(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

(i) Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ - }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 0 }{ 0dx } +\int _{ 0 }^{ \infty }{ k{ e }^{ -2x }dx } =1\) 
\(0+k\left( \frac { { e }^{ -2x } }{ -2 } \right) =1\Rightarrow k\left( \frac { { e }^{ -\infty }-{ e }^{ 0 } }{ -2 } \right) =1\Rightarrow k=2\) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} 2{ e }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx \end{matrix}\le 0 \end{cases}\)
 (ii) Distribution function
By definition the distribution function \(F(x)=P\left( x\le x \right) =\int _{ -\infty }^{ x }{ f(u) } du\) 
When x≤0 \(F(x)=\int _{ -\infty }^{ x }{ F(u) } du=\int _{ -\infty }^{ x }{ odu=0 } \) 
When x > 0 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du\int _{ -\infty }^{ x }{ 0du } +\int _{ 0 }^{ x }{ { 2e }^{ -2x }du\left( \frac { { e }^{ -2x } }{ -2 } \right) } =1-{ e }^{ 2x }\) 
This gives \(F(x)=\begin{cases} \begin{matrix} 0 & forx\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ 2x } & forx>0 \end{matrix} \end{cases}\) 
(iii) P(X < ) = P(X ≤2 ) = F(2 ) = 1-e^2\(\times\)2 (since F(x) is continuous)
(iv) The probability that X is at least equal to four unit of time is
P(X ≥ 4 ) = 1 - P(X < 4 ) = 1- F( 4) = 1 - ( 1-e^{-2\(\times\)4}) = e⁸
(v) In the continuous case, f (x) at x = a is not the probability that X takes the value a, that is f (x) at x = a is not equal to P( X ) a.  If X is continuous type, P(X = a) = 0 for a ∈ R. Therefore P(x = 3) = 0.

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

We have,
\(f'(x)=2x>0, \forall x\in(2,7)\) and
\(f'(x)=2x>0, \forall x\in(-2,0)\)
and hence the proof is completed.

Let the two positive numbers be x and y.
Given xy = 20
\(\Rightarrow y=\frac { 20 }{ x } \) 
Let f(x) = x+y
\(f(x)=x+\frac { 20 }{ x } \) 
\(f'(x)=1-\frac { 20 }{ { x }^{ 2 } } \)
f'(x) = 0
\(\Rightarrow 1-\frac { 20 }{ { x }^{ 2 } } =0\) 
\(\Rightarrow 1=\frac { 20 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=20\) 
\(\Rightarrow x=\pm \sqrt { 20 } \) 
\(\Rightarrow x=\pm 2\sqrt { 5 } \)
∴ The critical number are \(2\sqrt { 5 } -2\sqrt { 5 } \) 
\(f''(x)=\frac { 40 }{ { x }^{ 3 } } \) 
When \(x=2\sqrt { 5 } \)
\(f''\left( x \right) =\frac { 40 }{ \left( 2\sqrt { 5 } \right) ^{ 3 } } >0\) 
ஃ f(x) is minimum when \(x=2\sqrt { 5 } \) 
When \(x=2\sqrt { 5 } ,y=\frac { 20 }{ 2\sqrt { 5 } } \) 
= \(\frac { 10 }{ \sqrt { 5 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\frac { 10\sqrt { 5 } }{ 5 } =2\sqrt { 5 } \) 
Hence the required positive numbers are \(2\sqrt { 5 } \)
Minimum sum = \(2\sqrt { 5 } \) + \(2\sqrt { 5 } \)  = \(4\sqrt { 5 } \)

Equation of the given curve is y² - 4xy = x² + 5..(1)
Differentiating with respect to 'x' we get,
\(2y\frac { dy }{ dx } -4\left[ x\frac { dy }{ dx } +y(1) \right] =2x\)
⇒ \(2y\frac { dy }{ dx } -4x\frac { dy }{ dx } -4y=2x\)
⇒ \(\frac { dy }{ dx } \) (2y -4x) = 2x + 4y
⇒ \(\frac { dy }{ dx } \) = \(\frac { x+2y }{ y-2x } \)
Since the tangent to the curve is horizontal, \(\frac { dy }{ dx } \) = 0 
ஃ \(\frac { x+2y }{ y-2x } \) = 0
⇒ x+ 2y = 0
⇒ x = -2y .....(2)
Substituting (2) in (1) we get,
y² - 4(-2y)y = (-1y)² + 5
⇒ y² + 8y² = 4y² + 5
⇒ y² = 4y² + 5
⇒ 5y² = 5
⇒ y² = 1
⇒ y = 土 1
From (2), When y = 1, x = - 2
When y = -1, x = 2
∴ The required points are (2, -1) and (-2,1)

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  |3z - 5| = 3 |z + 1
⇒ |3(x+iy)-5| = 3|x+iy+1|
⇒ |(3x-5)+3y| = 3|(x+1)+iy|
⇒ \(\sqrt { (3x-5)^{ 2 }+3^{ 2 } } =3\left[ \sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \right] \)
Squaring both sides we get,
(3x - 5)² + 9 = 9 [(x + 1)² + y²]
⇒ 9x² - 30x + 25 + 9 = 9 [x² + 2x + 1 + y²]
⇒ 48x - 16 = 0
⇒ 3x-1 = 0

\(1+i\sqrt { 3 } \)
\(r=||z|=\sqrt { { 1 }^{ 2 }+\left( \sqrt { 3 } \right) ^{ 2 } } \)
|\(\theta ={ tan }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 3 } \)
Hence \(ang(z)=\frac { \pi }{ 3 } \)
Therefore, the polar form of \(1+i\sqrt { 3 } \) can be written as
\(1+i\sqrt { 3 } =2\left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) \)
\(=2\left( cos\left( \frac { \pi }{ 3 } +2k\pi \right) +isin\left( \frac { \pi }{ 3 } +2k\pi \right) \right) ,k\varepsilon z\).

Here, \(\vec { a } =(2\hat { i } -\hat { j } +2\hat { k } ),\vec { b } =(3\hat { i } +4\hat { j } +2\hat { k } )\).
The vector form of the given plane is \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\). Then \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\) and p = 5
We know that the position vector of the point of intersection of the line  \(\vec { r } =\vec { a } +t\vec { b } \) and the plane
\(\vec { r } .\vec { d } =p\vec { u } =\vec { a } +\left( \frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } \right) \vec { b } \), where \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Clearly, we observe that \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Now, \(\frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } =\frac { 5-(2\hat { i } -\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) }{ (3\hat { i } +4\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) } =0\). Therefore, the position vector of the point of intersection of the given line and the given plane is
\(\hat { r } =(2\hat { i } -\hat { j } +2\hat { k } )+(0)(3\hat { i } +4\hat { j } +2\hat { k } )=2\hat { i } -\hat { j } +2\hat { k } \)
That is, the given straight line intersects the plane at the point (2, −1, 2)
Aliter:
The Cartesian equation of the given straight line is \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 2 } =t\)(say)
We know that any point on the given straight line is of the form (3t+2, 4 t−1, 2 t+2). If the given line and the plane intersects, then this point lies on the given pane x−y+z−5 = 0.
So, (3t + 2)−(4t − 1) + (2t + 2) − 5 = 0 ⇒ t = 0.
Therefore, the given line intersects the given plane at the point (2, -1, 2)

Let p(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x
p(x) has no sign change.
p(-x)⇒ (-x)⁹ + 9(-x)⁷ + 7(-x)⁵ + 5(-x)³ + 3(-x)
= -x⁹ - 9x⁷ - 7x⁵- 5x³ -3x
p(-x) also has no sign change
∴ p(x) has no positive and no negative root.

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

Existence of identity:
We have to find an element a' ∈ A such that
a*a' = a'*a = e
⇒ a + e-ae = a
⇒ a + e-ae = a
⇒ e-ae = 0
⇒ e(1-a) = 0
⇒ e = \(\frac{0}{1-a}\) = 0∈A
∴ A has an identity under *.
Existence of inverse:
For every a ∈ A, there exist a' ∈ A such that a*a' = a' * a = e
⇒ a+a'-aa' = 0  [∵e=0]
⇒ a+a'(1-a) = 0
⇒ a'(1-a) = -a
⇒ a' = \(\frac{-a}{1-a}\)
To prove that \(\frac{-a}{1-a}\)≠ 1
Suppose \(\frac{-a}{1-a}\) = 1
⇒ -a = 1-a
⇒ -a + a = 1 ⇒ 0 ≠ 1
∴ Our assumption is wrong
⇒ \(\frac{-a}{1-a}\) ≠ 1
∴ A has inverse for every element x ∈ A

(( p V q)∧ ~p)) ➝ q

The statement (( p V q)∧ ~p) ➝ q is a tautology.

Equation of the given curve is y = e^-2x
Required Volume = \(\pi \int _{ 0 }^{ 1 }{ { { (e }^{ -2x }) }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { e }^{ -4x } } dx=\pi { \left[ \frac { { e }^{ -4x } }{ -4 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { -\pi }{ 4 } \left[ { e }^{ -4 }-{ e }^{ -0 } \right] =-\frac { \pi }{ 4 } \left( { e }^{ -4 }-1 \right) \)
\(V=\frac { \pi }{ 4 } (1-{ e }^{ -4 })\) cubic units

Equation of AB is \(\frac { y-1 }{ 2-1 } =\frac { x+1 }{ 3+1 } or\quad y=\frac { 1 }{ 4 } (x+5)\)
Equation of BC is \(\frac { y-5 }{ 2-5 } =\frac { x-0 }{ 3-0 } or\quad y=-x+5\)
Equation of AC is \(\frac { y-1 }{ 5-1 } =\frac { x+1 }{ 0+1 } or\quad y=4x+5\)
\(\therefore\) Area of \(\Delta\)ABC = Area DACO+ Area of OCBE − Area of DABE
\(=\int _{ -1 }^{ 0 }{ (4x+5)dx+\int _{ 0 }^{ 3 }{ (-x+5)dx-\frac { 1 }{ 4 } \int _{ -1 }^{ 3 }{ (x+5)dx } } } \)
\(\\ \\ \\ ={ \left[ \frac { { 4x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 0 }+{ \left[ -\frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ 0 }^{ 3 }-\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 3 }\)
\(=0-(+2-5)+\left( -\frac { 9 }{ 2 } +15 \right) -0-\frac { 1 }{ 4 } \left[ \frac { 9 }{ 2 } +15 \right] +\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } -5 \right] =\frac { 15 }{ 2 } \)

Given u (x, y, z) = xyz; x = e^-t y = e^-t cas t; z = sin t
\(\frac { \partial u }{ \partial x } \) = yz; \(\frac { \partial u }{ \partial y } \) = xz; \(\frac { \partial u }{ \partial z } \) = xy
⇒ \(\frac { \partial u }{ \partial x } \) = e^t cas t sin t
\(\frac { \partial u }{ \partial y } \) = e^t sin t
\(\frac { \partial u }{ \partial z } \) = e^-2t cas t and
\(\frac{dx}{dt}=-e^{-t}\)
⇒ \(\frac{dy}{dt}\) = e^-t (- sin t) - cas t e^-t
⇒ \(\frac{dz}{dt}\) = cos t
∴ By chain rule;
\(\frac { dw }{ dt } =\frac { \partial w }{ \partial x } .\frac { dx }{ dt } +\frac { \partial w }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } +\frac { \partial u }{ \partial z } .\frac { dz }{ dt } \)
= e^-1 cos t sin t (-e^-t) + e^-t sin t (-e^-t) + e^-t sin t (-e^-t sin t - e^-t cos t)+ (-e-^2t) cos t(cos t)
= -e^-2t [(sin t cos t + sin² t + sin t cos t - cos² t]
= -e^-2t [2 sin t cos t - (cos² t - sin² t)]
\(\frac { du }{ dt } \) = - e^-2t, [sin 2t - cos2t]
[∵ cas 2 t = cos² t - sin² t and sin 2t = 2 sin t cos t]

(1) The domain of f (x) is R \{−1,1}.
(2) Since f (−x,−y) = f (x, y) , the curve is symmetric about the origin.
(3) Putting y = 0, we get x = 0 . Hence the x -intercept is (0, 0).
(4) Putting x = 0, we get y = 0. Hence the y -intercept is (0, 0).
(5) To determine monotonicity, we find the first derivative as \(f'(x)\frac { -3\left( { x }^{ 2 }+1 \right) }{ { ({ x }^{ 2 }-1) }^{ 2 } } \)
Hence, f'( x) does not exist at x = −1,1. Therefore, critical numbers are x = −1, 1. The intervals of monotonicity is tabulated

6) Since there is no sign change in f'( x) when passing through critical numbers. There is no local extrema.
(7) To determine the concavity, we find the second derivative as \(f"(x)=\frac { 6x({ x }^{ 2 }-3) }{ { ({ x }^{ 2 }-1) }^{ 3 } } \) f"( x) = 0\(\Rightarrow\) x = 0 and f"(x) does not exist at x = −1, 1.
The intervals of concavity is tabulated.

(8) As x = −1 and 1are not in the domain of f(x) and at x = 0, the second derivative is zero and f"(x) changes its sign from positive to negative when passing through x = 0. Therefore, the point of inflection is (0, f (0)) = (0,0) .

(9) \(\underset { x\rightarrow \pm \infty }{ lim } f(x)=\underset { x\rightarrow \pm }{ lim } \cfrac { 3x }{ { x }^{ 2 }-1 } =\underset { x\rightarrow \pm \infty }{ lim } \cfrac { 3 }{ x\frac { 1 }{ x } } =0\) Therefore y = 0 is a horizontal asymptote.
Since the denominator is zero, when x = 士1.\(\underset { x{ \rightarrow -1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow -1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty ,\underset { x{ \rightarrow 1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow 1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty .\)
Therefore x = −1 and x = 1 are vertical asymptotes.
The rough sketch of the curve is shown on the right side.

The given function is defined and is differentiable at all \(x \in(0, \infty)\) 
f(x) = x log x + 3x.
Therefore f'(x) = log x+1+3 = 4 + log x.
The stationary points are given by 4 + log x = 0
That is x = e^-4
Hence the intervals of monotonicity are (0, e^-4) and \((e^{-4}, \infty)\) 
At \(x=e^{-5}\in(0,e^{-4})\), f'(e^-5) = -1<0 and hence in the interval (0, e^-4) -the function is strictly decreasing.
At \(x=e^{-3}\in(0,e^{-4})\), f'(e^-3) = 1>0 and hence strictly increasing in the interval \((e^{-4}, \infty)\). 
Since f′(x) changes from negative to positive when passing through x = e⁻⁴, the first derivative test tells us there is a local minimum at x = e^-4 and it is f(e-⁴) = -e^-4.

Let x (t) denote the amount of salt in the tank at time t. Its rate of change is \(\frac{dx}{dt}\) = in flow rate − out flow rate
Now, 5 grams times 10 litres gives an inflow of 50 grams of salt. Also, the out flow of brine is 10 litres per minute. This is 10 /1000 = 0.01of the total brine content in the tank. Hence, the out flow of salt is 0.01 times x(t) , that is 0.01x(t).
Thus the differential equation for the model is \(\frac{dx}{dt}\)= 50 − 0.01x = −0.01(x − 5000)
This can be written as \(\frac{dx}{x-5000}=-(0.01)dt\)
Integrating both sides, we obtain log |x − 5000| = −0.01t + log C
or x-5000 = Ce^-0.01t or x = 5000+Ce^-0.01t
Initially, whent = 0, x = 100 , so 100 = 5000 +C .Thus, C = −4900 .
Hence, the amount of the salt in the tank at time t is x = 5000 -4900e^-0.01

The given linear differential euation is of the form
\(\frac{d y}{d x}+P y=Q  \)
 where \(\mathrm{P}=\frac{1}{(1-x) \sqrt{x}} ; \mathrm{Q}=1-\sqrt{x}  \)
Take \(\sqrt{x}=\mathrm{t} \Rightarrow \mathrm{t}^2=x \)
\( x^{1 / 2}=\mathrm{t}  \)
\( \frac{1}{2} x^{1 / 2-1} \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{2} x^{-1 / 2} d x=\mathrm{dt} \)
\( \frac{1}{2 x^{1 / 2}} d x=\mathrm{dt} \Rightarrow \frac{1}{2 \sqrt{x}} d x=\mathrm{dt}  \)
\( \frac{d x}{\sqrt{x}}=2 \mathrm{dt} \)
\( \text { I.F }=e^{\int P d x}=e^{\int \frac{1}{(1-x) \sqrt{x}} d x}  \)
\( =e^{\int \frac{1}{1-t^{t^2} 2 d t}}=e^{\int \frac{2 d t}{1-1^2}} \)
\(\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 x} \log \left|\frac{a+x}{a-x}\right|  \) 
\(\text { Here } a=1 ; x=t\)
\(=e^{\log \left(\frac{1+1}{1-t}\right)} \)
\(\mathrm{I} . \mathrm{F}=\frac{1+t}{1-t} \)
\({ I.F }=\frac{1+\sqrt{x}}{1-\sqrt{x}} \)
The solution is y\(\times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times \mathrm{I}.Fd x+c\)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=\int(1-\sqrt{x}) \frac{1+\sqrt{x}}{1-\sqrt{x}} d x+c  \)
\( =\int(1+\sqrt{x}) d x+c  \)
\(  y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{x^{3 / 2}}{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x^{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x \sqrt{x}+c \quad \because x^{3 / 2}=x \sqrt{x} \)
Which is the required solution

Let f(x) = 8x³- 2x² - 7x +3 = 0
Here sum of the co-efficients of odd terms = 8 - 7 = 1
and sum of the co-efficients of even terms = -2 +3 = 1
Hence, x = - 1 is a root of f(x)
Let us divide f(x) by (x + 1)

∴ The other factor is 8x² - 10x + 3
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-4(8)(3) } }{ 2\times 8 } \)
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-96 } }{ 16 } \Rightarrow x=\frac { 10\pm 2 }{ 16 } \)
\(\Rightarrow x=\frac { 12 }{ 16 } \ or \)
\(x=\frac { 8 }{ 16 } \Rightarrow x=\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } \)
∴ The roots are -1, \(\frac{1}{2},\frac{3}{4}\)

(x-4)(x-7)(x-2)(x+1) = 16
Given (x - 4)(x -7)(x - 2)(x + 1) = 16
Rearrange the terms as
(x - 4)(x - 2)(x -7)(x + 1) = 16
⇒ (x² - 6x + 8)(x2 - 6x -7) = 16
Put x²-6x = y
⇒ (y + 8) (y - 7) = 16
⇒ y² +y - 56 - 16 = 0
⇒ (y+9)(y-8) = 0
⇒ y = -9, 8

case (i)
 When y = -9
⇒ x²-6x = -9
⇒ x²-6x+9 = 0
⇒ (x-3)² = 0
⇒ x = 3, 3

case ii
When y = 8,
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-4(1)(-8) } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-32 } }{ 2 } \Rightarrow x=\frac { 6\pm \sqrt { 68 } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm 2\sqrt { 17 } }{ 2 } \Rightarrow x=\frac { 2(3\pm \sqrt { 17 } ) }{ 2 } \)
\(\Rightarrow x=3\pm \sqrt { 17 } \)
∴ The roots are 3, 3, 3 +\(\sqrt{17}\), and 3 -\(\sqrt{17}\)

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

Given |z₁| = 1, |z₂|= 2, |z₃| = 3, |z₁ + z₂ +  z₃| = 1
|z₁|² = 1² ⇒ z₁ \(\overline { { z }_{ 1 } } \) = 1 ⇒ z₁ = \(\frac { 1 }{ { z }_{ 1 } } \)
|z₂|² = 4 ⇒ z₂ \(\overline { { z }_{ 2 } } \) = 1 ⇒ z₂ = \(\frac { 4 }{ { z }_{ 2 } } \)
|z₃|² = 9 ⇒ z₃ \(\overline { { z }_{ 3 } } \) = 1 ⇒ z₃ = \(\frac { 9 }{ { z }_{ 3 } } \)
∴ \(\left| 9,\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 4 }{ \overline { { z }_{ 2 } } } +4.\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 9 }{ \overline { { z }_{ 3 } } } +\frac { 4 }{ \overline { { z }_{ 2 } } } .\frac { 9 }{ \overline { z_{ 3 } } } \right| \)
\(\left| \frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } } +\frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 3 } } } +\frac { 36 }{ \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right| =\left| 36\left( \frac { \overline { { z }_{ 3 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 1 } } }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right) \right| \)
\(\\ \left[ \because |\overline { z_{ 1 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 3 } } |=|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | \right] \)
=\(\frac { 36|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } =36\frac { |\overline { \overline { { z }_{ 1 } } +\overline { z_{ 2 } } +\overline { z_{ 3 } } | } }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } \)
\(\left[ \because |\overline { { z }_{ 1 } } |=|{ z }_{ 1 }|,|\overline { { z }_{ 2 } } =|{ z }_{ 21 }|,|\overline { { z }_{ 3 } } |=|\overline { { z }_{ 3 } } | \right] \)
\(=\frac{36(1)}{1(2)(3)}=\frac{\not 36}{\not 6}=6\)
∴ |9z₁ + z₂ + 4z₁z₃ + z₂z₃| = 6

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.

Completing the square on x and y of 4x²+36y²+40x−288y+532 = 0,
4(x² + 10x + 25 − 25) + 36(y² − 8y + 16 − 16) + 532 = 0 , gives
4(x² + 10x + 25) + 36(y² − 8y + 16) = −532 + 100 + 576
4(x + 5)² + 36(y − 4)² = 144.
Dividing both sides by 144, the equation reduces to \(\frac { { \left( x+5 \right) }^{ 2 } }{ 36 } \frac { { \left( y-4 \right) }^{ 2 } }{ 4 } =1\)
This is an ellipse with centre (-5, 4), major axis is parallel to x-axis, length of major axis is 12 and length of minor axis is 4. Vertices are (1, 4) and (-11, 4).
Now, c² = a²−b² = 36 − 4 = 32
and c = ±4 \(\sqrt { 2 } \)
Then the foci are (−5 − 4\(\sqrt { 2 } \), 4) and (−5 + 4\(\sqrt { 2 } \) , 4).
Length of the major axis = 2a = 12 units and
the length of the minor axis = 2b = 4 units.

Equation of the circle is x² + y² − 6x + 6y − 8 = 0
∴ Equation of the tangent at (x₁, y₁) is
xx₁ + yy₁ - \(\frac 62\) (x + x₁) + \(\frac 62\) (y + y₁) - 8 = 0
Given (x₁, y₁) is (2, 2)
Equation of the tangent at (2, 2) is
x(2) + y(2) - 3(x + 2) + 3(y + 2) - 8 = 0
\(\Rightarrow 2 x-2 y-3 x+\not 6+3 y+\not 6-8=0\)
⇒ -x + 5y - 8 = 0
⇒ x − 5y + 8 = 0
Equation of the normal is
yx₁ - xy₁ + g(y - y₁) - f(x - x₁) = 0
⇒ y(2) -x(2) - 3(y - 2) -3(x - 2) = 0
∵ 2g = -6
⇒ g = -3
2f = 6
⇒ f = 3
⇒ 2y - 2x - 3y + 6 - 3x + 6 = 0
⇒ -5x - y + 12 = 0
Another method for Normal:
Equation of tangent is perpendicular to normal
x - 5y + 8 = 0
Perpendicular equation be 5x + y + k = 0
At (2, 2)
10 + 2 + k = 0
k = -12
Therefore 5x + y - 12 = 0 be equation of normal.

Given f(x, y) = x²y + 6x³ + 7
f(λx, λy) = λ²x²λy + 6λ³x³ + 7
= λ³ x² y + 6λ³ x³ + 7
≠ λf(x, y)
There is no common λ in this equation.
It is not homogeneous

\(\therefore Mean=E(x)=\Sigma xf(x)=0\left( \frac { 1 }{ 5 } \right) +1\left( \frac { 1 }{ 5 } \right) +2\left( \frac { 1 }{ 10 } \right) +3\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 1 }{ 5 } \right) +5\left( \frac { 1 }{ 10 } \right) \) 
\(\frac { 1 }{ 5 } +\frac { 1 }{ 5 } +\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { 1 }{ 2 } \) 
= \(\frac { 2+2+6+8+5+ }{ 10 } =\frac { 23 }{ 10 } =2.3\) 
= \(f({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\)
= \({ 0 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 1 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +2^{ 2 }\left( \frac { 1 }{ 10 } \right) +{ 3 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 4 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 5 }^{ 2 }\left( \frac { 1 }{ 10 } \right) \) 
= \(0+\frac { 1 }{ 5 } +\frac { 4 }{ 10 } +\frac { 9 }{ 5 } +\frac { 16 }{ 5 } +\frac { 25 }{ 10 } \) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\)
Variance = E(X²) - [E(X)]²
= 8.1- (2.3)2
= 8.1- 5.29
= 2.81

Let the equation of the ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given, length of LR = \(\frac { 1 }{ 2 } \) (Length of major axis)
⇒ \(\frac { { 2b }^{ 2 } }{ a } =\frac { 1 }{ 2 } \) (2a) ⇒ \(\frac { 2{ b }^{ 2 } }{ a } \) = a ⇒ 2b² = a²
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { { b }^{ 2 } }{ { 2a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { 1 }{ \sqrt { 2 } } \)

Given A³ = I ⇒ IA³I = |I|
⇒ |A.A.A| = 1
⇒ |A|. |A|·|A| = 1
⇒ |A|³ = 1
∴ |A| ≠ 0
Hence, A is non-singular.

Consider cos\(\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right]\).
Let \( { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) =\theta .\)
Then \( cos\theta =\frac { 1 }{ 8 } and\quad \theta \in \left[ 0,\pi \right] \)
Now, cos\(\theta=\frac{1}{8}\) implies 2 cos² \(\frac{\theta}{2}-1=\frac{1}{8}\).
Thus, cos\((\frac{\theta}{2})\) is positive
Thus, \(cos\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right] =cos\left( \frac { \theta }{ 2 } \right) =\frac { 3 }{ 4 } \)

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24