow\) c = 3,00,000
\(\therefore\) P = 3,00,000 e^kt ...(2)
Again when t = 40, P = 4,00,000
\(\therefore\) (2) \(\Rightarrow\) 4,00,000 = 3,00,000e^40k
\(\Rightarrow \frac { 4 }{ 3 } ={ e }^{ 40k }\)
\(\Rightarrow log\left( \frac { 4 }{ 3 } \right) =40K\)
\(\Rightarrow K=\frac { 1 }{ 40 } log\left( \frac { 4 }{ 3 } \right) \)
\(\Rightarrow k=log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } }...(3)\)
\(\therefore\) (2) becomes, P = 3,00,000\({ e }^{ log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } t } }\)
\(\Rightarrow\) P = 3,00,000 \(^{ { \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } } }\)

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.