\( \mathrm{F}(\mathrm{x}) =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right) \)
\(\mathrm{P}(0 \leq \mathrm{x} \leq 1) =\mathrm{F}(1)-\mathrm{F}(0) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+0\right) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}-\frac{\pi}{2}\right)=\frac{1}{4}\)

Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

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

To find the probabilities, the values of the parameters n and p must be known.
Given that
Mean = np = 2 and variance = npq = 1.5
This gives \(\frac { npq }{ np } =\frac { 1.5 }{ 2 } =\frac { 3 }{ 4 } \) 
\(q=\frac { 3 }{ 4 } \) and \(p=1-q=1-3\frac { 4 }{ 4 } =\frac { 1 }{ 4 } \)
np = 2  gives \(n=\frac { 2 }{ p } =8\) . Therefore \(X\sim B\left( 8,\frac { 1 }{ 4 } \right) \) 
Therefore probability distribution is
\(P(X=x)=f(x)=\left( \begin{matrix} 8 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ x }\left( \cfrac { 3 }{ 4 } \right) ^{ 8-x }\) 
(i) \(P(X=0)=f(0)=\left( \begin{matrix} 8 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 0 }\left( \cfrac { 3 }{ 4 } \right) ^{ 8-0 }=\left( \cfrac { 3 }{ 4 } \right) ^{ 8 }\) 
(ii) \(P(X=1)=f(1)=\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) \left( \cfrac { 3 }{ 4 } \right) ^{ 8-1 }=2\left( \cfrac { 3 }{ 4 } \right) ^{ 2 }\) 
(iii)  P(X≥1) = 1-P(X<1) = 1-P(X = 0) = \(1-\left( \frac { 3 }{ 4 } \right) ^{ 8 }\)

\(\mathrm{n}=9, \mathrm{p}=\frac{1}{2}, \mathrm{k}=7 \)
\( \mathrm{P}(X=x)={ }^{n} C_{x} p^{x} q^{n-x}, x=0,1,2, \ldots, n \)
\(p =\frac{1}{2} \)
\(q =1-p=\frac{1}{2} \)
\(P(X=7) ={ }^{9} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{2} \)
\( =\frac{9 \times 8}{2} \times \frac{1}{2^{9}} \)
\( =36 \times \frac{1}{512}=\frac{9}{128} \)

The family of straight lines passing through the origin is y = mx, where m is an arbitrary constant.
Differentiating both sides with respect to x, we get \(\frac{dy}{dx}=m\) 
From (1) and (2), we get y = x\(\frac{dy}{dx}\). This is the required differential equation. 
Observe that the given equation y = mx contains only one arbitrary constant and thus we get the differential equation of order one.

\({ { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 2 }=xsin\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \)
The highest derivative is 2
∴ Order 2
The given differential equation is not a polynomial equation in its derivative and so its degree is not defined.

Given f(x) = 4k (x-1)², 1 ≤ x ≤ 3.
Since f(x) is a p.d.f\(\int_{1}^{3} f(x) d x=1\)
\( \Rightarrow \int_{1}^{3} 4 k(x-1)^{3} d x =1 \)
\(\Rightarrow 4 k\left[\frac{(x-1)^{4}}{4}\right]_{1}^{3} =1 \)
\(\Rightarrow \ k\left[2^{4}-0^{4}\right] =1 \)
\(\Rightarrow \ 16 k =1 \)
\(\Rightarrow \ k =\frac{1}{16} \)
\(\therefore \mathrm{P}(1 \leq \mathrm{X} \leq 2) =\int_{+1}^{2} f(x) d x \)
\(\Rightarrow \ =\int_{1}^{2} \frac{1}{4}(x-1)^{3} d x \)
\(\Rightarrow 1{ }^{2} =\frac{1}{4}\left[\frac{(x-1)^{4}}{4}\right]_{1}^{2} \)
\( =\frac{1}{16}\left(1^{4}-0^{4}\right)=\frac{1}{16} \)

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

(i) Given that five fair coins are tossed once. Since the coins are fair coins the probability of getting an head in a single coin is
 \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 1 }{ 2 } \) 
Let X denote the number of heads that appear in five coins. X is binomial random variable that takes on the values 0, 1, 2,  3, 4 and 5 and \(p=\frac { 1 }{ 2 } \) That is \(X\sim B\left( 5,\cfrac { 1 }{ 2 } \right) \) 
Therefore the binomial distribution is
\(f(x)=\left( \begin{matrix} n \\ x \end{matrix} \right) p*\left( 1-p \right) ^{ n-x }\), x = 0, 1, 2,..,n
becomes
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ x }\left( \cfrac { 1 }{ 2 } \right) ^{ n-x }\), x = 0, 1, 2,..,5
That is 
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ n }\), x = 0, 1, 2...,n
(ii) A fair die is rolled ten times and X denotes the number of times 4 appeared. X is binomial
random variable that takes on the values 0, 1, 2, 3,...10 , with n = 10 and \(p=\cfrac { 1 }{ 6 } \). That is \(X\sim B\left( 10,\cfrac { 1 }{ 6 } \right) \) 
Probability of getting a four in a die is \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 5 }{ 6 } \)
Therefore the binomial distribution is 
\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 6 } \right) ^{ x }\left( \cfrac { 5 }{ 6 } \right) ^{ 10-x }\) x = 0, 1, 2,...,10

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

(i) Since f is a p.d.f. \(\int_{0}^{1} f(x) d x=1\)
\( k \int_{0}^{1}(1-x)^{3} d x=1 \)
\( k\left[\frac{(1-x)^{4}}{-4}\right]_{0}^{1}=1\)
\( -\frac{k}{4}\left[(1-x)^{4}\right]_{0}^{1} =1 \)
\(-\frac{k}{4}(0-1) =1 \)
\(\frac{k}{4} =1\)
(ii) \(P\left(X<\frac{2}{3}\right)\)
\( \mathrm{P}\left(\mathrm{X}<\frac{2}{3}\right) =\int_{0}^{3 / 2} k(1-x)^{3} d x \)
\( =4\left[\frac{(1-x)^{4}}{-4}\right]_{0}^{1 / 3} \)
\( =-\left[(1-x)^{4}\right]_{0}^{2 / 3} \)
\( =-\left(\left(1-\frac{2}{3}\right)^{4}-1^{4}\right) \)
\( =\left(\frac{1}{81}-1\right)^{3} \)
\( =\frac{80}{81} \)

(i) We know that 'e' is the identity element
if
a* e = e * a = a
a * 0 = a + 0 = a
0 * a = 0 + a = a
a * 0 = 0 * a = a
Hence 0 is the identity for this operation.
(ii) An element a \(\in\) A is said to be invertible with respect to the operation * if there exists an element b in A such that b is called the inverse of a.
a * (6-a) = a + (6-a) - 6 = 0
(6-a) * a = (6-a) + a-6 = 0
Hence, each 'a' of the set {0, 1 , 2, 3 , 4, 5} is invertible with (6 - a) being the inverse of 'a'.

Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.

Since the given function is a probability density function

\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ f(x) } dx+\int _{ 1 }^{ 4 }{ f(x) } dx+\int _{ 4 }^{ \infty }{ f(x) } dx=1\) 
From the given information
\(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 4 }{ { Cx }^{ 2 }dx } +\int _{ 4 }^{ \infty }{ 0dx } =1\) 
\(0+C\left[ \cfrac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 4 }+0=1\Rightarrow C\left[ \cfrac { 64-1 }{ 3 } \right] =1\Rightarrow 21C\Rightarrow C=\cfrac { 1 }{ 21 } \) 
Therefore the probability density function is
\(f(x)= \begin{cases}C x^{2} & 1
Since f (x) is continuous, the probability that X is equal to any particular value is zero. Therefore when the random variable is continuous, either or both of the signs < by ≤ and > by ≥ can be interchanged. Thus
(i) P(1.5 < X < 3.5) = P(1.5 ≤ X< 3.5)= P(1.5 < X ≤3.5) = P(1.5 ≤X ≤ 3.5)
Therefore
\(P(1.5
 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 21 } \left( \cfrac { \left( 3.5 \right) ^{ 3 }-\left( 1.5 \right) ^{ 3 } }{ 3 } \right) \) 
= \(\cfrac { 79 }{ 126 } \) 
(ii) \(P(X\le 2)=\int _{ -\infty }^{ 2 }{ f(x) } dx=\int _{ -\infty }^{ 1 }{ f(x)dx } +\int _{ 1 }^{ 2 }{ f(x)dx } \)
 
Therefore
\(P(X\le 2)=0+\cfrac { 1 }{ 21 } \int _{ 1 }^{ 2 }{ { x }^{ 2 }dx=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } ^{ 2 }_{ 1 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 2 }^{ 3 }-{ 1 }^{ 3 } }{ 3 } \right) =\cfrac { 7 }{ 63 } \) 
(iii) \(P(3
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

Let P be the Principal at time t.
Given ratio = 5%
\(\therefore \frac { dP }{ dt } =P\left( \frac { 5 }{ 100 } \right) =0.05P\)
\(\Rightarrow \frac { dP }{ dt } =0.05dt\)
[separating the variable]
Integrating, \(\int { \frac { dP }{ P } =0.05\int { dt } } \)
\(\Rightarrow logP=0.05t+logC\)
\(\Rightarrow logP-logC=0.05t\)
\(\Rightarrow log\left( \frac { P }{ C } \right) =0.05t\)
\(\Rightarrow \frac { P }{ C } ={ e }^{ 0.05t }\)
\(\Rightarrow P=C{ e }^{ 0.05t }..(1)\)
Given when t = 0, P = Rs.10,000 
Substituting in(1) we get,
\(\Rightarrow\) 10,000 = Ce₀
\(\Rightarrow\) C = 10,000 
\(\therefore\) (1) becomes, P = 10,000 e^0.05t
When t = 18months = 1 1/2years = \(\frac{3}{2}\) years we get
P = 10,000 e^{0.05\((\frac{3}{2})\)}
\(\therefore\) P = 10,000 e^0.075

x cos y dy = e^x(x log x + 1)dx
\(\Rightarrow cos\ y\ dy={ e }^{ x }\frac { (x\ log+1) }{ x } dx\)
\(=\left[ { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) \right] dx\)
\(\Rightarrow \int { cos\quad y\quad dy } =\int { { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) dx } \)
Taking integration on both sides, we get 
 \( \int \cos y d y=\int e^x\left[\log x+\frac{1}{x}\right] d x  \)
 RHS 
\( \int e^x\left[\log x+\frac{1}{x}\right] d x\)
 Take \(\mathrm{f}(x)=\log x \Rightarrow f^{\prime}(x)=\frac{1}{x} \)
This of the form  \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C \)
 \(\therefore \int e^x\left[\log x+\frac{1}{x}\right] d x=\mathrm{e}^x \log x+\mathrm{C}\)
Substituting in (1), we get 
\(\sin y=e^x \log x+C  \)