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i) Integrating factor of the differential of the form \frac{d x}{d y}+P_{1} x=Q_{1} is given by e^{\int P_{1}d y}

(ii) Solution of the differential equation of the type \frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}  is given by x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C

iii) Correct substitution for the solution of the differential equation of the type \frac{d y}{d x} f(x, y), where f(x, y) is a homogeneous function of zero degree is  y=v x 
(iv) Correct substitution for the solution of the differential equation of the type \frac{d x}{d y} g(x, y)where g(x, y) is a homogeneous function of the degree zero is x=v y
(V) Number of arbitrary constants in the particular solution of a differential
equation of order two is two.

(vi)The differential equation representing the family of circles x^{2}+(y-a)^{2}= a^{2} will be of order two.  
(vii) The solution of \frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3} is y^{\frac{2}{3}}-x^{\frac{2}{3}}=c
 
(viii) Differential equation representing the family of curves y=e^{x}(A \cos x+B \sin x ) is \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0

ix) The solution of the differential equation \frac{d y}{d x}=\frac{x+2 y}{x} is x+y=k x^{2}.

 

x) Solution of  \frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right) \text { is } \sin \frac{y}{x}=C x

xi) The differential equation of all non horizontal lines in a plane is  \begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}

i) Integrating factor of the differential of the form \frac{d x}{d y}+P_{1} x=Q_{1}$ is given by e^{\int P_{1}d y}$. Hence given statement is true.

 

(ii) Solution of the differential equation of the type \frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}$  is given by x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$. 

Hence given statement is true.

iii) Correct substitution for the solution of the differential equation of the type \frac{d y}{d x} f(x, y),$ where $f(x, y)$ is a homogeneous function of zero degree is  y=v x. 

Hence given statement is true.

(iv) Correct substitution for the solution of the differential equation of the type \frac{d x}{d y} g(x, y)where g(x, y) is a homogeneous function of the degree zero is x=v y.

Hence given statement is true.

(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is Flase.

(vi) In thegiven equation x^{2}+(y-a)^{2}=$ $a^{2}$ the number of arbitrary constant is one. So the order order will be one. 

Hence given statement is False.

 
(vii) \frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3}

\frac{d y}{y\frac{1}{3}}=\left(\frac{dx}{x^\frac{1}{3}}\right)

\int \frac{d y}{y\frac{1}{3}}=\int \left(\frac{dx}{x^\frac{1}{3}}\right)

\begin{array}{l} \frac{3}{2} y^{2 / 3}=\frac{3}{2} x^{2 / 3}+C^{\prime} \\ y^{2 / 3}-x^{2 / 3}=C \end{array}

Hence the given statement is true.
(viii)  y=e^{x}(A \cos x+$$B \sin x$ )

\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + e^x(A\sin x + B \cos x)

\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + y

\frac{d^2 y}{d x^2}=e^x(-A\sin x + B \cos x) + e^x(-A\cos x - B \sin x)+\frac{dy}{dx}

\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0.

Hence the given statement is true.

ix) Given:  \frac{d y}{d x}=\frac{x+2 y}{x}

\frac{d y}{d x}-\frac{2}{x} y=1

Compare with \frac{d y}{d x}+P_{1} y=Q_{1}$

Here P_{1} = \frac{-2}{x}, Q_{1} = 1

I.F. = e^{\int \frac{-2}{x}dx} = e^{\log \frac{1}{x^2}} = \frac{1}{x^2}

General solution 

y. \frac{1}{x^2} = \int \frac{1}{x^2}dx

\frac{y}{x^2} = \frac{-1}{x}+c

y+x= cx^2

Hence the given statement is true.

x) Given: \frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right)

\frac{ d y}{d x}=\frac{y}{x}+ \tan \left(\frac{y}{x}\right)

Let y =vx

\frac{ d y}{d x}=v+x\frac{dv}{dx}

v+x\frac{dv}{dx} = v+\tan v

x\frac{dv}{dx} = \tan v

\int \frac{dv}{\tan v} = \int \frac{dx}{x}

\log \sin v = \log x + \log c

\sin v = xc

\sin \frac{y}{x} = cx

Hence the given statement is true.

xi) Assume equation of a non-horizontal line in the plane

y = mx +c

\frac{dy}{dx} = m

\frac{dx}{dy} = \frac{1}{m}

 \begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}

Hence the given statement is true.

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State True or False for the statements in the Exercise.

 If A, B and C are three independent events such that P(A) = P(B) = P(C) = p, then P (At least two of A, B, C occur) =3p^2 - 2p^3

True

Let A, B,C be the occurrence of events A,B and C and A’,B’ and C’ not occurrence.
 P(A) = P(B) = P(C) = p and P(A’) = P(B’) = P(C’) = 1-p
P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C) + P(A ∩ B ∩ C)
events are independent:

\mathrm{P}$ (At least two of $\mathrm{A}, \mathrm{B},$ C occur $)=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{A}\right) \mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})=3 \mathrm{p}^{2}(1-\mathrm{p})+p^3$
=3 p^{2}-2p^{3}$
Hence, statement is true.

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State True or False for the statements in the Exercise.
If A and B are two events such that P(A) > 0 and P(A) + P(B) >1, then
\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \geq 1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}

True
\operatorname{As}, \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$ \\$\Rightarrow P(B \mid A)=\frac{P(A)+1-P\left(B^{\prime}\right)-P(A \cup B)}{P(A)}$ \\$\Rightarrow \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}+\frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$
The above equation means that:
\\P(B \mid A) \geq 1-\frac{P\left(B^{\prime}\right)}{P(A)}$ \\$\frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$
we need to add \mathrm{P}(\mathrm{A}) \quad$ to get the equal term
$\Rightarrow$ LHS is greater.

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State True or False for the statements in the Exercise.
If A and B are independent, then
P (exactly one of A, B occurs) = P(A)P(B′)+P(B) P(A′)

TRUE   

If A and B are independent events, that means
\\ P(A \cap B)=P(A) P(B) \\ P\left(A^{\prime} \cap B\right)=P\left(A^{\prime}\right) P(B) \\ \text { And } P\left(A \cap B^{\prime}\right)=P(A) P\left(B^{\prime}\right) \\ P(\text { exactly one of } A, B \text { occurs })=P\left(A^{\prime} \cap B\right)+P\left(A \cap B^{\prime}\right) \\ \Rightarrow P(\text { exactly one of } A, B \text { occurs })=P\left(A^{\prime}\right) P(B)+P(A) P\left(B^{\prime}\right)
∴ Statement is true.

 

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State True or False for the statements in the Exercise.
If A and B are independent events, then P(A′ ∪ B) = 1 – P (A) P(B′)

 

TRUE

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)
P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)
and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.

\\ \therefore P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B) \\ \Rightarrow P\left(A^{\prime} \cup B\right)=P\left(A^{\prime}\right)+P(B)-P(B)+P(A \cap B) \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A)+P(A) P(B)\{\text { independent events }\} \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A)\{1-P(B)\} \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A) P\left(B^{\prime}\right)

Hence Proved

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State True or False for the statements in the Exercise.
Another name for the mean of a probability distribution is expected value.

TRUE

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

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State True or False for the statements in the Exercise.
If A and B are two independent events then P(A and B) = P(A).P(B)

TRUE

If A and B are independent events it means that

P(A ∩ B) = P(A)P(B)
Thus, from the definition of independent event we say that statement is true.

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State True or False for the statements in the Exercise.
Two independent events are always mutually exclusive.

 

False

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)
From the equation it cannot be proved that

P(A∪B) = P(A) + P(B)
It is only possible if either P(A) or P(B) = 0, which is not given in question.

Hence, it is a false statement.

 

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State True or False for the statements in the Exercise.
If A and B are mutually exclusive events, then they will be independent also.

False

If A and B are mutually exclusive, that means

P(A∪B) = P(A) + P(B)
From this equation it cannot be proved that
P(A ∩ B)= P(A)P(B).
Hence, it is a false statement.

 

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State True or False for the statements in the Exercise.
If A and B are independent events, then A′ and B′ are also independent.

TRUE

As A and B are independent

\begin{aligned} &\Rightarrow P({A} \cap {B})=P(A) P(B)\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}({\mathrm{A}} \underline{\cup} \underline{\mathrm{B}})^{\prime}\{\text { using De morgan's law }\}\\ &P(A \cup B)^{\prime}=1-P(A \cup B)\\ &\text { We know } P(A \cup B)=P(A)+P(B)-P(A \cap B)\\ &\Rightarrow P(A \cup B)^{\prime}=1-[P(A)+P(B)-P(A \cap B)]\\ &\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \text { as } \mathrm{A} \& \mathrm{~B} \text { are independent }\}\\ &=[1-\mathrm{P}(\mathrm{A})]-\mathrm{P}(\mathrm{B})(1-\mathrm{P}(\mathrm{A})]\\ &\Rightarrow P\left(A^{\prime} \cap B^{\prime}\right)=(1-P(A))(1-P(B))\\ &=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \end{aligned}
hence proved

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