Given n = 5 and
P(X = 2) = P(X = 3)
P(X = x) = \(n \mathrm{C}_{x} p^{x} q^{n-x}\)
x = 0, 1, 2,...n
Therefore (1) becomes,
\( 5 \mathrm{C}_{2} p^{2} q^{3} =5 \mathrm{C}_{3} p^{3} q^{2} \)
\(\Rightarrow \frac{5 \times 4}{1 \times 2} q =\frac{5 \times 4 \times 3}{3 \times 2 \times 1} p \ \)
\(\Rightarrow \text { Since } p+q =p \)
\(\Rightarrow p+p =1 \)
\(\Rightarrow 2 p =1 \)
\(\Rightarrow p=\frac{1}{2} \)

Given = 10
\( \mathrm{P}(\mathrm{H})=\frac{1}{2} \Rightarrow p=\frac{1}{2} \)
\(\therefore q-1-p \)
\(=1-\frac{1}{2}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_{x} p^{x}(1-p)^{n-x} \)
\(x=0,1,2, \ldots n \)
\(\therefore \mathrm{P}(\mathrm{X}=5)=10 \mathrm{C}_{5} p^{5} q^{5} \)
\(=10 \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{5} \)
\(=10 C_{5}\left(\frac{1}{2}\right)^{10}\)

Comparing the given equations with
\( \vec{r}=\vec{a}+s \vec{b} \ \text { and } \ \vec{r}=\vec{c}+t \vec{d} \)
\(\vec{a}=\hat{i}-\hat{j} \quad \vec{b}=2 \hat{i}+\hat{k} \)
\( \vec{c}=2 \hat{i}-\hat{j} \quad \vec{d}=\hat{i}+\hat{j}+\hat{k} \)
\( \vec{c}-\vec{a}=\hat{i} \)
\( \vec{b} \times \vec{d}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
\( = \hat{i}(0-1)-\hat{j}(-2-1)+\hat{k}(2) \)
\(= -\hat{i}+3 \hat{j}+2 \hat{k} \)
\((\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d}) =\hat{i}(-\hat{i}+3 \hat{j}+2 \hat{k}) \)
\( =-1 \neq 0\)
They are skew lines
\( |(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d})| =|-1|=1 \)
\(|(\vec{b} \times \vec{d})| =\sqrt{1+9+4}=\sqrt{14} \)
Distance \( =\frac{|(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} \)
\( =\frac{1}{\sqrt{14}} \text { units } \)

\(R(x)=3 x^{2}+36 x+5\)
Marginal Revenue
\(\mathrm{MR}=\frac{d R}{d x}=6 \mathrm{x}+36\)
When x = 5, MR = 6(5) + 36 = 66

The probability distribution

\( \mathrm{E}(\mathrm{X}) =\sum p \mathrm{x} \)
\( =1 \cdot \mathrm{p}+2 \cdot \mathrm{qp}+3 \cdot \mathrm{q}^{2} \mathrm{p}+\ldots .+\mathrm{n} \mathrm{q}^{\mathrm{n}-1} \mathrm{p}+\ldots \)
\(=\mathrm{p}\left(1+2 \mathrm{q}+3 \mathrm{q}^{2}+\ldots . .+\mathrm{nq}^{\mathrm{n}-1}+\ldots \ldots\right) \)
\( =\mathrm{p}(1-\mathrm{q})^{-2} \)
\( =\mathrm{p}(\mathrm{p})^{-2}=\frac{p}{p^{2}}=\frac{1}{p} \) 

From the given data the transverse axis is along y-axis and hence the equation is of the form
\(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\)
The centre C (h, &) is the midpoint of F₁ and F_{2
\(i.e. C\ is \left(\frac{0+0}{2}, \frac{5-5}{2}\right)=(0,0) \)
\(F_{1} F_{2}=2 a e=10 \)}
The length of the transverse axis = 2a = 6
\( \Rightarrow a =3 \text { and } e=\frac{5}{3} \)
\(b^{2} =a^{2}\left(e^{2}-1\right) \)
\( =9\left(\frac{25}{9}-1\right)=16\)
\(\therefore\) The required equation is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\)

Let \( \cos ^{-1}\left(\frac{1}{2}\right) =y \)
\(\left(\frac{1}{2}\right) =\cos y \)
\(\cos \frac{\pi}{3} =\cos y \)
\(y =\frac{\pi}{3}\)
Let \( \sin ^{-1}\left(-\frac{1}{2}\right) =x \)
\(-\frac{1}{2} =\sin x \)
\(\sin \left(\frac{-\pi}{6}\right) =\sin x \)
\(x =\frac{-\pi}{6} \)
\( \cos ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1}\left(-\frac{1}{2}\right) =\frac{\pi}{3}-2\left(-\frac{\pi}{6}\right) \)
\( =\frac{\pi}{3}+\frac{\pi}{3} =\frac{2 \pi}{3} \)

f(x) is strictly increasing \(\left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] \)

c = 3

Given equation is x² + 3y² = a²
\(\div \) a² we get, \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ \frac { { a }^{ 2 } }{ 3 } } =1\)
Here a² and b2 = \(\frac{a^2}{3}\) ⇒ b = \(\frac{a}{\sqrt3}\)
Length of major axis is 2a and
Length of minor axis is 2b = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

\((i)\ \mathrm{P}(0.5<x<0.75) \)
\( =\mathrm{F}(0.75)-\mathrm{F}(0.5)=(0.75)^{2}-(0.5)^{2} \)
\( =0.3125 \)
\((ii) \ \mathrm{P}(\mathrm{X} \leq 0.5) \)
\( =\mathrm{P}(-\infty<x \leq 0.5)=\mathrm{F}(0.5)-\mathrm{F}(-\infty) \)
\( =(0.5)^{2}-0=0.25 \)
\((iii) \ \mathrm{P}(\mathrm{X}>0.75) \)
\( =\mathrm{P}(0.75 \leq x<\infty)=\mathrm{F}(\infty)-\mathrm{F}(0.75) \)
\( =1-(0.75)^{2}=0.4375 \)

The required plane Passes through the point  A(4, -2, -5) and is  Perpendicular to \(\overrightarrow{O A}\)
\(\therefore \vec{a}=4 \hat{i}-2 \hat{j}-5 \hat{k} \text { and } \vec{n}=\overrightarrow{O A}=4 \hat{i}-2 \hat{j}-5 \hat{k}\)
The required equation of the plane is \(\vec{r} \cdot \overrightarrow{\mathrm{n}}=\vec{a} \cdot \vec{n}\)
\(\vec{r} \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=(4 \hat{i}-2 \hat{j}-5 \hat{k}) \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})\)
= 16 + 4 + 25
\(\vec{r} \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=45\)
Cartesian form:
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=45\)
4x -2y -5z = 45

Given vertices are (9, 2) & (1, 2)
A (9, 2), A' (1, 2)
Centre is midpoint of AA'
\(C(h, k)=\left(\frac{9+1}{2}, \frac{2+2}{2}\right)\)
=(5, 2)
AA' = 2a = 8
a = 4
FF' = 2c = 10
c = 5
\(b^{2}=c^{2}-a^{2}\)
= 25 - 16
b² = 9
The transverse axis is parallel to x axis
Equation of hyperbola \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \)
\( \frac{(x-5)^{2}}{16}-\frac{(y-2)^{2}}{5}=1 \)
Length of Latus rectum \( =\frac{2 b^{2}}{a}=\frac{2(9)}{4} \)
\( =\frac{9}{2} \text { units } \)

\( 4 x^{2}+3 y^{2}+8 x+12 y+4 =0 \)
\(\left(4 x^{2}+8 x\right)+\left(3 y^{2}+12 y\right)+4 =0 \)
\(4\left(x^{2}+2 x\right)+3\left(y^{2}+4 y\right) =-4 \)
\(4\left\{(x+1)^{2}-1\right\}+3\left\{(y+2)^{2}-4\right\} =-4 \)
\(4(x+1)^{2}+3(y+2)^{2} =12 \)
\(\div \text { by } 12 \Rightarrow \quad \frac{(x+1)^{2}}{3}+\frac{(y+2)^{2}}{4} =1 \)
The major axis is parallel to y axis
\( a^{2}=4\ \ b=3 \)
\(a=2 \ \ b=\sqrt{3} \)
\(c^{2}=a^{2}-b^{2} \)
 = 4 - 3 
\(c^{2}=1 \) c = 1
\( a e =1 \Rightarrow 2(e)=1 \)
\(e =\frac{1}{2} \)
Centre (h, k) = (-1 , -2)
Equation of directrix y \( =\pm \frac{a}{e}+k \)
\( =\pm \frac{2}{\frac{1}{2}}-2 \)
\( =\pm 4-2 \)
y = 4 - 2 & y = -4-2
y = 2 & y = -6
Equation of latus rectum
\( \mathrm{y}=\pm \mathrm{ae}+\mathrm{k} \)
\( =\pm(2)\left(\frac{1}{2}\right)-2=\pm 1-2 \)
\( y=1-2 \ \ \quad y=-1-2 \)
\( y=-1 \ \ \quad y=-3 \)
Length of latus rectum \(=\frac{2 b^{2}}{a}=\frac{2(3)}{2}=3\)

The equation is type 1 even degree reciprocal equation.
Dividing by x² 
\(60 x^{2}-736 x+1433-\frac{736}{x}+\frac{60}{x^{2}}=0\)
\(60\left(x^{2}+\frac{1}{x^{2}}\right)-736\left(x+\frac{1}{x}\right)+1433=0\) ............(1)
Putting \(\mathrm{y}=x+\frac{1}{x} ; \quad \mathrm{y}^{2}-2=x^{2}+\frac{1}{x^{2}}\)
\(\text {(1) } \Rightarrow 60 y^{2}-736 y+1313=0\)
On solving, we get
\( y =\frac{101}{10} \text { or } \frac{13}{6} \)
\(\text {When } y =\frac{101}{10} \)
\(x+\frac{1}{x} =\frac{101}{10}\)
\( \Rightarrow 10 x^{2}-101 x+10 =0 \)
\(\text { i.e., } x =10, \frac{1}{10} \)
Similarly when y \(=\frac{13}{6}\), We get \(x=\frac{3}{2}, \frac{2}{3}\)
Thus the roots of the given equation are
\(10, \frac{1}{10}, \frac{3}{2}, \frac{2}{3}\)

(−∞,−2)concave downward
(−2,1)concave upward
(1,∞)concave downward
Points of inflections are (−2,48),(1,9)

Given 2y² = x³
⇒ y² = \(\frac { { x }^{ 3 } }{ 2 } \) ....(1)
⇒ y² = 32x  ...(2)
From (1) & (2),
\(\frac { { x }^{ 3 } }{ 2 } \) = 32x
⇒ x³ = 64x
⇒ x(x²-64) = 0
⇒ x = 0, 8, -8
when x = 0, y = 0
when x = 8, y²= 32(8)
⇒ y = ±16
when x = -8, y² = 32(-8)
which is not possible
∴ The point are (0, 0) (8, 16) (8, -16)
Differentiating 2y² = x³, with respect to 'x'
⇒ 4y\(\frac { dy }{ dx } \) = 3x²
⇒ \(\frac { dy }{ dx } =\frac { 3x^{ 2 } }{ 4y } \)
∴ m₁ = \(\left( \frac { dy }{ dx } \right) _{ (8,16) }=\frac { 3\times 8\times 8 }{ 4\times 16 } \)= 3
Differentiating y² = 32x with respect to 'x',
2y\(\frac { dy }{ dx } \) = 32
⇒ \(\frac { dy }{ dx } =\frac { 32 }{ 2y } =\frac { 16 }{ y } \)
∴ m₂ =\(\left( \frac { dy }{ dx } \right) _{ (8,16) }=\frac { 16 }{ 16 } \) = 1
Let θ be the angle between the given curves at (8, 16)
∴ tan θ = \(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\frac { 3-1 }{ 1+3(1) } \)
=\(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
∴ θ = \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \).

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

\( \mathrm{P}(\mathrm{H}) =\mathrm{P}(\mathrm{T})=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{X}=5) =\mathrm{P}(\mathrm{HH})=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \)
\(\mathrm{P}(\mathrm{X}=2) =\mathrm{P}(\mathrm{HT})+\mathrm{P}(\mathrm{TH}) \)
\( =\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{X}=1) =\mathrm{P}(\mathrm{TT})=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \)
Hence, the probability distribution function is

 \( \therefore \mathrm{E}(x) =1 \times \frac{1}{4}+2 \times \frac{1}{2}+5 \times \frac{1}{4} \)
\( =\frac{1}{4}+1+\frac{5}{4} \)
\( =\frac{1+4+5}{4}=\frac{10}{4}=2.50\)
Hence, the expected money to win is Rs. 2.50.

Given f(x) = 4k (x-1)², 1 ≤ x ≤ 3.
Since f(x) is a p.d.f\(\int_{1}^{3} f(x) d x=1\)
\( \Rightarrow \int_{1}^{3} 4 k(x-1)^{3} d x =1 \)
\(\Rightarrow 4 k\left[\frac{(x-1)^{4}}{4}\right]_{1}^{3} =1 \)
\(\Rightarrow \ k\left[2^{4}-0^{4}\right] =1 \)
\(\Rightarrow \ 16 k =1 \)
\(\Rightarrow \ k =\frac{1}{16} \)
\(\therefore \mathrm{P}(1 \leq \mathrm{X} \leq 2) =\int_{+1}^{2} f(x) d x \)
\(\Rightarrow \ =\int_{1}^{2} \frac{1}{4}(x-1)^{3} d x \)
\(\Rightarrow 1{ }^{2} =\frac{1}{4}\left[\frac{(x-1)^{4}}{4}\right]_{1}^{2} \)
\( =\frac{1}{16}\left(1^{4}-0^{4}\right)=\frac{1}{16} \)

Let A be the point (0, -2, 3)
\( \overrightarrow{\mathrm{r}} =\overrightarrow{\mathrm{OA}}=-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \)
\(\overrightarrow{\mathrm{F}} =\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \)
\(\overrightarrow{\mathrm{t}} =\overrightarrow{\mathrm{r}} \times \vec{F}=\left|\begin{array}{lll} \hat{\mathrm{i}} \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{array}\right| \)
\( =\hat{\mathrm{i}}(-2-3)-\hat{\mathrm{j}}(0-3)+\hat{\mathrm{k}}(0+2) \)
\(\overrightarrow{\mathrm{t}} =-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Magnitude of moment \(=\sqrt{(25)+(9)+(4)}=\sqrt{38}\)
Direction cosines \(=\left(\frac{-5}{\sqrt{38}}, \frac{3}{\sqrt{38}}, \frac{2}{\sqrt{38}}\right)\)

Let \(\theta\) be the angle between \(\vec{a}\ and\ \vec{b}\)
\( \therefore \vec{a} \times \vec{b} =|\vec{a}||\vec{b}| \sin \theta \hat{n} \)
\(|\vec{a} \times \vec{b}|^{2} =|\vec{a}||\vec{b}| \sin \theta \)
\(|\vec{a} \times \vec{b}|^{2} =|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \)
\((\vec{a} \cdot \vec{b})^{2} =|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \)
\(|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2} =|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \)
\( =|\vec{a}|^{2}|\vec{b}|^{2} \)

Let 'x' cm be the length of a side of an equilateral triangle and 'A' be its area
\( \mathbf{A} =\frac{\sqrt{3}}{4} x^{2} \)
\(\frac{d A}{d t} =\frac{\sqrt{3}}{2} x \cdot \frac{d x}{d t} \quad\left[\because \frac{d x}{d t}=2\right] \)
\({\left[\frac{d A}{d t}\right]_{x=20} } =\frac{\sqrt{3}}{2}(20)(2)=20 \sqrt{3}\)
Area is increasing at the rate of \(20 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}\)

\( \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}} \)         \(\left[1^{\infty}\right. \) form
\(\text {Let } y =(\cos x)^{\frac{1}{x}} \)
\(\log y =\frac{1}{x} \log \cos x \)
\(\lim _{x \rightarrow 0} \log y =\lim _{x \rightarrow 0} \frac{\log \cos x}{x}\)       \(\left[\frac{0}{0}\right. \) form
Applying L' Hopital's rule
\( =\lim _{x \rightarrow 0}-\frac{\sin x}{\frac{\cos x}{1}} \)
\( =\lim _{x \rightarrow 0}-\tan x=0 \)
\(\therefore \log \left(\lim _{x \rightarrow 0} y\right) =0 \)
\(\lim _{x \rightarrow 0} y =e^{0}=1 \)
\(ie., \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=1\)

Given \( f(x) =x^{4 / 5}(x-4)^{2} \)
\(f^{\prime}(x) =\frac{4}{5} x^{-1 / 5}(x-4)^{2}+x^{4 / 5} 2(x-4) \)
\( =\frac{4(x-4)^{2}}{5 x^{1 / 5}}+2 x^{4 / 5}(x-4) \)
\( =\frac{(x-4)}{5 x^{1 / 5}}(4 x-16+10 x) \)
\(f^{\prime}(x) =\frac{(x-4)}{5 x^{1 / 5}}(14 x-16) \)
\(f^{\prime}(x) =0 \Rightarrow \mathrm{x}=4, \frac{8}{7}\)
f'(x) does not exists at x = 0
Critical numbers are 0, 4, \(\frac{8}{7}\)

From the given data the major axis is parallel to y axis.
\(\therefore\) The equation is of the form 
\(\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1\)
The centre of the ellipse is midpoint of F₁, F₂
C is \( \left(\frac{1+1}{2}, \frac{3+9}{2}\right) =(1,6) \)
\(F_{1} F_{2} =2 a e=6 \)
\(a e =3 \)
But \( e =\frac{1}{2} \therefore a=6 \)
\(b^{2} =a^{2}\left(1-e^{2}\right)=36\left(1-\frac{1}{4}\right) =27 \)
Thus the required equation is \(\frac{(x-1)^{2}}{27}+\frac{(y-6)^{2}}{36}=1\)

The equation of the tangent at (x₁, y₁) to the given circle is
\( x x_{1}+y y_{1}-4\left(\frac{x+x_{1}}{2}\right)+4\left(\frac{y+y_{1}}{2}\right)-8=0 \)
\( x x_{1}+y y_{1}-2\left(x+x_{i}\right)+2\left(y+y_{1}\right)-8=0 \)
At (1, -2), the equation of the tangent is
-2x - 2y - 2(x - 2) + 2(y - 2) - 8 = 0
\(\Rightarrow\) -4x - 8 = 0
\(\Rightarrow\) x + 2 = 0 is the required equation of the tangent.

We have 
\( \sin ^{-1} \left(\sin \left(-600^{\circ}\right)\right)=\sin ^{-1}\left[-\sin \left(600^{\circ}\right)\right] \)
\( =\sin ^{-1}\left[-\sin \left(360^{\circ}+240^{\circ}\right)\right]=\sin ^{-1}\left(-\sin 240^{\circ}\right) \)
\(=\sin ^{-1}\left[-\sin \left(180^{\circ}+60^{\circ}\right)\right] \)
\( =\sin ^{-1}\left[\left(+\sin 60^{\circ}\right)\right] \)
\( =\sin ^{-1}\left(\sin 60^{\circ}\right)=60^{\circ} \)

Let the given equation be
\(x^{4}+P_{1} x^{3}+P_{2} x^{2}+P_{3} x+P_{4}=0\)
Here P₁ = P₂ = 0, P₃ = 2 and P₄ = 3
By Newton's theorem, \(S_{3}+S_{2} P_{1}+S_{1} P_{2}+3 P_{3}=0\)
\(\text { i.e., } \mathrm{S}_{3}+0+0+3.2=0 \)
\(\Rightarrow \mathrm{S}_{3}=-6 \)
i.e., sum of the cubes of the roots of \(x^{4}+2 x+3=0\) is -6