\(\frac { 2log5 }{ lof2 } hours\)

\(y=\frac { 2x-{ x }^{ 2 } }{ 2(1+sinx) } ,x\neq n\pi +(-1)^{ n }\frac { \pi }{ 2 } ,\)∀ n ε Z

\(y=-\frac { 1 }{ 2 } { cos }^{ 2 }x+\frac { { cos }^{ 5 }x }{ 5 } +{ xe }^{ x }-{ e }^{ x }+c\)

\(y=\frac { { x }^{ 2 } }{ 2 } +2x-13log|x+2|+c,x\varepsilon R-\{ 2\} \)

Let z = x+iy be the given complex number
∴ \(\bar { z } \) = x-iy
z\(\bar { z } \) = (x+iy) (x-iy) = x²+y²
∴ z\(\bar { z } \) -(2+3i)z -(2-3i)\(\bar { z } \)+9
⇒ x²+y²-(2+3i)(x+iy)-(2-3i)(x-iy)+9 = 0
⇒ x²+y²-[2x+2iy+3ix+i²y] - [2x-2iy-3ix+3i²y]+9 = 0
\(\Rightarrow x^{2}+y^{2}-2 x -\not 2 i y-\not 3i x +3 y-2 x+\not 2 i y+\not 3 i x+3 y+9=0 \)
⇒ x²+y²-4x+6y+9 = 0
Here 2u = -4 ⇒ u = -2
2v = 6 ⇒ v = 3 and d = 9
∴ Centre of the circle is (-u, -v) = (2, -3)
Radius =\(\sqrt { { u }^{ 2 }+{ v }^{ 2 }-d } =\sqrt { 4+9-9 } \)
=\(\sqrt { 4 } \) = 2 units
Hence, the centre of the circle is (2, -3) and radius is 2 units.

Let 2-2i = r(cosθ + isinθ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
The principal value α =tan^-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -z }{ z } \right| |\)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
Since the complex number 2 - 2i lies in the quadrant
θ = -α = -\(\frac { \pi }{ 4 } \)
∴ 2-2i = \(2\sqrt { 2 } \left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
∴ \((2\sqrt { 2 } )^{ \frac { 1 }{ 3 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) +isin\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) \right] \)
k = 0, 1, 2
The roots are
∴ When k = 0, \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } \right) \)
when k = 1, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
when k = 2, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 15\pi }{ 12 } \right) \)
∴ The product of the root
= \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } +\frac { 7\pi }{ 12 } +\frac { 15\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 21\pi }{ 12 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( 2\pi -\frac { \pi }{ 4 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 4 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ \frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \right] =2^{ 3\times \frac { 1 }{ 6 } }\left( \frac { 1-i }{ \sqrt { 2 } } \right) =2^{ 1/2 }\left( \frac { 1-i }{ \sqrt { 2 } } \right) \)
= 1-i

(1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)
= (1+5ω²+ω)(1+5ω+ω²)(5+ω+ω²)
[∴ ω⁴ = ω³.ω¹= ω]
= (1+ω+5ω²)(1+ω²+5ω)(5+ω+ω²)
= (-ω²+5ω²)(-ω+5ω)(5-1)
= (4ω²)(4ω)(4) = 64 ω³
= 64(1) = 64   [∴ ω³ = 1]
RHS
Hence proved

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the ellipse.
a² = 2⁵ and b² = 9 and c² = a² - b²
⇒ c² = 25 - 9 = 16 ⇒ c = 4
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) foci are (h - c, k), (h + c, k)
⇒ (0 - 4, 0), (0 + 4, 0)
⇒ (-4, 0) and (4, 0)
(c) Vertices are (h - a, k) and (h + a, k)
⇒ (0 - 5, 0) and (0 + 5, 0)
⇒ (-5, 0) and (5, 0)
(d) Directrices are x = \(\pm \frac { a }{ e } \)
⇒ x = \(\pm \frac { 5 }{ e } \)
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
∴ Directrice are x = \(\pm \frac { 5 }{ \frac { 4 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 4 } \)

Given that \(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =r\Rightarrow { z }_{ 1 }\bar { { z }_{ 1 } } ={ z }_{ 2 }\bar { { z }_{ 2 } } ={ r }^{ 2 }\)
\(\Rightarrow { z }_{ 1 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 1 } } } ,{ z }_{ 2 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } ,{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 3 } } \)
Therefore \({ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 1 } } +\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } +\frac { { r }^{ 2} }{ \bar { { z }_{ 3 } } } \)
= \({ r }^{ 2 }\left( \frac { \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } { \overline { z } }_{ 2 } }{ \overline { { z }_{ 1 } } \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } } \right) \) 
\(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| =\left| { r }^{ 2 } \right| \left| \frac { \overline { { z }_{ 2}{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } } }{ \overline { { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } } } \right| \) \(\left(\because \bar{z}_{1}+\bar{z}_{2}=\overline{z_{1}+z_{2}}\right)\)
= \({ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| } \)  \(\left( \because |z|=|\bar { z } |and\ \left| { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| \right) \)
= \(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| ={ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ { r }^{ 3 } } =\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ r } \)
\(\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| } \) = r   (given that \(z_{1}+z_{2}+z_{3} \neq 0\))
Thus,\(\left| \frac { { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } }{ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } \right| \) = r

Given * on A is commutative
Given b * a = c ⇒ a * b = c
Given c * a = a ⇒ a * c = a
Given b * c = a ⇒ c * b = a
Hence

Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

Given y = e^−x + mx + n .... (1) 
Derentiating cquation (1) wr.t 'x', we get
\(\frac { dx }{ dx } =-e^{ -x }(-1)+m\)
\(\frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-e^{ -x }+m\)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ e^{ x } }+0\)
\(\Rightarrow \left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) =e^x\)
Substituting the value of \(\frac{d^2y}{dx^2}\) in the given differential equation, we get
\(\Rightarrow e^{ x }\left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) -1=e^x(e^x)-1\\= e^{x-x}-1\\ e^0-1= 1-1=0\)
Thus the solution of the given differential equation e^x \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) -1 = 0

\(x={ e }^{ xy\left( \frac { dy }{ dx } \right) }\)
Taking log on both sides, log x = xy \(\frac{dy}{dx}\)
In this equation, the highest-order derivative is \(\frac{dy}{dx}\) so its power is 1.
The highest derivative is 1.
∴ its Order = 1, Degree = 1

Given A (adj A) =\(\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix} \right] =10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)...(1)
We know A (adj A) = (adj A) A = |A|. I₂ ...(2)
Comparing (1) and (2), we get |A| = 10

\(\sqrt { 3 } -i\)

r = 2 and \(-\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =-\alpha =-\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of
\(-\sqrt { 3 } -i\) are 2 and \(\frac { \pi }{ 6 } \)

\(\overset { 12 }{ \underset { n-1 }{ \Sigma } } \)iⁿ
\(\overset { 12 }{ \underset { n-1 }{ \Sigma } } \)iⁿ = (i¹+i²+i³+i⁴)+(i⁵+i⁶+i⁷+i⁸)+(i⁹+i¹⁰+i¹¹+i¹²)
= (i-1-i+1)+(i⁴⁺¹+i⁴⁺²+i⁴⁺³+(i⁴)²+(ⁱ⁸⁺¹+i⁸⁺²+i⁸⁺³+(i⁴)³)
= 0+(i+i²+i³+i⁴)+(i¹+i²+i³+i⁴)
[∴ i² = -1, i³ = -i, i⁴ = 1]
= 0 +(i-1-i+1)+(i-1-i+1)
= 0+0+0 = 0

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

\(\frac { 343 }{ 6 } \)

2

(i) 1
(ii) 2
(iii) 1

Clearly 1 is the identity element of (G₁)
Inverse of 1 is 1 [∴ (1)(1) = 1]
Inverse of i is -i [∴ (i) (-i) = -i² = 1]
Inverse of -1 is -1 [∴ (-1)(-1) = 1]
Inverse of is i [∴ (-i)(i) = -i² = 1]

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

From the line \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \)
(x₁, y₁, z₁) is (1, -1, 1)
(l₁, m₁, n₁) is (3, 2, 5)
From the line \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \)
we get, (x₂, y₂, z₂) is (-2, 1, -1)
(l₂, m₂, n₂) is 4, 3, -2
The Condition for intersecting lines is 
\(\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } \\ { l }_{ 1 } \\ { l }_{ 2 } \end{matrix}\begin{matrix} { y }_{ 2 }-{ y }_{ 1 } \\ { m }_{ 1 } \\ { m }_{ 2 } \end{matrix}\begin{matrix} { z }_{ 2 }-{ z }_{ 1 } \\ { n }_{ 1 } \\ { n }_{ 2 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} -2-1 \\ 3 \\ 4 \end{matrix}\begin{matrix} 1+1 \\ 2 \\ 3 \end{matrix}\begin{matrix} -1-1 \\ 5 \\ -2 \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} -3 \\ 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}\begin{matrix} -2 \\ 5 \\ -2 \end{matrix} \right| \)
= -3 (-4 -15) -2 (-6 -20) -2 (9 - 8)
= -3(-19) - 2(-26) -2 (1)
= 57 + 52 - 2 = 57 + 50
= 107 ≠ 0
Hence the given lines do not intersect

(1+i)¹⁸
Let 1+ i = \(r(cos\theta +isin\theta )\). Then , we get
\(r=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } ;\alpha ={ tan }^{ -1 }\left( \frac { 1 }{ 1 } \right) =\frac { \pi }{ 4 } \)
\(\theta =\alpha =\frac { \pi }{ 4 } \) (\(\because\) 1+i lies in the first Quadrant)
Therefore 1+ i = \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
Raising the power 18 on both sides
\(\left( 1+i \right) ^{ 18 }=\left[ \sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \right] ^{ 18 }=\sqrt { 12 } ^{ 18 }\left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
By de Moivre’s theorem
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \({ 2 }^{ 9 }\left( cos\left( 4\pi +\frac { \pi }{ 2 } \right) +isin\left( 4\pi +\frac { \pi }{ 2 } \right) \right) ={ 2 }^{ 9 }\left( cos\frac { \pi }{ 2 } +isin\frac { \pi }{ 2 } \right) \)
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }(i)=512i\)

In the given ellipse a² = 16, b² = 9
then c² = a² - b²
c² = 16 - 9
= 7
c = ±\(\sqrt { 7 } \)
Therefore the foci are F(\(\sqrt { 7 } \), 0) F¹(-\(\sqrt { 7 } \), 0). The focus of the parabola is (\(\sqrt { 7 } \), 0) \(\Rightarrow \) a = \(\sqrt { 7 } \)
Equation of the parabola is y² = 4\(\sqrt { 7 } \) x.

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.

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

Let \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) = r(cos θ + i sin θ)
r = \(\sqrt { \left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\sqrt { \frac { 3 }{ 4 } +\frac { 1 }{ 4 } } =\sqrt { \frac { 4 }{ 4 } } \)=1
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \right| =tan^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)
Since \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) lies is the I quadrant, θ = α
∴ \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } +isin\frac { 5\pi }{ 6 } \)  ....(1) [De moivres theorem]
Similarly \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left[ cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right] ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } -isin\frac { 5\pi }{ 6 } \) ....(2)
Adding (1) and (2) we get,
\(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }\)

= \(2cos\frac { 5\pi }{ 6 } =2cos\left( \pi -\frac { \pi }{ 6 } \right) \)
= \(-2cos\ \frac { \pi }{ 6 } \) [∵ \(\frac { 5\pi }{ 6 } \) lies in the II quard]
= \(-2\left( \frac { \sqrt { 3 } }{ 2 } \right) =-\sqrt { 3 } \).