Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Let [{:(sin^4theta,,-1,-sin^2theta,),(1+cos^2theta,,,cos^4theta,):}]=alphaI+betaM^-1, where alpha=(theta) and beta=beta(theta) are real numbers, and I is then 2xx2 identify matrix. If alpha is the minimum of the set {alpha(theta):thetain[0,2pi)}and beta is the minimum of the set {beta(theta):thetain[0,2pi)}, then the value of alpha^**+beta^** is

Answer»

`-(17)/(16)`
`-(31)/(16)`
`(37)/(16)`
`(29)/(16)`

SOLUTION :It is given matrix
`M=[{:(sintheta,,-1,-,sin^2theta),(1+cos^2theta,,cos^4theta,,):}]=alphaI+betaM^-1`, where `alpha=alpha(THETA) and beta =beta(theta)` are REAL NUMBER andI is the `2xx2` identify matrix.
Now,
det `(M)=|M|=sin^4theta cos^4theta +1sin^2theta +cos^2theta +sin^2theta cos^2theta =sin^4theta cos^4theta+sin^2theta+2`
`and [{:(sintheta,,-1,-,sin^2theta),(1+cos^2theta,,cos^4theta,,):}]=[{:(alpha,0),(0,alpha):}]+(beta)/(|M|)(adj M)[because M^-1=(adjM)/(|M|)]`
`rArr beta =-|M|` and `alpha =sin^4theta+cos^4theta`
`rArr alpha=alpha (theta) =1-(1)/(2)sin^2(2theta)`, and
`beta =beta(theta) =- {(sin^2theta cos ^2theta +(1)/(2))^2+(7)/(4)}=-{((sin^2(2theta))/(4)+(1)/(2))^2+(7)/(4)}`
Now, `alpha^**=.^alphamin=(1)/(2)and beta^**=.^betamin=-(37)/(16)`
`because alpha` is a minimum at `sin^2(2theta) =1and beta` is minimum at `sin^2(2theta)=1`
So, `alpha^**=(1)/(2)-(37)/(16)=-(29)/(16)`.
2.

If f(x) =ax^(2) +b, b ne 0, x le 1 =bx^(2) +ax + c, x gt 1 , then f(x) is continous and differentiable at x=1, if

Answer»

C=0, a=2b
a=b, c= ARBITRARY
a=b, c=0
a=b, `c NE 0`

ANSWER :A
3.

If L and L' are the ends of the latus rectum of the parabola x^(2)=6y find the equations of OL and OL' where 'O' is the origin.Also find the angle between them.

Answer»


ANSWER :`x+2y= 0, x-2y =0, PI - TAN^(-1) ( 4//3)`
4.

If the foot of the perpendicular drawn from the origin to a plane is (1,2,3), then a point on the plane is

Answer»

(3,2,1)
(7,2,1)
(7,3,-1)
(6,-3,4)

ANSWER :B
5.

Let f(x) = max (x, x^(3)) AA x in R. Calculate the area bounded by the curves y = f(x) and the x-axis between the ordinates x = - 1 and x = 1

Answer»


ANSWER :`(3)/(4)` SQ. UNITS
6.

The value of int_1^2(g(x) -f(x))dx is-

Answer»

2
3
4
4 ln2

Answer :B
7.

If the x-axis divides the area of the rigion bounded by the parabolas y= 4-x-^(2) and y^(2) -x^(2)-xin the ratio of a: b, then ab is equal to

Answer»


ANSWER :484
8.

If a,b,care distinctpositivenumbersthen the3expression( b+c-a) (c+a-b)(a+b-c)-abcis

Answer»

positive
negative
nonpositive
nonnegative

Answer :C
9.

A circle whose diameter is major aixs of ellipe (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (agtbgt0) meets axis at point P. Ifthe orthocentre of DeltaPF_(1)F_(2)lies on ellipse where F_(1)and F_(2) are foci of ellipse , then find the eccenricity of theellipse

Answer»


Solution :In the FIGURE, B is orthocentre.
Slope of altitude `S B,m_(1)=(b)/(ae)`
Slope of `SA,m_(2)=(a)/(ae)`
SINCE `m_(1)xxm_(2)=1`, we have

`(b)/(be)((1)/(-e))=-1`
`RARR 1-e^(2)=e^(4)`
`rArre^(4)+e^(2)-1==0`
`rArre^(2)=(sqrt(5)-1)/(2)`
`rArre=sqrt((sqrt(5)-1)/(2))`
10.

Let f (x) = {{:( x ^(2) - 3x + 2 "," , x lt 2 ), ( x ^(3) - 6x ^(2) + 9x + 2 "," , xge 2 ):}Then

Answer»

`LIM _( X to 2) f (x) ` does not exist
f is continuous at `x =2`
f is continuous but not differentiable at `x =2`
f is continuous and differentiable at `x =2`

ANSWER :C
11.

"Two curves "C_(1)equiv[f(y)]^(2//3)+[f(x)]^(1//3)=0 and C_(2)equiv[f(y)]^(2//3)+[f(x)]^(2//3)=12," satisfying the relation "(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^(2)-y^(2)) The area bounded by C_(1) and C_(2) is

Answer»

`2pi-sqrt(3)` SQ. units
`2pi+sqrt(3)` sq. units
`pi+sqrt(6)` sq. units
`2sqrt(3)-pi` sq. units

Solution :Given
`(x-y)F(x+y)-(x+y)f(x-y)=4zy(x^(2)-y^(2))`
`=(x^(2)-y^(2))[(x+y)^(2)-(x-y)^(2)]`
`=(x-y)(x+y)^(3)-(x+y)(x-y)^(3)`
`rArr""f(x+y)=(x+y)^(3)RARRF(x)=x^(3),f(y)=y^(3)`
Now equations of given curves are
`y^(2)+x=0"...(1)"`
`x^(2)+y^(2)=12"...(2)"`

Solving equations (1) and (2), we get `x=-3,y = pm sqrt(3)`
The area bounded by curves
`A=2[|underset(-2sqrt(3))overset(-3)intdx|+|underset(-3)overset(0)intsqrt(-x)dx|]`
`I_(1)=2overset(-3)underset(-2sqrt(3))intsqrt(12-x^(2))dx=2overset(-pi//3)underset(-pi//2)int12 cos ^(2) theta d""theta`
`=12[overset(-pi//3)underset(-pi//2)INT(1+cos 2theta)d""theta]`
`=12[theta+(SIN theta)/(2)]_(-pi//2)^(-pi//3)=12[-(pi)/(3)-(sqrt(3))/(4)+(pi)/(2)]`
`=12[(pi)/(6)-(sqrt(3))/(4)]=2pi-3sqrt(3).`
`I_(2)=2overset(0)underset(-3)intsqrt(-x)dx=(2[(-x)^(3//2)]_(-3)^(0))/(-3//2)=-(4)/(3)[10-3^(3//2)]=4sqrt(3).`
`A=2pi-3sqrt(3)+4sqrt(3)=2pi+sqrt(3)` sq. units.
12.

"Two curves "C_(1)equiv[f(y)]^(2//3)+[f(x)]^(1//3)=0 and C_(2)equiv[f(y)]^(2//3)+[f(x)]^(2//3)=12," satisfying the relation "(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^(2)-y^(2)) The area bounded by C_(1) and x+y+2=0 is

Answer»

`5//2` sq. units
`7//2` sq. units
`9//2` sq. units
None of these

Solution :
`"Required area "=OVERSET(2)underset(-1)int(-y^(2)-(-y-2))DY`
`=[(y^(2))/(2)+2Y-(y^(3))/(3)]_(1)^(2)`
`=[(4)/(2)+4-(8)/(3)-((1)/(2)-2+(1)/(3))]`
`=9//2` sq. units.
13.

"Two curves "C_(1)equiv[f(y)]^(2//3)+[f(x)]^(1//3)=0 and C_(2)equiv[f(y)]^(2//3)+[f(x)]^(2//3)=12," satisfying the relation "(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^(2)-y^(2)) The area bounded by the curve C_(2) and |x|+|y|=sqrt(12) is

Answer»

`12pi-24` SQ. units
`6-sqrt(12)` sq. units
`2sqrt(12)-6` sq. units
None of these

Solution :REQUIRED area = Area of circle- Area of SQUARE
`=12pi-24` sq. units.
14.

Determine the sign of the expression i) x^(2)+x+1 ii) -x^(2)+x-1 for x in R.

Answer»


ANSWER :i) `GT 0`; II) `LT 0`
15.

If the sum of the distance from a variable point P to the given points A(1,0) and B(0,1) is 2, then the locus of P is

Answer»

`3x^(2) +3Y^(2)-4x - 4Y = 0 `
`16X^(2) +7y^(2) - 64x - 48y = 0 `
`3x^(2) = 2xy +3y^(2) - 4x - 4y = 0 `
`16x^(2) +38xy +7y^(2) - 64x-48y = 0 `

Answer :C
16.

If lim_(xto oo){(x^2+1)/(x+1)-(ax+b)}to oo, then

Answer»

`a in (1,oo)`
`a NE 1 , b in R`
`a in (-oo,1)`
NONE of these

Solution :We have,
`lim_(n to oo) {(X^2+1)/(x+1)-ax-b}=oo`
`rArr lim_(x to oo) (x^2(1-a)-x(a+b)+1-b)/(x+1)=oo`
The limit of the given expression will be infinity, if degree of numberator is more than that of denominaotor.
`THEREFORE 1-a gt 0rArr alt 1`.
HENCE, `a in (- oo,1)` can assume any real value.
17.

The vector equation of the line passing through (2,3,4) and parallel to Z-axis is

Answer»

`BARR=(2hati+3hatj+4hatk)+lambdahatk`
`barr=2hati+lambda(hati+hatj)`
`barr=(2hati+3hatj+4hatk)+4LAMBDA(hati-hatj)`
`barr=hatk+lambda(2hati+3hatj+4hatk)`

ANSWER :A
18.

Ify = tan^(-1) ((sqrt(1+x^(2)) -sqrt(1-x^(2)))/(sqrt(1+x^(2)) + sqrt(1-x^(2)))) then dy/dx =

Answer»

`(X^(2))/(SQRT(1-x^(4))`
`(x^(2))/(sqrt(1-x^(4))`
`(x)/(sqrt(1-x^(4))`
`(x)/(sqrt(1-x^(4))`

ANSWER :D
19.

Thetransformedequationwithintegerco-effcientswhoserootsaremultipledby someconstantof thoseofx^4 -1/2 x^3 +3/4x^2 -5/4 x+(1)/(16 )=0 is

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`y^4 -y^3+3y^2 - 10y+1=0`
`y^4 - 24y^2 +9y -24 =0`
`y^4 -2y^3 +6Y -6=0`
`y^4 -5y^3 +3y^2-9y +27 =0`

ANSWER :A
20.

Prove that the following functions do not have maxima or minima h(x)=x^3+x^2+x+1

Answer»

SOLUTION :`h(x)=3x^2+2x+1ne0` for any REAL x `therefore` The discreminant of `3x^2+2x+1=0` is `4-4xx3xx1=-8lt0)thereforeh(x)` does.t have MAXIMA or MINIMA
21.

Evaluate the following integrals. int((1+logx)^n)/(x)dx

Answer»


ANSWER :`(1)/(n+1)(1+logx)^(n+1)+C`
22.

If a,b in R^(+) then find Lim_(nrarroo)sum_(k=1)^(n) ( n)/((k+an)(k+bn)) is equal to

Answer»

`(1)/(a-b) LN'(b(b+1))/(a(a+1))` if `a ne b`
`(1)/(a-b) ln'(a(b+1))/(b(a+1))` if `a ne b`
non EXISTENT if `a = b`
`(1)/(a(1+a))` if `a ne b`

Solution :N//A
23.

int (x dx)/(1-x cotx)=.....

Answer»

`LOG(cosx-X sinx)+C`
`log(x sinx-cosx)+c`
`log(sinx-xcosx)+c`
NONE of these

Answer :C
24.

If D=|{:(,a^(2)+1,ab,ac),(,ba,b^(2)+1,bc),(,ca,cb,c^(2)+1):}| then D=

Answer»

`1+a^(2)+B^(2)+C^(2)`
`a^(2)+b^(2)+c^(2)`
`(a+b+c)^(2)`
NONE of these

Answer :A
25.

Obtain the differental equations of the family of parabolas having their focus at the origin and the axis along the x-axis.

Answer»


ANSWER :`y y_(1)^(2) + 2xy_(1) - y = 0`
26.

The non-zero vectors a, b and c are related by a = 8b and c = - 7b. Then the angle between a and c is

Answer»

`pi`
`0`
`(pi)/(4)`
`(pi)/(2)`

Solution :Since, a = 8B and c = - 7b
`therefore` a is PARALLEL to b and c is anti-parallel to b.
`implies `a and c are anti-parallel
`implies` ANGLE between a and c is `pi`.
27.

If the letters of the 'AJANTA' are permutated in all possible ways and the words thus formed are arranged in dictionary order. Find the rank of 'JANATA'

Answer»


ANSWER :68
28.

If x+y=1, then sum_(r=6)^(n) r ^n C_r x^r . Y^(n-r)=

Answer»

1
n
nx
ny

Answer :C
29.

A manufacturer has three machines A,B and C installed in his factory. Machines A and B are capable of being operated for atmost 12 hours whereas machine C must be operated for atleast 5 hours a day. He produces only two items X and Y each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of X and Y on three machines are given in the following table: He makes a profit of Rs. 600 and Rs. 400 on each item of X and Y respectively. How many of each item should he produce so as to maximize his profit assuming that he can sell all the items that he produced?

Answer»


SOLUTION :NA
30.

Equation of the plane containing the lines. r=i+2j-k+lambda(i+2j-k) and r=i+2j-k+mu(i+j-3k) is

Answer»

`R.(7i-4j-k)=0`
`7(x-1)-4(y-1)-(z+3)=0`
`r.(1+2j-k)=0`
`r.(i+j+3k)=0`

ANSWER :A
31.

Evaluate: int(dx)/(cos^(3)xsqrt(2sin2x))

Answer»


ANSWER :`SQRT(TANX)+1/5(tanx)^(5/2)+C`
32.

If alpha+beta=gamma.then cos^(2)alpha+cos^(2)beta+cos^(2)gamma-2 cos alpha cos beta cos gamma=

Answer»

1
`0`
`-1`
`2`

ANSWER :A
33.

sec(x+y)= xy

Answer»


SOLUTION :We have, `SEC(x+y) = xy`
On differentiatingboth sidesw.r.t. we get
` (d)/(DX) sec (x+y) = d/(dx) (xy)`
`RARR sec (x+y).tan(x+y).(d)/(dx)(x+y)=x.d/dx y+y.(d)/(dx) x`
`rArr sec(x+y). tan(x+y).(1+(DY)/(dx)) = x '(dy)(dx) +y`
`rArr sec(x+y)tan(x+y)+sec(x+y).tan(x+y).(dy)/(dx)=x'(dy)/(dx) + y`
`rArr (dy)/(dx) [sec (x+y).tan(x+y)-x] = y - sec(x+y).tan(x+y)`
`:. (dy)/(dx) = (y-sec(x+y).tan(x+y))/(sec(x+y).tan(x+y)-x)`
34.

The sum (2^(1))/(4^(1) - 1) + (2^(2))/(4^(2) - 1) + (2^(4))/(4^(4) - 1) + (2^(8))/(4^(8) - 1) +... oo is equal to

Answer»


SOLUTION :SUM `=sum_(k=0)^(OO)(2^(2^k))/(4^(2^(k))-1) = sum((2^(2^(k)) +1)/(4^(2^(k))-1)-(1)/(4^(2^(k))-1))`
`sum((1)/(2^(2^(k))-1)-(1)/(4^(2^(k))-1))=sum((1)/(2^(2(2^(k-1))))-(1)/(4^(2^(k))-1))`
`= sum_(k=0)^(oo)((1)/(4^(2^(k))-1)-(1)/(4^(2^(k))-1))`
`= ((1)/(4^(2^(1))-1)-(1)/(4^(2^(0))-1))+((1)/(4^(2^(0))-1)-(1)/(4^(2^())-1))+((1)/(4^(2^(1))-1)-(1)/(4^(2^(2))-1))`.....{X}+{-x}={{:(0 if x in I), (1 if x !in I):}`
`= (1)/(4^(-2^(1))-1) = (1)/(2-1) = 1`
35.

Let vec(alpha)=hat(i)+hat(j)+hat(k),vec(beta)=hat(i)-hat(j)-hat(k)andvec(gamma)=-hat(i)+hat(j)-hat(k) be three vectors. A vector vec(delta), in the plane of vec(alpha)andvec(beta), whose projection on vec(gamma)" is "(1)/sqrt(3), is given by

Answer»

`-hat(i)-3HAT(J)-3hat(k)`
`hat(i)-3hat(j)-3hat(k)`
`-hat(i)+3hat(j)+3hat(k)`
`hat(i)+3hat(j)-3hat(k)`

ANSWER :C
36.

Consider the following information regarding the number of men and women workers in three factories I,II and III : {:("""Men workers","Women workers",),(I,30,25),(II,25,31),(III,27,26):} Represent the above information in the form of a3xx2 matrix . What does the entry in the third row and second column represent ?

Answer»


ANSWER :`=[{:(30,25),(25,31),(27,26):}]`
37.

A manufacturing company makes two types of teaching aids A and B of Matematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finighing. Each type of B requires 12 labour orse for fabricating and 3 labour hours for finighsing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs. 80 on ech piece of type A and Rs. 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as a LPP and solve graphically. What is the maximum profit per week ?

Answer»


Solution :Let the NUMBER of pieces of type A and type B manufactured per week be x and y RESPECTIVELY.
Then, we have to MAXIMIZE `P=80x+120y` subject to the constraints
`9x+12yle180implies3x+4yle60,""...(i)`
`x+3yle30""...(ii)`
and `x ge0,y ge0.`ltbegt We leave it to the reader to DRAW th graphs.
For maximum profit, we shall have 12 pieces of type A and 6 PICES of type B, and the maximum profit is Rs. 1680.
38.

If veca=hati+6hatj+3hatk, vecb=3hati+2hatj+hatk and vec c=(alpha+1)hati+(beta-1)hatj+hatkare linearlydependentvectors and |vec c|=sqrt(6), then the possible value(s) of (alpha+beta) can be :

Answer»

1
2
3
4

Answer :A::C
39.

A coin is tossed 2n times. During the toss any person can't get head and tail with equal numbers. Its probability is ........

Answer»

`((2N)!)/((N!)^(2))((1)/(2))^(2n)`
`1-((2n)!)/((n!)^(2))`
`1-((2n)!)/((n!)^(2))(1)/(4^(n))`
None of these

Answer :C
40.

Let a and b be natural numbers and let q and r be the quotient and remainder respectively when a^2+b^2 is divided by a + b . Determine the number q and r if q^2+ r = 2000.

Answer»


ANSWER :64
41.

int (1)/(2 + 3 cos x) dx =

Answer»

`(1)/(sqrt(5))log |(sqrt(5) + TAN""(x)/(2))/(sqrt(5) -tan""(x)/(2)) |`+ c
`(1)/(sqrt(3)) log |(sqrt(3)+ tan""(x)/(2))/(sqrt(3) - tan""(x)/(2)) | + c`
`(1)/(sqrt(5)) log |(sqrt(3)+ tan""(x)/(2))/(sqrt(5) - tan""(x)/(2)) | + c`
`(2)/(sqrt(5)) log |(SQRT5)+ tan""(x)/(2))/(sqrt(5) - tan""(x)/(2)) | + c`

Answer :A
42.

If the curves x^2+py^2=1 and qx^2+y^2=1 are orthogonal to each other, then

Answer»

`p-q=2`
`1/p-1/q=2`
`1/p+1/q=-2`
`1/p+1/q=2`

ANSWER :D
43.

Integrate the function is (xcos^-1x)/sqrt(1-x^2)

Answer»


ANSWER :`-(1)/(3)[1+(1)/(x^(2))]^((3)/(2))[LOG(1+(1)/(x^(2)))-(2)/(3)]+C`
44.

lim_(xto0) |x|

Answer»

Solution :We see that when `xto0, |x|to0`
Let US establish this using `EPSILON-delta` technique. Let `epsilongt0` we seek a `deltagt0` DEPENDING on
`epsilons.t. |x-0|epsilon implies||x|-0|let epsilon`
Now `||x|-o|=||x||=|x|ltdelta`
Choosing `epsilon=delta` we have
`|x|topsoil implies||x|-0|ltepsilon`
`therefore lim_(xto0)|x|=0`
45.

The value of Cos(tan^(-1)(3/4))is:

Answer»

`4/3`
`4/5`
`3/4`
NONE of these

Answer :A
46.

If the letters of the word 'QUESTION' are arranged at random. What is the probability that there are exactly two letters between Q and U.

Answer»


ANSWER :`(5)/(28)`
47.

Find the values of (dy)/(dx), if y = x^(tanx)+sqrt((x^(2)+1)/(2)).

Answer»


Solution :We have `y=x^(tanx)+SQRT((x^(2)+1)/(2))"...."(i)`.
Taking ` u=x^(tanx)` and `v = sqrt((x^(2)+1)/(2))`
`logu = tanxlogx"...."(ii)`
and `v^(2) = (x^(2)+1)/(2)"....."(III)`
On differentiating Eq. (ii) w.r.t.x, we get
`1/u.(du)/(dx)= tanx.(1)/(x)+logx.sec^(2)x`
`rArr (du)/(dx)=u[(tanx)/(x)+logx.sec^(2)x]`
`=x^(tanx)[(tanx)/(x)+logx.sec^(2)x]"....."(iv)`
ALSO, differentiatingEq. ( ii) w.r.t.x, we get
`2v.(dv)/(dx)=1/2(2x)rArr (dv)/(dx)= (1)/(4V).(2x)`
`rArr (dv)/(dx) = (1)/(4.sqrt((x^(2)+1)/(2))).2x= (x.sqrt(2))/(2sqrt(x^(2)+1))`
`rArr (dv)/(dx)=(x)/(sqrt(2(x^(2)+1)))"...."(v)`
Now, `y=u+v`
`:. (dy)/(dx)=(du)/(dx)+(dv)/(dx)`
`=x^(tanx)[(tanx)/(x)+logx.sec^(2)x]+(x)/(sqrt(2(x^(2)+1)))`
48.

If (1+x)^n = C_0 + C_1x+ C_2 x^2 + ….....+ C_n x^n,then C_0+2. C_1 +3. C_2 +….+(n+1) . C_n=

Answer»

`2^n + n.2^(n-1)`
`2^(n-1) + n.2^n`
`2^n + (n+1)2^(n-1)`
`2^(n+1)`

ANSWER :A
49.

Resolve into Partial Fractions (vi) (x^(2)+13x+15)/((2x+3)(x+3)^(2))

Answer»

`(1)/(2x+3)-(1)/(x+3)+(5)/((x+3)^(2))`
`(-1)/(2x+3)+(1)/(x+3)+(5)/((x+3)^(2))`
`(-1)/(2x+3)-(1)/(x+3)+(5)/((x+3)^(2))`
`(1)/(2x+3)+(1)/(x+3)+(5)/((x+3)^(2))`

ANSWER :B
50.

Prove that .^(n)C_(0) + .^(n)C_(5) + .^(n)C_(10) + "….." = (2^(n))/(5) (1+2cos^(n)'(pi)/(5)cos'(npi)/(5)+2cos'(pi)/(5)cos'(2npi)/(5)).

Answer»

Solution :Here JUMP in the series is `'5'`. So, we use fifth roots of unity.
`(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2)x^(2)+.^(n)C_(3)x^(3)+"..."+.^(n)Cx^(n)`
Now, we put `x = 1, alpha, alpha^(2), alpha^(3), alpha^(4)`
where `alpha = cos 'a(2pi)/(5)+ISIN'(2pi)/(5)`.
Puttingthese values and then adding,we get
`(1+1)^(n) + (1+alpha)^(n) + (1+alpha^(2))^(n) + (1+alpha^(3))^(n) +(1+alpha^(4))^(n)`
`= 5(.^(n)C_(0) + .^(n)C_(10) + "....")`
`:. 5(.^(n)C_(0)+.^(n)C_(5) +.^(n)C_(10)+".....")`
`= 2^(n)+(1+alpha)^(n)+(1+bar(alpha))^(n)+(1+alpha^(2))^(n)+(1+bar(alpha^(2)))^(n)`
`=2^(n)+2Re(1+cos'(2pi)/(5)+isin'(2pi)/(5))^(n)+2Re(1+cos'(4PI)/(5)+isin'(4pi)/(5))^(n)`
`= 2^(n) + 2Re(2cos^(2) '(pi)/(5)i2sin'(pi)/(5)cos'(pi)/(5))^(n) + 2Re(2cos^(2)'(2pi)/(5)+i2sin'(2pi)/(5)cos'(2pi)/(5))^(n)`
`=2^(n)+2xx2^(n)cos^(n)'(pi)/(5)cos'(npi)/(5)+2xx2^(n)cos^(n)'(2pi)/(5) cos'(2npi)/(5)`
`:. (.^(n)C_(0)+.^(n)C_(5)+.^(n)C_(10)+"......")`
` = (2^(n))/(5)(1+2cos^(n)'(pi)/(5)cos'(npi)/(5)+2cos^(n)'(2pi)/(5)cos'(2npi)/(5))`