Here the component statements p and q are given by,
p: Today is Monday; q: 4 + 4 = 8.
The truth value of p→q is T because the conclusion q is T.
An important point is that p→q should not be treated by actually considering the meanings of p and q in English. Also it is not necessary that p should be related to q at all.

Given R (x, y) = 80 x + 90 y + 0.04xy - 0.05 x² + 0.05 y² and
(x,y) = 8x + 6y + 2000
Profit function P (x,y) = Revenue - cost
P (x,y) = R (x,y) - C (x,y)
= -80 x + 90 y + 0.04xy - 0.05 x² - 0.05y² - 8x - 6y - 2000
P (x, y) = 72x + 84y + 0.04 xy - 0.05 x² - 0.05y² - 2000
(ii) \(\frac { { \partial P } }{ \partial { x } } \) = 72 + 0 + 0.04y - 0.05(2x) - 0 - 0
= 72 + 0.04y- 0.1x
∴ \(\frac { { \partial P } }{ \partial { x } } \) (1200, (1800)
= 72+ 0.04 (1800) - 0.1(1200)
= 72 + 72 - 120 = 24 .......(1)
\(\frac { \partial v }{ \partial y} \) = 0 + 84+ 0.4x-0-0.5(2y) - 0
= 84 + 0.04x - 0.1y
= 84 + 0.04 (1200) - 0.1(1800)
∴ \(\frac { \partial v }{ \partial y} \)_(1200,1800) = 84 + 48 - 180 = - 48 .......(2)
From (1) and (2), keeping y constant and 4 increasing x then increases profit.

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

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) Differentiating F(x) with respect to x at continuity points of f(x), we get
\(f(x)={ F }^{ 1 }(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & x\ge 1 \end{matrix} \end{cases}\) 
The pdf f(x) is not continuous at x = 0, or at x = 1. We can define f(0) and f(1) in any manner. Choosing f(0) = 1, and f(1) = 0 .
Therefore the probability density function f(x) is
\(f(x)=\begin{cases} \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(ii) P(0.2 ≤ X ≤ 0.7) = F(0.7) − F(0.2)
= 0.7-0.2 = 0.5
\(P(0.2\le X\le 0.7)=\int _{ 0.2 }^{ 0.7 }{ f(x) } dx=\int _{ 0.2 }^{ 0.7 }{ 1dx } =0.5\)

Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ 2\pi }{ { x }^{ 2 } } sin\quad nx\quad dx={ \left[ \left( { x }^{ 2 } \right) \left( -\frac { cos\quad nx }{ n } \right) -(2x)\left( -\frac { sin\quad nx }{ { n }^{ 2 } } \right) +(2)\left( \frac { cos\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ 2x }\)
\(=\left[ (4{ \pi }^{ 2 })\left( -\frac { 1 }{ n } \right) -0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] -\left[ 0-0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] \) since cos 2n\(\pi\) = 1 and sin 2n \(\pi\) = 0
\(=-\frac { 4{ \pi }^{ 2 } }{ n } +\frac { 2 }{ { n }^{ 3 } } -\frac { 2 }{ { n }^{ 3 } } =\frac { { 4\pi }^{ 2 } }{ n } \)

Given a = edge of the circle
= 10 cm and da
= 0.2 cm
Volume of cube = a³
Approximate amount of cubic centimeters of paint is used to paint this cube = 3a² da
= 3(10²)(0.2)
= 300 \(\left( \frac { 2 }{ 10 } \right) \) = 60 cm³
Exact amount of paint used = f(x + ∆x) - f(x)
= f(10.2) - f(10)
= 10.2³ -10³
= 1061.208 - 1000
= 61.208 cm²

Actual radius of the circular plate = 12.5 cm
Measured radius of the circular plate = 12.65
dr = 12.65-12.5
= 0.15
\( \mathrm{A} =\pi \mathrm{r}^{2} \\ \mathrm{dA} =2 \pi \mathrm{rdv} \)
Change in Area
A(12.65)-A(12.5) = dA
\( =2 \pi \times 12.5 \times 0.15 \)
\(=3.75 \pi \)
Absolute error = 3.7725\(\pi\) - 3.75\(\pi\)
:0.0225\(\pi\) cm²

\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2 }{ { x }^{ 2 }-1 } -\frac { x }{ x-1 } \right) =\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-x(x+1) }{ { x }^{ 2 }-1 } \right) \)
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } \)
Form which is indeterminate,
Applying L' Hopital rule we get,
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } =\frac { -2-1 }{ 2(1) } =\frac { -3 }{ 2 } \)

Let f (t) represents the distance covered at 't' hour.
Given f(0) = 20 and f(2) = ?
Also speed = f' (t) ≥ 150
The distance function is continuous as well as differentiable.
By Lagrange's mean value theorem, there exists c such that
⇒ f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ f'(c) = \(\frac{f(2)-20}{2-0}\) ≥ 150 [Man speed is 150 km/hr and f' (c) represents speed]
⇒ \(\frac{f(2)-20}{2-0}\) ≥ 150
⇒ f(2) - 20 ≥ 300
⇒ f(2) ≥ 320
Hence, in the next two hours he can cover 320km.

\(f'\left( x \right) =6\times \frac { 4 }{ 3 } { x }^{ \frac { 4 }{ 3 } -1 }-3\times \frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } -1 }\) 
= \({ 8x }^{ \frac { 1 }{ 3 } }-x^{ \frac { -2 }{ 3 } }\) 
f'(x) = 0
\(\Rightarrow{ 8 }x^{ \frac { 1 }{ 3 } }-\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow \frac { 8x-1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow x=\frac { 1 }{ 8 } \)
Thus, the critical number is \(x=\frac { 1 }{ 8 } \) 
Evaluating f(x) at the end points = -1
x = 1 and at the critical number x = \(\frac { 1 }{ 8 } \) 
we get
\(f(-1)=6(-1)^{ \frac { 4 }{ 3 } }-3\left( -1 \right) ^{ \frac { 1 }{ 3 } }\) 
= 6( 1) - 3 (-1) = 6 + 3 = 9
\(f(1)=6(1)^{ \frac { 4 }{ 3 } }-3(1)^{ \frac { 1 }{ 3 } }=6-3=3\)
\(f\left( \frac { 1 }{ 8 } \right) =6\left( \frac { 1 }{ 8 } \right) ^{ \frac { 4 }{ 3 } }-3\left( \frac { 1 }{ 8 } \right) ^{ \frac { 1 }{ 3 } }\) 
=\(6\left( { 2 }^{ -3 } \right) ^{ \frac { 4 }{ 3 } }-3\left( 2^{ -3 } \right) ^{ \frac { 1 }{ 3 } }\) 
= \(\frac { 6 }{ 16 } -\frac { 3 }{ 2 } =\frac { 3 }{ 8 } -\frac { 3 }{ 2 } \) 
= \(\frac { 3-12 }{ 8 } =\frac { -9 }{ 8 } \)
From these values, the absolute maximum is 9 which occurs at x = -1 and the absolute minimum is \(-\frac { 9 }{ 8 } \) which occurs at x = \(\frac { 1 }{ 8 } \)

The given equation is rewritten as \(\frac { dy }{ dx } =\frac { { y }^{ 2 } }{ xy-{ x }^{ 2 } } \)
This is a homogeneous differential equation 
Put y = vx . Then, we have \(x\frac { dv }{ dx } =\frac { v }{ v-1 } \)
By separating the variables, \(\frac { v-1 }{ v } dv=\frac { dx }{ x } .\)
Integrating, we obtain v − log |v| = log |x| + log |C| or v = log |vxC|.
Replacing v by \(\frac{y}{x}\), we get, \(\frac{y}{x}\) = log |Cy| = e^y/x or y = ke^y/x (how!) which is the required solution.

The given function is y = a cos(log x) + bsin (log x)   ...(1)
where a, b are two arbitrary constants. In order to eliminate the two arbitrary constants, we have to differentiate the given function two times successively
Differentiating equation (1) with respect to x , we get 
y' = -a sin (log x).\(\frac{1}{x}\)+ b cos (log x).\(\frac{1}{x}\Rightarrow\)xy' = -a sin(log x)+b cos (log x).
Again differentiating this with respect to x, we get
xy" + y' = -a cos(log x).\(\frac{1}{x}-b\) sin (log x).\(\frac{1}{x}\Rightarrow\)x²y" + xy'+ y = 0
Therefore, y = a cos(log x) + bsin (log x) is a solution of the given differential equation.

Given m (x) = \(\sqrt { 3 } x\) = \(\sqrt { 3 } .{ x }^{ \frac { 1 }{ 2 } }\)
Differentiating with respect to 'x' we get,

when x = 3, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 3 } } =\frac { 1 }{ 2 } \) Kg/m
when x = 27, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 27 } } \)
\(=\frac{\sqrt{\not 3}}{2(3) \sqrt{\not 3}}=\frac{1}{6} \mathrm{Kg} / \mathrm{m}\)

\(cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right) \)
\(\Rightarrow \frac { 4 }{ 5 } =sinx\)

\(\therefore cosx=\frac { adj }{ hyp } =\frac { 3 }{ 5 } \)
Let \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) =y\)
\(\Rightarrow tany=\frac { 3 }{ 4 } \)

\(\Rightarrow siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \ cosy=\frac { adj }{ hup } =\frac { 4 }{ 5 } \)
\(\therefore cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
= cos (x - y) = cos x cos y + sin x sin y
= \(\frac { 3 }{ 5 } .\frac { 4 }{ 5 } +\frac { 4 }{ 5 } .\frac { 3 }{ 5 } \)
= \(\frac { 12 }{ 25 } +\frac { 12 }{ 25 } =\frac { 24 }{ 25 } \)
= \(\frac { 24 }{ 25 } \)

(z₁ + z₂)z₃ = z₁z₃ + z₂ + z₃
LHS = (z₁ + z₂)z₃
= (3-7i)(5 + 4i)
15+12i-35i-28i²
= 15-23i + 28
= 43 - 23i
RHS = z₁z₃ + z₂z₃
3(5+4i)+(-7i)(5+4i)
= 15+12i -35i -28i²
= 15-23i + 28
= 43 - 23i
LHS = RHS
∴ (z₁+z₂)z₃ = z₁z₃ + z₂z₃

Equation of the given plane is 2x - 3y + 5z +7 = 0
Equation of the plane parallel to the given plane
is 2x - 3y + 5z + k = 0  ..(1)
Since this plane passes through the point (3, 4, -1). we get
2(3) - 3(4) + 5(-1) + k = 0
\(\Rightarrow\) 6 - 12 - 5 + k = 0
\(\Rightarrow\) 11 +k = 0  

k = 11
\(\therefore\) (1) becomes, 2x - 3y + 5z + 11 = 0 which is the equation of the required plane. Distance between two parallel planes.
= \(\frac { \left| { d }_{ 1 }-{ d }_{ 2 } \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \)
\(d=\frac { \left| 7-11 \right| }{ \sqrt { { 2 }^{ 2 }+\left( -3 \right) ^{ 2 }+{ 5 }^{ 2 } } } \)
= \(\cfrac { \left| -4 \right| }{ \sqrt { 4+9+25 } } \)
= \(\cfrac { 4 }{ \sqrt { 38 } } \) units.

Given lines are \(\frac { x-5 }{ 5m+2 } =\frac { 2-y }{ 5 } =\frac { 1-z }{ -1 } \)
⇒ \(\frac { x-5 }{ 5m+2 } =\frac { y-2 }{ -5 } =\frac { z-1 }{ 1 } \)
∴ \(\vec { b } =(5m+2)\hat { i } -5\hat { j } +\hat { k } \) ...(1)
and x = \(\frac { 2y+1 }{ 4m } =\frac { 1-z }{ -3 } \)
⇒ \(\frac { x }{ 1 } =\frac { y+\frac { 1 }{ z } }{ 2m } =\frac { z-1 }{ 3 } \)
∴ \(\vec { d } =\hat { i } +2m\hat { j } +3\hat { k } \) ...(2)
since \(\vec { b } \bot \vec { d } \Rightarrow \vec { b } .\vec { d } \)= 0
⇒ \((\hat { i } +2m\hat { j } +3\hat { k } ).((5m+2)\hat { i } -5\hat { j } +\hat { k } )=0\)
⇒ (5m+2)1+2m(-5)+3(1) = 0
⇒ 5m+2-10m+3 = 0
⇒ 5-5m = 0
⇒ 5 = 5m
∴ m = 1

Let the vertex be (0, 0) .
The equation of the parabola is
y² = 4ax
(1) Since the diameter is 40cm and the depth is 30 cm , the point (30, 20) lies on the parabola.
20² = 4a × 30
4a = \(\frac { 400 }{ 30 } \) = \(\frac { 40 }{ 3 } \)
Equation is y² = \(\frac { 40 }{ 3 } \)x.
(2) The bulb is at focus (0, a).
Hence the bulb is at a distance of \(\frac { 10 }{ 3 } \)cm from the vertex.

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

Let z = x + iy
Then \(\frac { 1+z }{ 1-z } \) = cos2θ + i sin 2θ
⇒ \(\frac { 1+x+iy }{ 1-x-iy } \) = cos 2θ + i sin 2θ ....(1)
Taking modulus,
\(\left| \frac { 1+x+iy }{ 1-x-iy } \right| \) = |cos 2θ+i sin 2θ| ⇒ \(\frac { |1+x+iy| }{ |1-x-iy| } \)
=\(\sqrt { cos^{ 2 }2\theta +sin^{ 2 }2\theta } \) = 1
⇒ |1 + x + iy| = |1 - x - iy|
⇒ \(\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } =\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } \)
⇒ (1 + x)²+ y² = (1-x)²+ y²

⇒ 4x = 0 ⇒ x = 0
From (1) \(\frac { (1+x)+iy }{ (1-x)-iy } \times \frac { (1-x)+iy }{ (1-x)+iy } \)
= cos2θ + isin 2θ
Choosing the imaginary part alone we get,
\(\frac { y(1+x)+y(1-x) }{ (1-x)^{ 2 }+{ y }^{ 2 } } \)= sin 2θ

\(\frac { 2y }{ 1+y^{ 2 } } \) = sin 2θ
⇒ \(\frac { 2tan\theta }{ 1+tan^{ 2 }\theta } \) = sin 2θ
∴ y must be equal to tan θ
⇒ y = tan θ
z = x+ iy
∴ z = 0 + i tan θ
⇒ z = tan θ

Let OACB be a rhombus. Taking O as the origin, let the position vectors of A and B be \(\vec { a } \) and \(\vec { b } \) respectively.
Then \(\vec { OA } =\vec { a } \) and \(\vec { OB } =\vec { b } \)  [∵ \(\vec { AC } =\vec { OB } \)]
So, the p.v. of C is \(\vec { a } +\vec { b } \)
∴ Position vector O f the miid-point of OC is \(\frac { \vec { a } +\vec { b } }{ 2 } \)
Similarly, the position vector of mid-point of AB is \(\frac { \vec { a } +\vec { b } }{ 2 } \).
Hence, the mid-point of OC coincides with the mid-point of AB.
Now, \(\vec { OC } .\vec { AB } =(\vec { a } +\vec { b } ).(\vec { b } -\vec { a } )=|\vec { b } |^{ 2 }-|\vec { a } |^{ 2 }\)
= OB²- OA² = 0   [∵ OB = OA]
⇒ \(\vec { OC } \bot \vec { AB } \).
Hence, the diagonals of a rhombus bisect each other at right angles.

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

Given A =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂

With usual notations in triangle ABC, let \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \)  and  \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Applying dot product, we get
\(\vec { BC } .\vec { BC } =-\vec { BC } .\vec { CA }-\vec { BC }. \vec { AB } \)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }=-\left| \vec { BC } \right| \left| \vec { CA } \right| \) cos(兀-c)-\(\left| \vec { BC } \right| \left| \vec { AB} \right| \)cos(兀-B)
⇒ a² = ab cos C + ac cos B
Therefore a = b cos C + c cos B
The results (ii) and (iii) are proved in a similar way

Given (3 -i) x - (2 - i) y + 2i + 5
= 2x + (-1 + 2i) y + 3 + 2i
⇒ 3x - ix - 2y + iy + 2i + 5 = 2x - y + 2iy + 3 + 2i
choosing the real and imaginary parts
(3x-2y + 5) + i (-x + y + 2) = 2x - y + 3 + i (2y+ 2)
Equating the real and imaginary parts both sides, we get
3x- 2y+ 5 = 2x-y+3
⇒ 3x - 2y + 5 - 2x +y - 3 = 0
⇒ x-y = -2... (1)
-x+y+2 = 2y+2
⇒ -x+y+2-2y-2 = 0
⇒ -x-y = 0 ⇒ x+y = 0.. (2)
(1)-(2) we get,

y = 1
Substituting y = 1 in (2) we get.
x+1 = 0 ⇒ x = -1
∴ x = -1 and y = 1

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

The ellipse is symmetric about both the axes. The major axis lies along x-axis. The region to be revolved is sketched
Hence, the required volume is given by
\(V=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx } =\pi \int _{ -a }^{ a }{ \frac { { b }^{ 2 } }{ { a }^{ 2 } } \left( { a }^{ 2 }-{ x }^{ 2 } \right) dx } \)
\(=\frac { 2\pi { b }^{ 2 } }{ { a }^{ 2 } } \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand is an even function
\(=\frac { { 2\pi b }^{ 2 } }{ { a }^{ 2 } } { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } ={ \left( { a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } \left( \frac { { 2a }^{ 3 } }{ 3 } \right) =\frac { 4\pi { ab }^{ 2 } }{ 3 } \)

Given
\(\\ \\ \\ \\ \\ f(x)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 }+1 }{ k } & forx=0,1,2 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\\ \\ \\ \\ \\ \\ \)
The random variable X take the values 0, 1, 2.
Probability mass function.

\(\sum _{ i=0 }^{ 2 }{ f(x_{ i })=1\Rightarrow f(0)+f(1)+f(2)=1 } \)
\(\Rightarrow \frac { 0+1 }{ k } +\frac { 1+1 }{ k } +\frac { 4+1 }{ k } \) 
\(\Rightarrow \frac { 1 }{ k } +\frac { 2 }{ k } +\frac { 5 }{ k } =1\) 
\(\Rightarrow \frac { 8 }{ k } =1\) 
\(\Rightarrow k=8\) 
(ii) Cumulative distribution function

\(F(0)=P(X<0)\\P(x=0)\\ =\frac { 1 }{ 8 } \) 
\(F(1)=P(X = 0)+ P(X = 1)\\ \frac { 1 }{ 8 } +\frac { 2 }{ 8 } =\frac { 3 }{ 8 } \)
\(F(2)=P(X= 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } +\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =1\) 
Cumulative distribution function is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & for & x\le 0 \end{matrix} \\ \begin{matrix} \frac { 2 }{ 8 } +\frac { 1 }{ 8 } & for & \frac { 3 }{ 8 } forx\le 1 \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } +\frac { 5 }{ 8 } =1 & for & x\le 2 \end{matrix} \end{cases}\)
(iii) p(x ≥ 1) = p(x = 1) + p(x = 2)
= \(\frac { 2 }{ 8 } +\frac { 5 }{ 8 } \)
\(p(x\ge 1)=\frac { 7 }{ 8 } \)

Taking \(\vec { p } =(\vec { a } \times \vec { b } )\) as a single vector and using the vector triple product expansion, we get 
\((\vec { a } \times \vec { b } )\times(\vec{c}\times\vec{d})=\vec{p}\times(\vec{c}\times\vec{d})\)
= \((\vec { p } .\vec { d } )\vec { c } -(\vec { p } .\vec { c } )\vec { d } \)
= \(((\vec { a } \times \vec { b } ).\vec { d } )\vec { c } -((\vec { a } \times \vec { b } ).\vec { c } )\vec { d } =[\vec { a } ,\vec { b } ,\vec { d } ]\vec { c } -[\vec { a } \vec { b } \vec { c } ]\vec { d } \)
Similarly, taking \(\vec { q } \) = \(\vec { c } \times \vec { d } \)
\((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=(\vec { a } \times \vec { b } )\times \vec { q } \)
= \((\vec { a } .\vec { q } )\vec { b } -(\vec { b } .\vec { q } )\vec { a } \)
\(= [\vec { a } ,\vec { c }, \vec d ]\vec { b } -[\vec { b } ,\vec { c },\vec d ]\vec { a } \)

Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 25 \\ 13 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 4 & 3 & 6 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 13 \\ 25 \\ 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & -1 & -13 \end{matrix}|\begin{matrix} 13 \\ -27 \\ -25 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 1 & 13 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 25 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 17{ R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 0 & 199 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 398 \end{matrix} \right] \).
The equivalent system is written by using the echelon form:
x + 5y + 7 = 13, … (1)
17y + 22z = 27, … (2)
199z = 398..... (3)
From (3), we get z = \(\frac { 398 }{ 199 } \) = 2.
Substituting z = 2 in (2), we get y = \(\frac { 27-22\times 2 }{ 17 } =\frac { -17 }{ 17 } \) = -1
Substituting z = 2, y = -1, in (1), we get x = 13 - 5 x (-1) - 7 \(\times\) 2 = 4.
So, the solution is (x = 4, y = -1, z = 2).

According to our notations, a_n= 2 and a₀ = 3.
If \(\frac{p}{q}\) is a zero of the polynomial, then as (p, q) = 1, p must divide 3 and q must divide 2.
Clearly, the possible values of p are 1, −1, 3, −3 and the possible values of q are 1, −1, 2, −2.
Using these p and q we can form only the fractions \(\pm \frac { 1 }{ 1 } ,\pm \frac { 1 }{ 2 } ,\pm \frac { 3 }{ 2 } ,\pm \frac { 3 }{ 1 } \). 
Among these eight possibilities, after verifying by substitution, we get \(\frac { -3 }{ 2 } \) is the only rational zero.
To find other roots, we divide the given polynomial 2x³+ 3x²+ 2x + 3 by 2x + 3 and get x²+1 as the quotient with zero remainder. Solving x²+1 = 0, we get i and −i as roots. Thus \(\frac { -3 }{ 2 } \), -i, i are the roots of the given polynomial equation.