~(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 multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

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
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

\((x+a)\frac { dy }{ dx } -2y={ (x+a) }^{ 4 }\)
\(\div (x+a)\)we get,
\(\frac { dy }{ dx } -\frac { 2 }{ x+a } y={ (x+a) }^{ 3 }\)
This is a linear differential equation
\(\therefore P=\frac { -2 }{ x+a } ;Q={ (x+a) }^{ 3 }\)
\(\int { pdx } =\int { \frac { -2 }{ x+a } } dx=-2log(x+a)\)
\(=log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 }\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 } }=\frac { 1 }{ { (x+a) }^{ 2 } } \)
The solution is \(\Rightarrow { ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y\left( \frac { 1 }{ { (x+a) }^{ 2 } } \right) =\int { { (x+a) }^{ 3 }\frac { 1 }{ (x+a{ ) }^{ 2 } } dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 } }{ 2 } +c\)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 }+2c }{ 2 } \)
\(2y={ (x+a) }^{ 4 }{ +2c(x+a) }^{ 2 }\)

Given p, q are the roots of lx² + nx + n = 0 
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S 
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S

Let p be the probability that indicates the defective rate of an electronic device
n = 10
\(P=5\%=0.05 \) 
q = 1 - p
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 - P(X < 1)
= 1-P(X = 0)
= 1-¹⁰C₀ (0.05)⁰ (0.95)¹⁰
P(X ≥1) = 1 - (0.95)¹⁰
(ii) Exactly two defective items  
P(X = 2) =¹⁰C₂(0.05)² (0.95)⁸

We have |2z-3-i| = 3
|2(x + iy)-3 - i| = 3
Squaring on both sides, we get
|(2x - 3) + (2y - 1)i|² = 9
\(\Rightarrow\) (2x - 3)² + (2y - 1)² = 9
\(\Rightarrow\) 4x² + 4y² -12x - 4y + 1 = 0, the locus of z in Cartesian form

Consider \(\frac { 19-7i }{ 9+i } =\frac { 19-7i }{ 9+i } \times \frac { 9-i }{ 9-i } \)
= \(\frac { 171-19i-63i+7i^{ 2 } }{ (9)^{ 2 }-{ i }^{ 2 } } =\frac { 171-82i-7 }{ 81+1 } \)
= \(\frac { 164-82i }{ 82 } =\frac { 82(2-i) }{ 82 } \) = 2- i
Also \(\frac { 20-5i }{ 7-6i } =\frac { 20-5i }{ 7-6i } \times \frac { 7+6i }{ 7+6i } \)
= \(\frac { 140+120i-35i-30^{ 2 } }{ { 7 }^{ 2 }-(6i)^{ 2 } } \)
\(=\frac{140+85 i+30}{49+36}=\frac{170+85 i}{85}=\frac{\not 85(2+i)}{\not 85}=2+i\)
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) = (2-i)12+(2+i)12
Let z = (2-i)¹²+(2+i)¹²
∴ \(\overline { z } \) = \(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
[∵ \(\overline { { z }_{ 1 }+{ z }_{ 2 } } =\overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } \)]
= (2 + i)¹²+(2 - i)¹²= z
∴ \(\overline { z } \) = z ⇒ z is purely real
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) is real.

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

order 2 ; degree 1

2

\(\frac { 1 }{ loga } -\frac { 1 }{ logb } \)

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 1\wedge 1 & 0\wedge 0 \end{matrix}\begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 0\wedge 0 & 1\wedge 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \) [∵ a៱b=max(a,b)]

(( p ∧ q) ∨ (¬r ∨¬s)) ∧ (¬ t ∧ v)) contains 6 variables p, q, r, s, t, and v. Hence the corresponding truth table will contain 2⁶ = 64 rows.

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

\(\theta =\frac { 3\pi }{ 2 } \)
When θ = \(\frac { 3\pi }{ 2 } \)
Rotation of z is \(ze^{ i\theta }=ze^{ i\frac { 3\pi }{ 2 } }\)
= \(2\sqrt { 2 } e^{ -\frac { \pi }{ 4 } }.e^{ i\frac { 3\pi }{ 2 } }=2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 2 } -\frac { \pi }{ 4 } \right) }\)
= \(2\sqrt { 2 } e^{ i5^{ \frac { \pi }{ 4 } } }\)