Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Integral Reduction Formula and Answers

Engineering Mathematics Multiple Choice Questions on “Integral Reduction Formula”.

1. Find (int_{0}^{frac{pi}{2}} sin^6(x)dx).
a) 0
b) π⁄8
c) π⁄4
d) (frac{15pi}{96})
Answer: d
Explanation: Using the formula for even n we have
(int_{0}^{frac{pi}{2}} sin^n(x)dx=frac{(n-1).(n-3)…3.pi}{n.(n-2)times …2.2})
We have
=(frac{5.3.pi}{6.4.2.2})
=(frac{15pi}{96})
.

2. Find (int_{0}^{frac{pi}{2}} sin^{10}(x)cos(x)dx).
a) 1
b) 0
c) (frac{13pi}{1098})
d) (frac{21pi}{2048})
Answer: d
Explanation: Rewriting the function as
=(int_{0}^{frac{pi}{2}} (sin^{10}(x)(1-sin^2(x)))dx=int_{0}^{frac{pi}{2}} sin^{10}(x)dx-int_{0}^{frac{pi}{2}} sin^{12}(x)dx)
We now apply the formula seperately for the two integrals
(frac{9.7.5.3.pi}{10.8.6.4.2.2}-frac{11.9.7.5.3.pi}{12.10.8.6.4.2.2})
(frac{9.7.5.3.pi}{10.8.6.4.2.2}times (1-frac{11}{12})=frac{21pi}{2048})

3. Find (int_{0}^{frac{pi}{4}} tan^3(x)dx).
a) 0
b) 1
c)-1
d) None of the mentioned
Answer: b
Explanation: Using the formula we have
(int_{0}^{frac{pi}{4}} tan^n(x)dx=frac{1}{n-1}-int_{0}^{frac{pi}{2}} tan^{n-2}(x)dx)
=(frac{1}{2}-int_{0}^{frac{pi}{4}} tan(x)dx)
=(frac{1-ln(2)}{2})

4. Find the value of (int_{0}^{frac{pi}{2}} cos^{11}(x).sin^9(x), dx).
a) 1⁄10!
b) 5!6!⁄11!
c) 10!⁄5!6!
d) 0
Answer: b
Explanation: Using the definition of beta function we see that the integral is equal to the beta function at (6,5)
Now using the relation between the Beta and the Gamma function we have
(beta(m, n)=frac{Gamma (m).Gamma (n)}{Gamma (m+n)})
(beta(6, 5)=frac{Gamma (6).Gamma (5)}{Gamma (11)}=frac{6!.5!}{11!})

5. Find (int_1^{sqrt{2}} (x^{frac{8}{5}}-frac{1}{x^{frac{2}{5}}})^{frac{5}{2}}dx).
a) -1
b) 1
c) 0
d) 1⁄5 – 1⁄3 + 1⁄1 – π⁄4
Answer: d
Explanation: Simplifying we have
(int_1^{sqrt{2}} (frac{x^2-1}{x})^{frac{5}{2}}dx)
Substitute x=sec(t)
(int_{0}^{frac{pi}{4}} tan^6(t)dt)
Now using the formula
(int_{0}^{frac{pi}{4}} tan^n(x)dx=frac{1}{n-1}-int_{0}^{frac{pi}{2}} tan^{n-2}(x)dx)
We have(=frac{1}{5}-frac{1}{3}+frac{1}{1}-frac{pi}{4})

6. Find (int_{0}^{frac{pi}{4}} x^4.sin(x)dx).
a) -1
b) 1
c) 0
d) 4((π⁄2)3 – 3π + 1)
Answer: b
Explanation: Using the formula
(int_{0}^{frac{pi}{2}} x^a.sin(x)dx=left ((frac{pi}{2})^{a-1}-(a-1)int x^{a-2}sin(x)right))
We have
(int_{0}^{frac{pi}{2}} x^4.sin(x)dx=4.left ((frac{pi}{2})^{3}-3int_0^{frac{pi}{2}} x^{2}sin(x)dxright))
4((π/2)3-3π+1)

7. Find (int_{-infty}^{0}x^5.e^x dx).
a) 1
b) 199
c) -5!
d) 5!
Answer: c
Explanation: Using the formula
(int_{-infty}^{0}x^a.e^x ,dx=-aint_{-infty}^{0} x^{a-1}.e^x dx)
We have
(int_{-infty}^{0}x^5.e^x =-5.-4.-3.-2.-1=-5!)

8. Find (int_{0}^{frac{pi}{2}} cos^3(x).cos(2x)dx).
a) 0
b) 5
c) 87
d) -16⁄105
Answer: d
Explanation: Rewriting the function as
(int_{0}^{frac{pi}{2}} cos^3(x).cos(2x)dx=2int_{0}^{frac{pi}{2}} cos^7(x)dx-int_{0}^{frac{pi}{2}} cos^5(x)dx)
We now use the formula
(int_0^{frac{pi}{2}} cos^n(x)dx=frac{(n-1).(n-3)…2}{n.(n-2)…times 1})
(=2.frac{6.4.2}{7.5.3.1}-frac{4.2}{5.3.1}=frac{4.2.-2}{5.3.1.7})
=( frac{16}{105})

250+ TOP MCQs on Separable and Homogeneous Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Separable and Homogeneous Equations”.

1. Solution of the differential equation (frac{dy}{dx} = frac{y(x-y ln⁡y)}{x(x ln x-y)}) is _____________
a) (frac{x ln⁡x+y ln⁡y}{xy} = c)
b) (frac{x ln⁡x-y ln⁡y}{xy} = c)
c) (frac{ln⁡x}{x} + frac{ln⁡y}{y} = c)
d) (frac{ln⁡x}{x} – frac{ln⁡y}{y} = c)
Answer: a
Explanation: (frac{dy}{dx} = frac{y(x-y ln⁡y)}{x(x ln x-y)})
–> x2 ln⁡x dy-xy dy=xy dx – y2 ln⁡y dx …….dividing by x2 y2 then
(frac{ln⁡y}{y^2} ,dy, – frac{1}{xy} ,dy, = frac{1}{xy} ,dx, – frac{ln⁡y}{x^2} ,dx)
((ln x(frac{1}{-y^2}dy) + frac{1}{xy} ,dx) + (ln y(frac{1}{-x^2}dx) + frac{1}{xy} ,dy) = 0 )
(d(frac{ln⁡x}{y}) + d(frac{ln⁡y}{x}) = 0)
on integrating we get
(int d(frac{ln⁡x}{y}) + int d(frac{ln⁡y}{x}))
( frac{ln⁡x}{y} + frac{ln⁡y}{x}) = c…. where c is a constant of integration.

2. Solution of the differential equation (frac{dy}{dx} = e^{3x-2y} + x^2 e^{-2y}) is ______
a) ( frac{e^{2y}}{3} = frac{e^{3x}}{3} + frac{x^2}{2} + c)
b) ( frac{e^{3y} (e^{2x}+x^3)}{6} + c)
c) (frac{e^{2y} (e^{3x}+x^3)}{6} + c)
d) ( frac{e^{2y}}{2} = frac{e^{3x}}{3} + frac{x^3}{3} + c)
Answer: d
Explanation: (frac{dy}{dx} = e^{3x-2y} + x^2 e^{-2y})
(frac{dy}{dx} = e^{-2y} (e^{3x} + x^2))
separating the variable
e2y dy = (e3x+x2)dx…..integrating
∫ e2y dy =∫ (e3x+x2)dx
( frac{e^{2y}}{2} = frac{e^{3x}}{3} + frac{x^3}{3} + c).

3. Solution of the differential equation sec2 x tan⁡y dx + sec2 y tan⁡x dy=0 is _______
a) (sec x. sec y)=k
b) (sec x .tany)=k
c) (tan x. tany)=k
d) (sec x .tan x)+(sec y .tan y)=k
Answer: c
Explanation: sec2 x tan⁡y dx + sec2 y tan⁡x dy=0
dividing throughout by tan y.tan x we get
(frac{sec^2 x}{tan⁡x} ,dx + frac{sec^2 y}{tan⁡y} ,dy=0)……separating the variable
now integrating we get (int frac{sec^2 x}{tan⁡x} ,dx + int frac{sec^2 y}{tan⁡y} ,dy=c)
substituting tan x = t & tan y=p→sec2 x dx=dt & sec2 y dy=dp
( rightarrow int frac{1}{t} ,dt + int frac{1}{p} ,dp=c)
log t + log p = c –>log(tan x)+log(tan y) = c = log k….since it is an unknown constant
log(tan x .tan y) = log k
(tan x tan y) = k is the solution.

4. Solution of the differential equation (frac{dy}{dx} = (4x+2y+1)^2) is ______
a) (frac{1}{2sqrt{2}} tan^{-1}⁡(frac{4x+2y+1}{sqrt{2}})=x+c )
b) (frac{1}{sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c )
c) (frac{1}{sqrt{2}} tan^{-1}⁡⁡(frac{4x+2y+1}{sqrt{2}})=c )
d) cot-1⁡⁡⁡(4x+2y+1)=x+c
Answer: a
Explanation: (frac{dy}{dx} = (4x+2y+1)^2)
here we use substitution for ( 4x+2y+1 = t→4 + 2frac{dy}{dx} = frac{dt}{dx} rightarrow frac{dy}{dx} = frac{1}{2} frac{dt}{dx} – 2)
(frac{1}{2} frac{dt}{dx} – 2 = t^2)
(frac{dt}{dx}=2t^2+4)
separating the variable and integrating
(int frac{1}{2t^2+4} ,dt = int dx)
(frac{1}{2sqrt{2}} tan^{-1}⁡ ⁡frac{t}{sqrt{2}} = x + c)
(frac{1}{2sqrt{2}} tan^{-1}⁡(frac{4x+2y+1}{sqrt{2}})=x+c ) is the solution.

5. Solution of the differential equation (xy frac{dy}{dx} = 1+x+y+xy) is ______
a) (y-x)-log(x(1+y))=c
b) log(x(1+y))=c
c) (y+x)-log(x)=c
d) (y-x)-log(y(1+x))=c
Answer: a
Explanation: (xy frac{dy}{dx} = 1+x+y+xy)
(xy frac{dy}{dx} = (1+x)+y(1+x)=(1+x)(1+y))
separating the variables & hence integrating
(frac{y}{1+y} ,dy = frac{1+x}{x} ,dx)
(intfrac{y}{1+y} ,dy = int frac{1+x}{x} ,x)
(intfrac{(1+y)-1}{1+y} ,dy = intfrac{1}{x} ,dx + int1 ,dx)
(int1 ,dy – intfrac{1}{1+y} ,dy – log ,x – x = c)
y – log(1+y) – log x – x = c
(y-x) – log(x(1+y)) = c is the solution.

6. Solve the differential equation (frac{dy}{dx} = frac{x^2+y^2}{3xy}) is _______
a) xp=(x2+2y2)-3
b) x2 p=(x2-2y2)3
c) x4 p=(x2-2y2)-3
d) x6 p=(x2+2y2)3
Answer: b
Explanation: (frac{dy}{dx} = frac{x^2+y^2}{3xy} = frac{1+frac{y^2}{x^2}}{3 frac{y}{x}}) we can clearly see that it is an homogeneous equation
hence substituting (y = vxrightarrow frac{dy}{dx} = v + x frac{dv}{dx} = frac{1+v^2}{3v})
separating the variables and integrating
(x frac{dv}{dx} = frac{1+v^2}{3v} – v = frac{1-2v^2}{3v})
(int frac{3v}{1-2v^2} ,dv = int frac{1}{x} dx)…….substituting 1-2v2=t→-4v dv=dt we get
(frac{-3}{4} log⁡ ,t = log⁡ ,x + log⁡ ,c rightarrow frac{-3}{4} log⁡(1-2v^2) = log ⁡,cx…..but ,v = frac{y}{x})
(-3log⁡(frac{x^2-2y^2}{x^2})=4log⁡ ,cx rightarrow log⁡(frac{x^2-2y^2}{x^2})^{-3} = log⁡ ,kx^4 )
(frac{x^6}{(x^2-2y^2)^3} = kx^4 rightarrow x^2 ,p = (x^2-2y^2)^3) is the solution where p is constant.

7. The solution of differential equation (frac{dy}{dx} = frac{y}{x} + tan⁡frac{y}{x}) is ______
a) (cot(frac{y}{x}) = xc)
b) (cos(frac{y}{x}) = xc)
c) (sec^2(frac{y}{x}) = xc)
d) (sin(frac{y}{x}) = xc)
Answer: d
Explanation: (frac{dy}{dx} = frac{y}{x} + tan⁡frac{y}{x}) we can clearly see that it is an homogeneous equation
substituting (y = vx rightarrow frac{dy}{dx} = v + x frac{dv}{dx} = v + tan ,v)
separating the variables and integrating we get
(int frac{1}{tan⁡v} ,dv = int frac{1}{x} dx)
log(sin v) = log x + log c
(sin ,v = xc rightarrow sin(frac{y}{x}) = xc) is the solution where c is constant.

8. Particular solution of the differential equation (frac{dy}{dx} = frac{y^2-2xy-x^2}{y^2+2xy-x^2}) given y=-1 at x=1.
a) y=x
b) y+x=2
c) y=-x
d) y-x=2
Answer: c
Explanation: (frac{dy}{dx} = frac{y^2-2xy-x^2}{y^2+2xy-x^2} = frac{frac{y^2}{x^2} – frac{2y}{x} – 1}{frac{y^2}{x^2} + frac{2y}{x} -1} )……. is a homogeneous equation
thus put (y = vx rightarrow frac{dy}{dx} = v + x frac{dv}{dx} = frac{v^2-2v-1}{v^2+2v-1})
separating the variables and integrating we get
(xfrac{dv}{dx} = frac{v^2-2v-1}{v^2+2v-1} – v = -frac{(v^3+v^2+v+1)}{v^2+2v-1} = -frac{(v^2+1)(v+1)}{v^2+2v-1})
(int frac{v^2+2v-1}{(v^2+1)(v+1)} ,dv = int frac{-1}{x} ,dx)
(intfrac{2v(v+1)-(v^2+1)}{(v^2+1)(v+1)} ,dv = log ,c – log ,x)
(int(frac{2v}{v^2+1}-frac{1}{v+1}) ,dv = log ,c – log ,x)
log(v2+1) – log(v+1) + log x = log c –> (frac{(v^2+1)x}{(v+1)}=c)
(rightarrow frac{x^2+y^2}{x+y} = c rightarrow k(x^2+y^2)=(x+y)) where k=1/c
at x=1, y=-1 substituting we get 2k=0→k=0
thus the particular solution is y=-x.

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Special Functions – 4 (Legendre) and Answers

Ordinary Differential Equations Questions and Answers for Campus interviews focuses on “Special Functions – 4 (Legendre)”.

1. Find the General Solution of the given differential equation.
((8x+7)frac{dy}{dx}+2y=x)
a) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{40}-frac{5}{16})
b) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{8})
c) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{40}-frac{7}{16})
d) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{16})
Answer: d
Explanation: This special case where D has a co-efficient is solved using Legendre’s method.
Let log⁡(8x+7)=z
Then, ez=8x+7
((8x+7)frac{dy}{dx}=8Dy), where D=(frac{d}{dz} )
Substituting this in the equation,
8Dy+2y=x
y(8D+2)=x
Thus the auxiliary equation is 8D+2=0
Thus, D = (frac{-1}{4})
C.F = (c_1 e^{frac{-z}{4}} = c_1(frac{1}{(8x+7)})^frac{1}{4})
P.I = (frac{1}{(8D+2)} × x)
= (frac{1}{(8D+2)} × (frac{e^z-7}{8}))
= (frac{1}{(8D+2)} × frac{e^z}{8} – frac{1}{(8D+2)} × frac{7}{8} × e^0 )(Solving by substituting powers of ez)
= (frac{e^z}{80}-frac{7}{16})
=(frac{(8x+7)}{80}-frac{7}{16})
The General solution is C.F+P.I = (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{16}.)

2. Find the Continuous Function and Particular Integral for the given differential equation.
((2x+3)frac{dy}{dx}-3y=8x)
a) (c_1(2x+3)^frac{5}{2}, -4(2x+3)-4)
b) (c_1(2x+3)^frac{3}{2}, -2(2x+3)-4)
c) (c_1(2x+3)^frac{3}{2}, -4(2x+3)-4)
d) (c_1(2x+3)^frac{3}{2}, -4(2x+3)-4)
Answer: c
Explanation: This is the special case of Legendre’s Function.
Assume log(2x+3)=z
Then, ez=2x+3
((2x+3)frac{dy}{dx}=2Dy), where D=(frac{d}{dz})
Substituting this in the equation,
2Dy-3y=8x
y(2D-3)=8x
Thus the auxiliary equation is 2D-3=0
Thus, D=(frac{3}{2})
C.F=(c_1 e^frac{3z}{2} = c_1(2x+3)^frac{3}{2})
P.I=(frac{1}{(2D-3)}×8x)
=(frac{1}{(2D-3)}×8(frac{e^z-3}{2}))
=(frac{1}{(2D-3)}×e^frac{z}{2}×8-frac{1}{(2D-3)}×8×frac{-3}{2}×e^0) (Solving by substituting powers of ez)
=-4ez-4
=-4(2x+3)-4
Thus, C.F is (c_1(2x+3)^frac{3}{2})
And P.I is -4(2x+3)-4.

3. Find the C.F of the following Differential Equation.
((2x+3)^2 frac{d^2 y}{dx^2} + (2x+3)frac{dy}{dx} – 2y = x )
a) (c_1(2x+3)+ c_2(2x+3)^frac{-1}{2})
b) (c_1(2x+3)+ c_2(2x+3)^frac{1}{2})
c) (c_1(2x+3)+ c_2(2x+3)^frac{-3}{2})
d) (c_1(2x+3)+ c_2(2x+3)^frac{-1}{2})
Answer: a
Explanation: Assume log⁡(2x+3)=z
Then, ez=(2x+3)
(2x+3)(frac{dy}{dx})=2Dy, Where D is (frac{d}{dz})
((2x+3)^2 frac{d^2 y}{dx^2}=2^2 D(D-1)y)
Substituting in the equation given
4D(D-1)y+2Dy-2y=x
y(4D(D-1)+2D-2)=x
y(4D2-2D-2)=x
Thus, the Auxiliary Equation is 4D2-2D-2
D=1 or D=(frac{-1}{2})
Thus, the C.F for the given equation is (c_1e^z + c_2e^frac{-z}{2}=c_1(2x+3) + c_2(2x+3)^frac{-1}{2}).

4. Find the P.I of the given Differential Equation.
((x+1)^2 frac{d^2 y}{dx^2}+(x+1) frac{dy}{dx}+y=sin⁡(log⁡(1+x)))
a) -log⁡(x+1)×(frac{cos⁡(log⁡(x+1))}{2})
b) -log⁡(x+1)×(frac{sin⁡(log⁡(x+1))}{2})
c) log⁡(x+1)×(frac{cos⁡(log⁡(x+1))}{2})
d) -log⁡(x+2)×(frac{cos⁡(log⁡(x+1))}{2})
Answer: a
Explanation: Assume log⁡(x+1)=z
Then, ez=(x+1)
((x+1)frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+1)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+Dy+y = sin⁡(log⁡(1+x))
y(D(D-1)+D+1) = sin⁡(log⁡(1+x))
y(D2+1) = sin⁡(log⁡(1+x))
Thus, the Auxiliary Equation is D2+1
P.I = (frac{1}{D^2+1})×sin⁡(z)
To find P.I, substitute D2=-(1)2
Since the denominator becomes zero, multiply the numerator by z and differentiate the denominator.
P.I = (frac{z}{2D})×sin⁡(z)
=(frac{-zcos(z)}{2})
=-log⁡(x+1)×(frac{cos⁡(log⁡(x+1))}{2}).

5. Solve this Differential Equation to find its General Solution.
((x+3)frac{d^2y}{dx^2}+2 frac{dy}{dx}+frac{y}{(x+3)}=4)
a) (frac{4x}{3}+2+frac{1}{(x+3)}×c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3)))
b) (frac{4x}{3}+4+frac{1}{(x+3)}×c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3)))
c) (x+4+frac{1}{(x+3)}×c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3)))
d) (frac{2x}{3}+4+frac{1}{(x+3)}×c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3)))
Answer: b
Explanation: Multiply both the sides with (x+3)
We get the equation,
((x+3)^2 frac{d^2 y}{dx^2}+2(x+3) frac{dy}{dx}+y=4(x+3))
This equation is in Legendre’s form.
Assume log⁡(x+3)=z
Then, ez=(x+3)
((x+3)frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+3)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+2Dy+y=4(x+3)
y(D(D-1)+2D+1)=4(x+3)
y(D2+D+1)=4(x+3)
Thus, the Auxiliary Equation is D2+D+1=0
(D=frac{-1}{2}+frac{sqrt{3}}{2}i ,or. D=frac{-1}{2}-frac{sqrt{3}}{2}i)
Thus, the C.F for the given equation is (e^{frac{-z}{2}} (c_1cos⁡(frac{sqrt{3}}{2}z)+c_2sin⁡(frac{sqrt{3}}{2}z)))
C.F = (frac{1}{(x+3)}× c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3)))
P.I = (frac{1}{D^2+D+1}×4e^z)
To find P.I, substitute D=1
P.I = (frac{4×e^z}{3})
= (frac{4×(x+3)}{3})
= (frac{4x}{3}+4)
Thus, the general solution is
(frac{1}{(x+3)}× c_1cos⁡(frac{sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(frac{sqrt{3}}{2} log⁡(x+3))+frac{4x}{3}+4)
.

6. Find the C.F for the following Differential Equation.
((3x+2)^3 frac{d^3y}{dx^3}+2(3x+2)^2 frac{d^2 y}{dx^2}+(3x+2) frac{dy}{dx}-y=(3x+2)^2)
a) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 cos⁡(0.22 log⁡(3x+2))+c_3 sin(0.22 log⁡(3x+2))))
b) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 cos⁡(0.22 log⁡(3x+3))+c_3 sin(0.22 log⁡(3x+2))))
c) (c_1(3x+2)^{0.026}+(3x+2)^{1.15}(c_2 cos⁡(0.22 log⁡(3x+2))+c_3 sin(0.22 log⁡(3x+2))))
d) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 sin⁡(0.22 log⁡(3x+2))+c_3 sin(0.22 log⁡(3x+2))))
Answer: c
Explanation: This equation is in the Legendre Form.
Assume log⁡(3x+2)=z
Then, ez=(3x+2)
((3x+2) frac{dy}{dx}=3Dy), Where D is (frac{d}{dz})
((3x+2)^2 frac{d^2 y}{dx^2}=3^2 D(D-1)y)
((3x+2)^3 frac{d^3 y}{dx^3}=3^3 D(D-1)(D-2)y)

Substituting in the equation given

27D(D-1)(D-2)y+18D(D-1)y+3Dy-y=(3x+2)2
y(27(D(D-1)(D-2))+18(D(D-1))+3D-1)=(3x+2)2
y(27D3-63D2+39D-1)
Thus the Auxiliary Equation is 27D3-63D2+39D-1=0
D=0.026 or D=1.15+0.22i or D=1.15-0.22i

Thus, the C.F of the equation is given by
C.F=(c_1e^{0.026z}+e^{1.15z} (c_2cos⁡(0.22z)+c_3sin(0.22z)))
C.F=( c_1(3x+2)^{0.026}+(3x+2)^{1.15} (c_2cos⁡(0.22 log⁡(3x+2))+c_3sin(0.22 log⁡(3x+2)))).

7. Find the solution for the given Higher Order Differential Equation.
(2(3x+5)^2 frac{d^2 y}{dx^2}+(3x+5) frac{dy}{dx}+y=sin⁡(log⁡(3x+5)))
a) (c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{64})
b) (c_1(3x+5)^{0.76}+ c_2(3x+7)^{0.073}+frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{64} )
c) (c_1(3x+5)^{0.76}+ c_2(3x+7)^{0.073}+frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{16} )
d) (c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{16} )
Answer: a
Explanation: The given equation is a Legendre’s Function.
Assume log⁡(3x+5)=z
Then, ez=(3x+5)
((3x+5) frac{dy}{dx}=3Dy), Where D is (frac{d}{dz})
((3x+5)^2 frac{d^2 y}{dx^2}=3^2 D(D-1)y)
Substituting in the equation given
(2×9)D(D-1)y+3Dy+y= sin⁡(log⁡(1+x))
y(18(D2-D)+3D+1) = sin⁡(log⁡(1+x))
y(18D2-15D+1) = sin⁡(log⁡(1+x))
Thus, the Auxiliary Equation is 18D2-15D+1
D=0.76 or D=-0.073
Thus, the C.F for the given equation is
(c_1e^{0.76}z+ c_2e^{0.073}z=c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073})
P.I = (frac{1}{(18D^2-15D+1)}×sin⁡(z))
To find P.I, substitute D2=-(1)2
=(frac{1}{(-17-15D)}×sin⁡(z))
Multiplying numerator and Denominator with factor(17-15D)
=(frac{(-15D+17)}{64}×sin⁡(z)), Using a2-b2=(a+b)(a-b) and substituting D2=-(1)2
=((15(-cos⁡(z)-17 sin⁡(z)))/64)
=(frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{64})
Thus, the general solution is
(c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{64})
.

8. Find the C.F for the given Higher Order Differential Equation.
(x^2 frac{d^2 y}{dx^2}+3(x+2) frac{dy}{dx}+4(1+x) frac{d^2 y}{dx^2}+2y=x)
a) (frac{1}{(x+1)}×(c_1cos⁡(log⁡(x+2))+c_2sin⁡(log⁡(x+2))))
b) (frac{1}{(x+2)}×(c_1cos⁡(log⁡(x+2))+c_2cos⁡(log⁡(x+2))))
c) (frac{1}{(x+1)}×(c_1cos⁡(log⁡(x+1))+c_2sin⁡(log⁡(x+1))))
d) (frac{1}{(x+2)}×(c_1cos⁡(log⁡(x+2))+c_2sin⁡(log⁡(x+2))))
Answer: d
Explanation: Bring all terms of (frac{d^2 y}{dx^2}) together.
((x+2)^2 frac{d^2 y}{dx^2}+3(x+2) frac{dy}{dx}+2y=x)
This is in Legendre’s Form.
Let, log⁡(x+2)=z
Then, ez=(x+2)
((x+2) frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+2)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+3Dy+2y=x
y((D2-D)+3D+2)=x
y(D2+D+2)=x
Thus, the Auxiliary Equation is y(D2+D+2)=0
D=-1+i or D=-1-i
C.F is
(e^{-z}(c_1cos⁡(z)+c_2sin⁡(z))=frac{1}{(x+2)}×(c_1cos⁡(log⁡(x+2))+c_2sin⁡(log⁡(x+2)))).

9. Find the P.I for the given Differential Equation.
(16x^2 frac{d^2 y}{dx^2}+(2x+4) frac{dy}{dx}+4x(x+4) frac{d^2 y}{dx^2}+2y=cos⁡(log⁡(2x+4)))
a) (frac{1}{4}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4))))
b) (frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4))))
c) (frac{1}{2}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4))))
d) (frac{1}{2}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4))))
Answer: b
Explanation: Bring all terms of ( frac{d^2 y}{ dx^2}) together.
The equation becomes-
((2x+4)^2 frac{d^2 y}{dx^2}+(2x+4) frac{dy}{dx}+2y= cos⁡(log⁡(2x+4)))
This is in Legendre’s Form.
Let, log⁡(2x+4)=z
Then, ez=(2x+4)
((2x+4) frac{dy}{dx}=2Dy), Where D is (frac{d}{dz})
((2x+4)^2 frac{d^2 y}{dx^2}=4D(D-1)y)
Substituting in the equation given
4D(D-1)y+2Dy+2y=cos⁡(log⁡(2x+4))
y(4(D2-D)+2D+2)=cos⁡(log⁡(2x+4))
y(4D2-2D+2)=cos⁡(log⁡(2x+4))
Thus, the Auxiliary Equation is y(4D2-2D+2)=0
P.I = (frac{1}{4D^2-2D+2}×cos⁡(log⁡(2x+4)))
= (frac{1}{4D^2-2D+2}×cos⁡(z))
In case of cos⁡() function, Substitute D2=-(1)2
= (frac{1}{-2-2D}×cos⁡(z)=frac{-1}{2}×frac{1}{(D+1)}×cos⁡(z))
Multiplying numerator and Denominator with factor(D-1)
=(frac{1}{4}×(D-1)×cos⁡(z))
=(frac{1}{4}×(-sin⁡(z)-cos⁡(z)))
=(frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4))))
Thus, the P.I for the given equation is =(frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))).

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on System of Equation using Gauss Elimination Method and Answers

Linear Algebra Questions and Answers for Experienced people focuses on “System of Equation using Gauss Elimination Method”.

1. Solve the following equations using Gauss Elimination Method and find the value of x and z.

x + y + 2z + 3w = 1
2x + 3y - 2z + 4w = 2
2x + 3y + z - w = 0
3x - 2y + z - 3w = 3

a) x=1.1 and z=-0.2
b) x=0.3 and z=-0.2
c) x=1.1 and z=0.3
d) x=0.3 and z=-0.6
Answer: a
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}1&1&2&3\2&3&-2&4\2&3&1&-1\3&-2&1&-3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\2\0\3end{bmatrix})
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R2=R2-R1 and R3=R3-2R1 and R4=R4-3R1
(begin{bmatrix}1&1&2&3\0&1&-6&-2\0&1&-3&-7\0&-5&-5&-12end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\0\-2\0end{bmatrix})
R3=R3-R2 and R4=R4+5R2
(begin{bmatrix}1&1&2&3\0&1&-6&-2\0&0&3&-5\0&0&-35&-22end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\0\-2\0end{bmatrix})

From this, we get
x+y+2z+3w=1 – (1)
y-6z-2w=0 – (2)
3z-5w=-2 – (3)
-35z-22w=0 – (4)
From equations (3) and (4) we get
z=-0.2 and w=0.3
Substituting these values in equation (2)
y=6(-0.2)+2(0.3)
y=-0.6
Substituting these values in equation (3)
x+(-0.6)+2(-0.2)+3(0.3)=1
x=1.1
Thus, the solution of the following equation is
x=1.1
y=-0.6
z=-0.2
w=0.3.

2. Solve the following equations using Gauss Elimination Method.

x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6

a) x=0.5 y=0 and z=1
b) x=-0.5 y=0 and z=-1.5
c) x=-0.5 y=0 and z=1.5
d) x=0.5 y=0 and z=1.5
Answer: c
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\5\6end{bmatrix})
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R2=R2-2R1 and R3=R3-3R1
(begin{bmatrix}1&2&3\0&-1&-2\0&-5&-4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\-3\-6end{bmatrix})
R3=R3-5R2
(begin{bmatrix}1&2&3\0&-1&-2\0&0&6end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\-3\9end{bmatrix})
From the above Matrix, we get –
x+2y+3z=4 -(2)
-y-2z=-3 -(1)
6z=9
Thus, z=1.5
Substituting in (1)
y=3-2(1.5)
y=0
Substituting in (2)
x=4-2(0)-3(1.5)
x=-0.5
Thus, x=-0.5 y=0 and z=1.5.

3. Find the value of z that satisfies all the given equations using Gauss Elimination Method.

12x + 4y + 3z = -1
3x + 8y + z = 2
2x + 3y + 4z = 5

a) z=(frac{53}{41})
b) z=(frac{-143}{287})
c) z=(frac{79}{287})
d) z=(frac{-53}{41})
Answer: a
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}2&3&4\3&8&1\12&4&3end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}5\2\-1end{bmatrix})
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R2=R2-2R1
(begin{bmatrix}1&5&-3\3&8&1\12&4&3end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-3\2\-1end{bmatrix})

R2=R2-3R1 and R3=R3-12R1
(begin{bmatrix}1&5&-3\0&-7&10\0&-56&39end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-3\11\35end{bmatrix})

R3=R3-8R2
(begin{bmatrix}1&5&-3\0&-7&10\0&0&-41end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-3\11\-53end{bmatrix})
From the above Matrix, we get-
x+5y-3z=-3-(2)
-7y+10z=11-(1)
-41z=-53
Thus, z=(frac{53}{41})
Substituting in (1)
y=(frac{79}{287})
Substituting in (2)
x=(frac{-143}{287}).

4. Find the value of x and y that satisfy the given sets of equation using Gauss Elimination Method.

2x - 3y + 2z - 2w = 4
3x + 2y - 3z + 5w = 1
4x + 5y + 11z + 3w = 3
5x + 3y + 2z + 7w = - 1

a) x=0.85 and y=-1.2
b) x=0.85 and y=0.38
c) x=0.27 and y=0.38
d) x=0.27 and y=-1.2
Answer: d
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}5&3&2&7\2&-3&2&-2\3&2&3&5\4&5&11&3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-1\4\1\3end{bmatrix})
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R1=R1-R4
(begin{bmatrix}1&-2&-9&4\2&-3&2&-2\3&2&3&5\4&5&11&3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\4\1\3end{bmatrix})

R2=R2-2R1 and R3=R3-3R1 and R4=R4-4R1
(begin{bmatrix}1&-2&-9&4\0&1&20&-10\0&8&30&-7\0&13&47&-13end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\12\13\19end{bmatrix})
R3=R3-8R2 and R4=R4-13R2
(begin{bmatrix}1&-2&-9&4\0&1&20&-10\0&0&-130&73\0&13&-213&117end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\12\-83\-137end{bmatrix})
From this, we get
x-2y-9z+4w=-4 – (1)
y+20z-10w=12 – (2)
-130z+73w=-83 – (3)
-213z-117w=-137 – (4)
From equations (3) and (4) we get
z=0.85 and w=0.38
Substituting these values in equation (2)
y=12-20(0.85)+10(0.38)
y=-1.2
Substituting these values in equation (3)
x+(-2×(-1.2))-9(0.85)+4(0.38)=-4
x=0.27
Thus, the solution of the following equation is
x=0.27
y=-1.2
z=0.85
w=0.38.

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250+ TOP MCQs on Volume Integrals and Answers

Linear Algebra Multiple Choice Questions on “Volume Integrals”.

1. Find the value of (int_0^1 int_0^2 int_1^2 xy^2 z^3 dxdydz.).
a) 2
b) 3
c) 4
d) 5
Answer: d
Explanation: (int_0^1 int_0^2 int_1^2 xy^2 z^3 dxdydz = (int_0^1 xdx)(int_0^2 y^2 dy)(int_1^2 z^3 dz) )
(= frac{1}{2}*frac{8}{3}*(frac{16}{4}-frac{1}{4}) )
(= 5. )

2. Evaluate ∫∫∫z2 dxdydz taken over the volume bounded by the surfaces x2+y2=a2, x2+y2=z and z=0.
a) (pi frac{a^8}{12} )
b) (pi frac{a^8}{4} )
c) (pi frac{a^4}{4} )
d) (pi frac{a^8}{8} )
Answer: a
Explanation: z: 0 to (x^2+y^2 )
( y: pi frac{a^8}{12} ) to ( sqrt{a^2-x^2} )
x: -a to a
(displaystyleintintint z^2 dxdydz = int_{-a}^a int_{-sqrt{(a^2-x^2 )}}^{sqrt{a^2-x^2}}int_0^{x^2+y^2}z^2 dxdydz )
(displaystyle = int_{-a}^a int_{-sqrt{a^2-x^2}}^{sqrt{a^2-x^2}}frac{z^3}{3} dxdy ) from 0 to ( x^2+y^2. )
(displaystyle = int_{-a}^a int_{-sqrt{a^2-x^2}}^{sqrt{a^2-x^2}} frac{(x^2+y^2 )^3}{3} dxdy)
( = frac{1}{3} int_0^π int_{-a}^a r^7 drdθ )
( = π frac{a^8}{12}. )

3. Find the volume of the cylinder bounded by x2+y2 = 4, y+z = 4 and z=0.
a) (16π-frac{32}{3} )
b) (32π-frac{32}{3} )
c) (16-32 frac{π}{3} )
d) (32-32 frac{π}{3} )
Answer: a
Explanation: (z: (4-y))
(x: -sqrt{4-y^2} ) to ( sqrt{4-y^2} )
y: -2 to 2
Volume = (displaystyleint_{-2}^2 int_{-sqrt{4-y^2}}^{sqrt{4-y^2}} (4-y)dxdy )
(= 2int_{-2}^2 (4y-y^2 )dxdy ) from 0 to ( sqrt{4-y^2} )
(= 4big{4(0 + π) – 4 + frac{4}{3}big} )
(= 16π-frac{32}{3}. )

4. Find the volume of sphere by triple integration.
a) (8a^3 frac{π}{3} )
b) (4a^3 frac{π}{3} )
c) (2a^3 frac{π}{3} )
d) (a^3 frac{π}{3} )
Answer: b
Explanation: The sphere is given by x2+y2+z2=a2.
θ : 0 to 2 π
φ : 0 to π
r : 0 to a
Volume = ( int_0^a int_0^π int_0^{2π} r^2 sinθdθdrdϕ )
(= 4(int_0^a r^2 dr)(int_0^π sinθdθ)(int_0^{π/2}dϕ) )
(= 4a^3 frac{π}{3}.)

5. Using polar coordinates, find the volume of the cylinder with radius a and height h.
a) (πa^2h )
b) (frac{πa^2h}{3} )
c) (2 frac{πa^2h}{3} )
d) ( 4 frac{πa^2h}{3} )
Answer: a
Explanation: r: 0 to a
θ : 0 to 2 π
z: 0 to h
Therefore, volume = (int_0^a int_0^{2π}int_0^h rdθdrdz )
(= (int_0^a rdr)(int_0^{2π}dθ)(int_0^h dz) )
(= frac{a^2}{2} * 2π * h )
(= πa^2h. )

6. In multiple integrals, if the limits depends on variable, then the order of integration can be anything.
a) True
b) False
Answer: b
Explanation: In multiple integration, if the limits depend on variable then the order of integration can’t be anything. First the dependent integration should be done and then the independent integration.

7. To find volume _________________ can be used.
a) single integration
b) double integration
c) triple integration
d) double & triple integration
Answer: d
Explanation: To find volume, triple integration should be used and proper limits should be given for each variable.
But volume integration can also be done using double integration by using 1D equation of the 3D object as the function.

8. Evaluate (int_{-1}^1 int_0^z int_{x-z}^{x+z}(x+y+z)dxdydz.)
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: (int_{-1}^1 int_0^z int_{x-z}^{x+z}(x+y+z)dxdydz=
int_{-1}^1 int_0^z (xy+frac{y^2}{2}+zy)dxdz ) from x-z to x+z
(= int_{-1}^1int_0^z(3x^2+z^2+2xz)dxdz= int_{-1}^1(x^3+z^2 x+x^2 z)dz ) from 0 to z
(= int_{-1}^1(z^3+z^3+z^3)dz= frac{z^4}{4} ) from -1 to 1
= 0.

9. Evaluate (int_0^1int_0^{sqrt{1-x^2}}int_0^{sqrt{1-x^2-y^2}}xyzdxdydz.)
a) (frac{1}{6} )
b) (frac{1}{12} )
c) (frac{1}{24} )
d) (frac{1}{48} )
Answer: c
Explanation: (int_0^1 int_0^{sqrt{1-x^2}}int_0^{sqrt{1-x^2-y^2}}xyzdxdydz=
int_0^1 int_0^{sqrt{1-x^2}} frac{xyz^2}{2} dxdy ) from 0 to ( sqrt{1-x^2-y^2} )
(= frac{1}{2} int_0^1 frac{xy^2}{2} – frac{x}{2} – frac{xy^4}{4} dx ) from 0 to (sqrt{1-x^2} )
(= frac{1}{2} int_0^1 frac{(x-x^3)}{2}-frac{(x^3-x^5)}{2}-frac{(x+x^5-2x^3)}{4} dx)
(= frac{1}{4}big{frac{1}{6} – frac{1}{2} + frac{1}{2}big} )
(= frac{1}{24}. )

10. What is the volume of a cube with side a?
a) (frac{a^3}{8} )
b) (a^2)
c) (a^3)
d) (frac{a2}{4} )
Answer: c
Explanation: (int_0^a int_0^a int_0^a dxdydz = (int_0^a dx)(int_0^a dy)(int_0^a dz)= a^3.)

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250+ TOP MCQs on DeMoivre’s Theorem and Answers

Complex Analysis Multiple Choice Questions on “DeMoivre’s Theorem”.

1. Find the value of (1+i)100.
a) 2100 (cos⁡100π+isin100π)
b) 2100 (cos⁡25π+isin25π)
c) 250 (cos⁡100π+isin100π)
d) 250 (cos⁡25π+isin25π)
View Answer

Answer: d
Explanation: We know that,
1+i=(sqrt 2 (frac{1}{sqrt 2}+frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}+isin frac{pi}{4}))
((1+i)^{100}=(sqrt 2 left (cos frac{pi}{4}+isin frac{pi}{4}right ))^{100}=2^{50}(left(cos frac{pi}{4}+isin frac{pi}{4}right))^{100})
By Applying the DeMoivre’s Theorem
((1+i)^{100}=2^{50} left (cos 100frac{pi}{4}+isin 100frac{pi}{4} right ))
((1+i)^{100}=2^{50} (cos⁡25pi+isin25pi)).

2. Find the value of (1-i)100.
a) 2100 (cos⁡100π-isin100π)
b) 2100 (cos⁡25π-isin25π)
c) 250 (cos⁡25π-isin25π)
d) 250 (cos⁡100π-isin100π)
View Answer

Answer: c
Explanation: We know that,
(1-i=sqrt 2 (frac{1}{sqrt 2}-frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4}))
((1-i) ^{100}=(sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4})) ^{100}=2^{50}((cos frac{pi}{4}-isin frac{pi}{4})) ^{100})
By Applying the DeMoivre’s Theorem
((1-i)^{100}=2^{50} (cos frac{100pi}{4}-isin frac{100pi}{4}))
((1-i)^{100}=2^{50} (cos⁡25pi-sin25pi)).

3. If (a=frac{1}{sqrt 2}+frac{i}{sqrt 2}), find the value of a5 + conjugate of a5=?
a) (cos⁡ frac{3pi}{4})
b) (2 sin ⁡frac{5pi}{4})
c) (2 cos frac{5pi}{4})
d) (cos frac{5pi}{4})
View Answer

Answer: c
Explanation: We know that,
(a=(frac{1}{sqrt 2}+frac{i}{sqrt 2})=(cos frac{pi}{4}+isin frac{pi}{4}))
Conjugate of a = ((frac{1}{sqrt 2}-frac{i}{sqrt 2})=(cos frac{pi}{4}-isin frac{pi}{4}))
(a^5) + (conj of a)5=((cos frac{pi}{4}+isin frac{pi}{4})^5+(cos frac{pi}{4}-isin frac{pi}{4})^5)
By Applying the DeMoivre’s Theorem
a5 + (conj of a)5 = (cos⁡ frac{5pi}{4} +isin frac{5pi}{4}+cos⁡ frac{5pi}{4} -isin frac{5pi}{4})
a5 + (conj of a)5 = (2 cos ⁡frac{5pi}{4}).

4. Evaluate (frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) ^{17}}).
a) (frac{1}{2}×(cos frac{pi}{6}+isin frac{pi}{6}))
b) (frac{1}{3}×(cos frac{pi}{6}+isin frac{pi}{6}))
c) (frac{1}{2}×(cos frac{pi}{3}+isin frac{pi}{3}))
d) (frac{1}{4}×(cos frac{pi}{6}+isin frac{pi}{6}))
View Answer

Answer: a
Explanation: We know that,
(1+sqrt 3 i=2(cos frac{pi}{3}+isin frac{pi}{3}))
(sqrt 3-i=2(cos frac{pi}{6}-isin frac{pi}{6}))
(frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) {^17}}=frac{2^{16}(cos frac{pi}{3}+isin frac{pi}{3}) ^{16}}{2^{17}(cos frac{pi}{6}-isin frac{pi}{6}) ^{17}})
By Demoivre’s Theorem
= (frac{1}{2}×(cos frac{16pi}{3}+isin frac{16pi}{3})(cos frac{17pi}{6}+isin frac{17pi}{6}))
= (frac{1}{2}×(cos(frac{⁡49pi}{6})+isin(frac{49pi}{6})))
= (frac{1}{2}×(cosfrac{pi}{6}+isinfrac{pi}{6})).

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