(i) If 6 + 2 = 5, then the milk is white.
Let p: 6 + 2 = 5 (F)
q: Milk is white (T)  
p ➝ q is having the truth value T
(ii) China is in Europe or \(\sqrt3\) is an integer.
p: China is in Europe (F)
q: \(\sqrt3\) is an integer (F)
p v q is having the truth value (F).
(iii) It is not true time 5 + 5 = 9 or Earth is a planet.
Let P: 5 + 5 = 9 is not true (T)
q: Earth is a planet (T
~p ∨ q is having the truth value T
(iv) 11 is a prime number and all the sides of a rectangle are equal.
p:11 is a prime number (T)
q: Allthe sides of arectangle areequal (F)
p _^ q is having the truth value F

(1) ¬p: It is not cold.
(2) p ∧ q: It is cold and raining.
(3) p ∨ q: It is cold or raining.
(4) q ∨ ¬p: It is raining or it is not cold
Observe that the statement formula ¬ p has only 1 variable p and its truth table has 2 = ( 2¹ ) rows. Each of the statement formulae p ∧ q and p ∨ q has two variables p and q. The truth table corresponding to each of them has 4 = (2² ) rows. In general, it follows that if a statement formula involves n variables, then its truth table will contain 2ⁿ rows.

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0

y''− 5y' + 6y = 0 ......(1)
Given y = e^mx  .....(2)
Differentiating cquation (2) w.r.t 'x', we get
\(\frac{dy}{dx} = em^x . m\)
To find the value of m:
Given y" - 5y' + 6y = 0 
e^mx . m² -5e^mx+ 6e^mx = 0
e^mx [m- 5m +6] = 0
m - 5m + 6 = 0
(m - 3) (m - 2) = 0
m = 3, 2

(i) If at any time t, The amount of Radium present is Q. The rate at which Q is decreasing \(\frac { dQ }{ dt } \).
This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt } = kQ\). where k is the dt constant of proportionality. Which is a required differential equation.
(ii) The rate of change of population Solution increases with respect to time t, is \(\frac { dp }{ dt } \) & the rate of population is proportional| the product of population is \(\frac { dp }{ dt } \) = kP & the also the difference between 5,00,000 & the population is \(\frac { dp }{ dt } \) = kP (5,00,000 - P) is a  required differential equation.
(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dp }{ dt } \)& the rate of dt increase vapor pressure is P at time T is proportional to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dp }{ dt } \)\(\infty\) P and \(\frac { dp }{ dt } ​​\infty\frac{1}{T^2}\)
Combining the two, we get
\(\frac { dp }{ dt } ​​\infty\frac{p}{T^2} \Rightarrow \frac { dp }{ dt }= k(\frac{p}{T^2})\), where 'k' is constantof proportionality
(iv) Let x be the amount. Amount varies from every year. (ie) Amount varies with respect to time t is \(\frac { dp }{ dt } \) & in addition the income from other source credited Rs. 400 continuously for every year.
\(\frac { dx }{ dt } = \frac{8}{100}\times x + 400\)
\(\Rightarrow\frac{dx}{dt} = \frac{2x}{25}+400\) is a required differential equation.

\({ { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 2 }=xsin\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \)
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.

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

We know that sin^-1:  [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
\({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 3 } \right) \right) \) = -\(\frac{\pi}{3}\), since -\(\frac{\pi}{3}\) \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

z w
= (5-2i)(-1+3i)
= -5+15i+2i-6i²
= -5+17i-6(-1)
= -5+17i+6
= 1+17i

y = 4 sin (-2x)
The amplitude of sin x is 1
\(\Rightarrow\) amplitude of sin (-2x) is 1
\(\therefore\) Amplitude 4 sin(-2x) is 4 \(\times\) 1 = 4.
The period of sin \((-2x)is2x=2\pi \Rightarrow =\frac { 2\pi }{ 2 } =\pi \)

The normal vectors of the two given planes \(\vec { r } .(2\hat { i } +2\hat { j } +2\hat { k } )\)= 11 and 4x+2y+2z = 15 are \(\vec { { n }_{ 1 } } =2\hat { i } +2\hat { j } +2\hat { k } \) and \(\vec { { n }_{ 2 } } =4\hat { i } -2\hat { j } +2\hat { k } \) respectively.
If θ is the acute angle between the planes, then we have
\(\theta =cos^{ -1 }\left( \frac { |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | }{ |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | } \right) =cos^{ -1 }\left( \frac { |((2\hat { i } +2\hat { j } +2\hat { k } ).(4\hat { i } -2\hat { j } +2\hat { k } ))| }{ |(2\hat { i } +2\hat { j } +2\hat { k } )||4\hat { i } -2\hat { j } +2\hat { k } | } \right) =cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 3 } \right) \).

Here, \(\vec { a } =\hat { i } +\hat { 2j } -\hat { 3k } \), \(\vec { b } =\hat { 2i } -\hat { j } +\hat { 2k } \), \(\vec { c } =\hat { 3i } +\hat { j } -\hat { k } \)
We know that  \(\vec { a } ,\vec { b } ,\vec { c } \)  are coplanar if and only if \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0. Now, \([\vec { a } ,\vec { b } ,\vec { c } ]\) = \(\left| \begin{matrix} 1 & 2 & -3 \\ 2 & -1 & 2 \\ 3 & 1 & -1 \end{matrix} \right| =0\)
Therefore, the three given vectors are coplanar.

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

(z+w)
= (5-2i) + (-1+3i)
= (5-1) + i(-2+3)
= 4+i(1)
= 4+i

Truth Table for ( p V q) ∧ ~q

p	q	p V q	~q	( p V q) ∧ ~q
T	T	T	F	T
T	F	T	T	T
F	T	T	F	T
F	F	F	T	T

\(f(x)= \begin{cases}\frac{1}{30} & 0
Mean =\(E(X)=\int _{ 0 }^{ 30 }{ x3f(x)dx } \) 
= \(\int _{ 0 }^{ 30 }{ x.\frac { 1 }{ 30 } dx } \) 
\(E(X)=\frac { 1 }{ 30 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 30 }\) 
= \(\frac { 1 }{ 30 } [ \frac{30\times 30}{2}-0]\)  
E(X) = 15 minutes
The average waiting time for the student is 15| minutes.

\(\underset { x\rightarrow \infty }{ lim } \frac { x }{ logx } =\frac { \infty }{ \infty } \)
Which is in indeterminate form
∴ By L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1 } }{ x } =\underset { x\rightarrow \infty }{ lim } x=\infty \)

a) f(x) is continuous in [0, 9]
b) f(x) is differentiable in (0, 9)
c) f(0) = 0
\(f(9)=\sqrt { 9 } -\frac { 9 }{ 3 } =3-3=0\)
∴ f(0) = f(9)
∴ By Rolle's theorem, there exists C ∈ [0, 9] such that f'(c) = 0
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } =0\)
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }=\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ 2\sqrt { c } } =\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ \sqrt { c } } =\frac { 2 }{ 3 } \)
⇒ \(\sqrt { c } =\frac { 2 }{ 3 } \)
Squaring both sides, c = \(\frac94\) ∈ [0, 9]

The curve y = sin x intersects the positive x -axis. When y = 0 which gives, x = 
\( x=n\pi , n=1,2,3,...\)
Now, \(\frac{dy}{dx}=cos x\). The slpoe \(x=n\pi\)  are \(cos(n\pi)=(-1)^{n}\).   
Hence, the required angle of intersection is m₂ = 0
\(tan \theta = \frac{(-1)^n - 0}{1+((-1)^n(0)} = 1 ∀ n\)

\(\overline { z } \) = z-1
⇒ \(\overline { z } \) =\(\frac{1}{z}\)
⇒ z\(\overline { z } \) = 1
⇒ |z|² = 1
⇒ x² + y² = 1 which is the required Cartesian equation.
Aliter : 
\( \bar{z} =z^{-1} \)
\(x-i y =\frac{1}{x+i y} \)
\(x-i y =\frac{1}{x+i y} \times \frac{x-i y}{x-i y} \)
\(x-i y =\frac{x-i y}{x^{2}+y^{2}} \)
\(x^{2}+y^{2} =1\)

Let the points be A (2, 3, 4), B (-1, 4, 5) and C (8, 1, 2)
Equation of the line joining A and B is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
⇒ \(\frac { x-2 }{ -1-2 } =\frac { y-3 }{ 4- } =\frac { z-4 }{ 5-4 } \)
⇒ \(\frac { x-2 }{ -3 } =\frac { y-3 }{ 1 } =\frac { z-4 }{ 1 } \)
Substitute the point C (8, 1, 2) in line (1),
\(\frac { 8-2 }{ -3 } =\frac { 1-3 }{ 1 } =\frac { 2-4 }{ 1 } \)
⇒ -2 = -2 = -2
Since the point C satisfies the equation of line joining A and B, all the three points lie on the same line.
Hence the given points are collinear.

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.

Area of the triangle formed by the vertices z, iz and Z+ iz is 50 sq. units
Let z = x + iy
Then iz = i(x + iy) = ix + i²y = -y + ix
z + iz = x + iy-y + ix
= (x - y) + i(x + y)
If A denotes the area of the triangle formed by z, iz and z + iz, then 
A = \(\frac { 1 }{ 2 }\ \left| \begin{matrix} x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1 \end{matrix} \right| \)
R₂ ⟶ R₂-R₁-R₃, we get
A = \(\frac { 1 }{ 2 }\ \left| \begin{matrix} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 1 \end{matrix} \right| \)
Expanding along R₂ we get
A = \(\frac { 1 }{ 2 }\ \left[ +1\left| \begin{matrix} x & y \\ -y & x \end{matrix} \right| \right] =\frac { 1 }{ 2 } \)(x²+y²)
Given A = 50 sq units
∴ 50 = \(\frac{1}{2}\)(x²+y²) ⇒ 100 = x²+ y²
Then \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 100 } \) = 10
∴ |z| = 10  [∵ |z| = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)]
AIiter :
Given area of triangle = 50 sq. unit
\(\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ -x-y & x+y & 1 \\ -y & x & 1 \end{array}\right|=50\)
\(\stackrel{R_{2} \rightarrow R_{2}-R_{3}}{\rightarrow} \frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 1 \end{array}\right|=50\)
\(\left.\frac{1}{2}\left[\begin{array}{cc} x & y \\ -y & x \end{array}\right]\right]\) = 50
\(\frac{1}{2}\left[x^{2}+y^{2}\right]=50\)
\(x^{2}+y^{2}=100\)
\(|z|^{2}=100\)
|z| = 10

\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) \).
The range of principal values of cos^-1x is [0, \(\pi\)].
Since \(\frac{13\pi}{3}\not \in[0,\pi]\), we write \(\frac{13\pi}{3} as \frac{13\pi}{3}=4\pi+\frac{\pi}{3}, whre \frac{\pi}{3}\in[0,\pi]\)
Now, cos\((\frac{13\pi}{3}=cos (4\pi+\frac{\pi}{3})=cos\frac{\pi}{3}\)
Thus,\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) ={ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 3 } \right) \right) ,\ since\frac { \pi }{ 3 } \in [0,\pi ]\).

Let A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 0 & 5 \end{matrix} \right] \). Applying elementary row operations, we get
A \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & -6 & -4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & 0 & 0 \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ(A) = 2.

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

cos x = 0
\(\Rightarrow x=\left( 2n+1 \right) \frac { \pi }{ 2 } ,n\varepsilon Z\)
But \(-6\pi \le x\le 6\pi \)
\(\therefore \) n can take values from  
\(x=(2n+1)\frac { \pi }{ 2 } ,n=0\pm 1,\pm 2,...\pm 5\), -6

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

Let X be the thrown random variable denotes the total in two the thrown a die.
Sample space S

I/II	1	3	3	5	5	5
1	2	4	4	6	6	6
3	4	6	6	8	8	8
3	4	6	6	8	8	8
5	6	8	8	10	10	10
5	6	8	8	10	10	10
5	6	8	8	10	10	10

n (S) = 36
X = {2, 4, 6, 8, 10}

Values of the random variable	2	4	6	8	10	Total
No. of elements in inverse images	1	4	10	12	9	36

\(p(x=2)=\cfrac { 1 }{ 36 } \)
\(p(x=4)=\cfrac { 4 }{ 36 } \)
\(p(x=6)=\cfrac { 10 }{ 36 } \)
\(p(x=8)=\cfrac { 12 }{ 36 } \)
\(p(x=10)=\cfrac { 9 }{ 36 } \)
(i) Probability mass function is

x	2	4	6	8	10
f(x)	\(\\ \cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12 }{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

(ii) Cumulative distribution function .
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)(X = x_i)
P(X<2) = 0 for \(\infty\) < x < 2
\(F(2)=\frac { 1 }{ 36 } \)
\(F(4)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 36 } \)
\(F(6)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(8)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 }\)
\(F(10)=\frac { 27 }{ 36 } +\frac { 9 }{ 36 } =\frac { 36 }{ 36 } =1\)
∵ The cumulative distribution function n
\(F(x)=\left\{\begin{array}{lll} 0 & \text { for } & x<2 \\ \frac{1}{36} & \text { for } & x \leq 2 \\ \frac{5}{36} & \text { for } & x \leq 6 \\ \frac{15}{36} & \text { for } & x \leq 8 \\ 1 & \text { for } & x \leq 10 \end{array}\right.\)
(iii) p(4≤ X < 10) = p(x = 4) + p(x = 6) + p(x = 8)
= \(\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } =\frac { 13 }{ 18 } \)  
(iv) p(x ≥ 6) = p(x = 6) + p(x = 8) + p(x = 10)
= \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)
Sample space = {4 childrens}

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0

Put \(\sqrt { \frac { x }{ a } } =y\Rightarrow 2y+\frac { 3 }{ y } =\frac { b }{ a } +\frac { 6a }{ b } \)
\(\Rightarrow \frac { { 2y }^{ 2 }+3 }{ y } =\frac { { b }^{ 2 }+{ 6a }^{ 2 } }{ ab } \)
\(\Rightarrow ab({ 2y }^{ 2 }+3)=\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) y\)
\(\Rightarrow 2ab{ y }^{ 2 }+3ab-\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) =0\)
\(\Rightarrow 2ab{ y }^{ 2 }-y\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) +3ab=0\)
\(\Rightarrow 2ab{ y }^{ 2 }-{ b }^{ 2 }y-{ 6a }^{ 2 }+3ab=0\)
\(\Rightarrow by(2ay-b)-3a(2ay-b)=0\)
\(\Rightarrow (2ay-b)(by-3a)=0\)
\(\Rightarrow 2ay=b,\ by=3a\)
\(\Rightarrow y=\frac { b }{ 2a } ,y=\frac { 3a }{ b } \)
Case (i) When \(y=\frac { b }{ 2a } \)
\(\Rightarrow \sqrt { \frac { x }{ a } } =\frac { b }{ 2a } \Rightarrow \frac { x }{ a } =\frac { { b }^{ 2 } }{ { 4a }^{ 2 } } \Rightarrow x=\frac { { b }^{ 2 } }{ 4a } \)
Case (ii) When \(y=\frac { 3a }{ b } \)
\(\sqrt { \frac { x }{ a } } =\frac { 3a }{ b } \Rightarrow \frac { x }{ a } =\frac { 9a^{ 2 } }{ { b }^{ 2 } } \Rightarrow x=\frac { { 9a }^{ 3 } }{ { b }^{ 2 } } \)
∴ The roots are \(\frac { { b }^{ 2 } }{ 4a } ,\frac { 9a^{ 3 } }{ b^{ 2 } } \)