\(\frac { 208{ a }^{ 2 } }{ 3 } \)

(i) 4 + 7 = 12
it s a proposition as its truth value is F
(ii) What are you doing?
It is a question and not a proposition 
(iii) 3ⁿ ≤ 81, n ∈ N
It is a proposition as it is true when
n = 1, 2, 3, 4
(iv) Peacock is our national bird. It is a proposition as its truth value is T.
(v) How tall this mountain is!
This is an exclamation, not a proposition.

(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.

Given \(U(x,y,z)=xy+sin\left( \frac { { y }^{ 2 }-2{ x }^{ 2 } }{ xy } \right) \)
\(u(\lambda x,\lambda y,\lambda z)=\lambda x\lambda y+sin\left( \frac { { \lambda }^{ 2 }{ y }^{ 2 }-2{ \lambda }^{ 2 }{ z }^{ 2 } }{ \lambda x\lambda y } \right) \)
\(={ \lambda }^{ 2 }xy+sin\left( \frac { { y }^{ 2 }-2{ x }^{ 2 } }{ xy } \right) \)
≠ λp. u (x, y, z)
There is no common λ
\(\therefore\) It is not homogeneous.

Given \(f(x)=\begin{cases} \begin{matrix} { 3e }^{ -3x } & x>0 \end{matrix} \\ \begin{matrix} 0 & elsewhere \end{matrix} \end{cases}\)
\(E(X)=\int _{ 0 }^{ \infty }{ x.f(x)dx } =\int _{ 0 }^{ \infty }{ x.3.{ e }^{ -3x }dx } \) 
= \(3\int _{ 0 }^{ \infty }{ x.{ e }^{ -3x }dx } \left[ \therefore \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(3\times \frac { 1! }{ { 3 }^{ 2 } } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
ஃ Expected life of the electronic equipment is = \(\frac { 1 }{ 3 } \) 

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

f(x) = |3x + 1|, x ∈ |-1, 3|

Since \(LHL \neq RHL, f'(-\frac{1}{3})\) does not exist.
Hence, Lagrange's mean value theorem is not applicable.

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

2x = 3y =  −z \(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{-z}{6}\) (Dividing by 6 all)
 \(\Rightarrow \frac{x-0}{3}=\frac{y-0}{2}= \frac{z-0}{-6}\)....(1)
6x = -y = -4z \(\Rightarrow \frac{x}{2}=\frac{-y}{12}=\frac{-z}{3} \) (Dividing by 6 all)
\(\Rightarrow \frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}\) ....(2)
From (1) & (2), we get
\(\vec b = 3\vec i+2\vec j- 6\vec k\)and \( \vec d = 2\vec i-12\vec j- 3\vec k\)
Angle between lines (1) and (2) = Angle between \(\vec b\ and\ \vec d\)
Acute angle between lines cos 0 = \(\frac{|\vec b . \vec d|}{|\vec b||\vec d|} \)
\(\vec b . \vec d \)= (\(\vec b = 3\vec i+2\vec j- 6\vec k\)). (\( 2\vec i-12\vec j- 3\vec k\))
6-24+18 = 0
\( \cos \theta=0 \)
\(\theta=\frac{\pi}{2} \text { or } 90^{\circ}\)

cosec^-1\((-\sqrt{2})\)
\(\Rightarrow -\sqrt { 2 } =cosex\theta \)
\(\Rightarrow sin\theta =\frac { -1 }{ \sqrt { 2 } } \)
\(\Rightarrow sin\theta =-sin\frac { \pi }{ 4 } \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow \theta =-\frac { \pi }{ 4 } \)
\(\therefore\) \(\theta cosec^{ -1 }\left( -\sqrt { 2 } \right) =-\frac { \pi }{ 4 } \)

Let z = -6+8i
|z| =\(\sqrt { (-6)^{ 2 }+8^{ 2 } } \)
= \(\sqrt { 36+64 } =\sqrt { 100 } \) = 10
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
[Here |z| = 10, a = -6, b = 8]
\(\sqrt { -6+8i } \pm \left( \sqrt { \frac { 10-6 }{ 2 } } +i\frac { 8 }{ |8| } \sqrt { \frac { 10+6 }{ 2 } } \right) \)
= \(\pm \left( \sqrt { \frac { 4 }{ 2 } } +i\sqrt { \frac { 16 }{ 2 } } \right) \)
= \(\pm (\sqrt { 2 } +i\sqrt { 8 } )\)
= \(\\ \pm (\sqrt { 2 } +i2\sqrt { 2 } )\)
Aliter :
Square root of -6 + 8i
Let a + ib = - 6 + 8i
a = -6, b = 8
\(|z|=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\)
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{10-6}{2}}+i \sqrt{\frac{10+6}{2}}\right]\)
\(=\pm[\sqrt{2}+i \quad 2 \sqrt{2}]\)

i¹⁺⁵⁰ + i²⁺⁵⁰ + ....+ i¹⁰⁺⁵⁰
= i⁵¹+ i⁵²+ ....+ i⁶⁰
Taking i⁵⁰ common we get,
i⁵⁰ [i + i²+ i³+ i⁴) + (i⁵+ i⁶+ i⁷+ i⁸) + i⁹+ i¹⁰]
= i⁵⁰[0 + (i⁴⁺¹+ i⁴⁺²+ i⁴⁺³+ i⁴⁺⁴) + (i⁸⁺¹ + i⁸⁺²)]
= i⁵⁰ [0 + 0 + i + i²] [∵ i + i²+ i³+ i⁴ = 0]
= i⁵⁰ [i-1] = i⁴⁸⁺²(i-1)
= i²(i-1)  [∴ i⁴⁸ = 1]
= -1(i-1) = -i+1 = 1-i

Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

 \(y=-sin\left( { \frac { 1 }{ 3 } x } \right) \)
 The amplitude sin x is 1
\(\Rightarrow \) amplitude of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is also 1.
The period of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is \(\frac { 1 }{ 3 } x=2\pi \Rightarrow x=6\pi \)

Vector form of the equation of the plane is
\(\vec { r } .(6\hat { i } +4\hat { j } -3\hat { k } )\) = 12
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(6\hat { i } +4\hat { j } -3\hat { k } )=12\)
⇒ 6x + 4y - 3z = 0
Dividing by 12, we get
\(\frac { 6x }{ 12 } +\frac { 4y }{ 12 } +\frac { 3z }{ 12 } =1\)
[\(\because \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) is the equation of the plane in intercept form]
⇒ \(\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { z }{ -4 } =1\)
∴ The x-intercepts of the plane is 2, y intercept is 3 and z-intercept is -4.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

Here A = 3, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

\(\left| \frac { 2+i }{ -1+2i } \right| =\frac { \left| 2+i \right| }{ \left| -1+2i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } }{ \sqrt { \left( -1 \right) ^{ 2 }+{ 2 }^{ 2 } } } =1\) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)

sec^-1\((\frac{2}{\sqrt3})\)
Let \({ sec }^{ -1 }\left( { \frac { 2 }{ \sqrt { 3 } } } \right) \)
\(\Rightarrow \frac { 2 }{ \sqrt { 3 } } =sec\theta \Rightarrow cos\theta =\frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow cos\theta =cos\left( \frac { \pi }{ 6 } \right) \)
\(\Rightarrow \theta =\frac { \pi }{ 6 } \)
\(\therefore { sec }^{ -1 }\left( \frac { 2 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

Let A be a non-singular matrix of order 2m+1, where m = 0, 1, 2,... Then, we get |A| ≠ 0 and, by property (ii), we have |adj A| = |A|^(2m+1) − 1 = |A|^2m.
Since |A|^2m is always positive, we get that |adj A| is positive.

\(Let\ I=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 2 }x{ cos }^{ 4 }xdx } \)
\({ I }_{ m,n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ m } } x{ cos }^{ n }xdx=\frac { n-1 }{ m+n } { I }_{ m,m-2 }n\ge 2\)
\(=\left( \frac { m-1 }{ n+m } \right) \left( \frac { m-3 }{ n+m-2 } \right) \left( \frac { m-5 }{ m+m-4 } \right) ...\frac { 2 }{ n+3 } .\frac { 1 }{ n+1 } \)
Here m = 2, n = 4
\(\therefore I=\frac { 3 }{ 6 } \times \frac { 1 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 32 } \)

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

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

General equation of a straight line is 
ax + by + c = 0 .......(1)
where a, b, c \(\in\) R.
Since, the lines are non - vertical,we have b \(\neq\) 0
Dividing b' by equation (1),
\(( \frac{a}{b})x+y+(\frac{c}{b}) = 0 \)
\(Ax+y=C = 0, where A = \frac{a}{b}, C = \frac{c}{b}\) .....(2)
Thus, eventhough 3 arbitrary constants (a, b, c) are present in (1), they can be considered as 2 constants only, as above (2).
Differentiating (1) with respect to x
a + b \(\\ \frac { dy }{ dx } =0\)
Differentiating again with respect to 'x' we get,
(b) \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\quad [\because b\neq 0]\) .....(3)
This is the differential equation of family of all non - vertical lines in a plane.

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

Let \(\hat { d } =6\hat { i } +2\hat { j } -3\hat { k } \) and p = 12
If \(\hat { d } \) is the unit normal vector in the direction of the vector \(6\hat { i } +2\hat { j } -3\hat { k } \)
then \(\hat { d } =\frac { \hat { d } }{ \left| \hat { d } \right| } =\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )\)
If \(\hat { r } \) is the position vector of an arbitrary point (x, y, z) on the plane, then using \(\vec { r } .\hat { d } =p\), the vector equation of the plane in normal form is \(\vec { r } .\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) in the above equation, we get \((x\hat { i } +y\hat { j } +z\hat { k } ).\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Applying dot product in the above equation and simplifying, we get 6x + 2y - 3z = 84,  which is the Cartesian equation of the required plane.

2 +i2\(\sqrt { 3 } \)
Let 2+i2\(\sqrt { 3 } \)  = x + iy = r (cosθ + i sinθ)
r = modulus =\(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
=\(\\ \sqrt { { 2 }^{ 2 }+(2\sqrt { 3 } )^{ 2 } } \)
= \(\sqrt { 4+12 } =\sqrt { 16 } \) = 4
α = tan-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 2\sqrt { 3 } }{ 2 } \right| \)
= \(tan^{ -1 }(\sqrt { 3 } )=\frac { \pi }{ 3 } \)
Since the complex number 2+i2 \(\sqrt { 3 } \) lies in the I quadrant, [x, y both +ve] its principal value θ = α = \(\frac { \pi }{ 3 } \)
∴ Its polar form is 2+i2\(\sqrt { 3 } \)
= 4\(\left[ cos\left( 2k\pi +\frac { \pi }{ 3 } \right) +isin\left( 2k\pi +\frac { \pi }{ 3 } \right) \right] ,k\in Z\).

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.

Since 2-\(\sqrt{3}\)i is a root of the required polynomial equation with real coefficients, 2 +\(\sqrt{3}\)i is also a root. Hence the sum of the roots is 4 and the product of the roots is 7. Thus x²-4x + 7= 0 is the required monic polynomial equation.

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)

\(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { x }^{ 5 }dx } +\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx\quad dx+ } \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { tan }^{ 3 }xdx+ } \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ dx } \)
\(=0+0+0+{ [x] }_{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }\)
\(=\frac { \pi }{ 2 } -\left( -\frac { \pi }{ 2 } \right) \)
\(=\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \)
\(=\pi \)
\(\\ \\ \\ \\ \\ \because \int { { x }^{ 5 } } dx\) is an odd function \(\int { { x } } \) cos xdx is an odd function \(\int { { tan }^{ 3 } } \)xdx is an odd function

Given probability mass function is

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

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

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

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

(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}

Let p(x) = 9x⁹ - 4x⁸ + 4x⁷ - 3x⁶ + 2x⁵ + x3 + 7x2 + 7x + 2 = 0
Clearly there are 4 sign changes for the given| polynomial P(x) and hence number of positive roots of P(x) can't be more than four.
hence the number of positive roots of p(x) cannot be more than 4.
p(-x) = 9(-x)⁹ - 4(-x)⁸ + 4(-x)⁷ - 3(-x)⁶+ 2(-x)⁵ + (-x)³ + 7 (-x)² + 7(-x) + 2
There are two sign changes. Hence the number of negative roots can't be more than two.
It has atmost 4 positive roots and atmnost two negative roots.