SOLVE THE EQUATION 6X^3-11X^2-3X+2=0 Given roots are in H.P

 

                             BS GREWAL SOLUTION

  QUESTION:-SOLVE THE EQUATION 6X^3-11X^2-3X+2=0 Given roots are in H.P

SOLUTION:-

GIVEN THAT THE EQUATION OF THE ARE IN HP, THE ROOTS OF THE EQUATION HAVING RECIPROCAL ROOTS WILL BE IN A.P.

THE EQUATION WITH RECIPROCAL ROOTS IS

6(1/X)^3-11(1/X)^2-3(1/X)+2=0

2X^3-3X^2-11X+6=0 ---------1 EQUATION

SINCE THE ROOTS OF THE EQUATION ARE IN HP, THEREFORE THE ROOTS OF EQUATION 1 IS IN AP. LET THE ROOTS BE a-d,  a+d then

By adding these roots we get,

3a=3/2

a=1/2

and by multiplying  we get

a(a^2-d^2)=-3

½(1/4d^2)=-3

¼-d^2=-6

¼+6=25/4

d=5/2

thus, the roots of equation 1 are -2,1/2,3

Hence the roots of the  given equation are -1/2,2,1/3

find the particular integral of (d^3+1)y=sin(2x+3)

                   BS GREWAL SOLUTION

QUESTION:-find the particular integral of  (d^3+1)y=sin(2x+3)

 AND HERE IS THE SOLUTION

solve p^3+2xp^2-(yp)^2-2(y^2)xp=0

 BS GREWAL SOLUTION 

QUESTION:-solve p^3+2xp^2-(yp)^2-2(y^2)xp=0

SOLUTION:- and here is the answer

  

if u= tan-1(y^2/x) prove that x2 d2u/dx2+2xy d2u/dxdy+y2d2u/dy-sin2u*sin2u

                              Bs Grewal solution

 question:-if u= tan-1(y^2/x) prove that x2 d2u/dx2+2xy d2u/dxdy+y2d2u/dy-sin2u*sin2u.

solution:-

if y=[x+(x^2+1)^1/2]^m prove that yn+2+(n^2-m^2)yn=0 at x=0 .hence find the yn(0)

                                 BS GREWAL   SOLUTIONS

  question:- if y=[x+(x^2+1)^1/2]^m prove that yn+2+(n^2-m^2)yn=0 at x=0 .hence find the yn(0)

SOLUTION:-

 hence proved 

==> REDUCE TAN-1(COSӨ+ISINӨ) TO THE FORM OF A+IB. HENCE SHOW THAT , TAN-1(E^IӨ) =NΠ/2+Π/4-I/2* LOG TAN(Π/4-Ө/2)

solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0

           BS GREWAL

        DIFFERENTIAL EQUATION SOLUTION

QUESTION:-solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0

SOLUTION:-

 => solve (x^2+y^2-a^2)xdx + (x^2-y^2-b^2)ydy=0

solve (x^2-ay)dx =(ax-y^2)dy

=> Reduce tan-1(cosӨ+isinӨ) to the form of a+ib. hence show that , tan-1(e^iӨ) =nπ/2+π/4-i/2* logtan(π/4-Ө/2)

=> complex function ,separate into real and imaginary parts 1.exp(z^2),z=(x+iy) 2. exp(5+iπ/2) 3. e^(16+30i)

Prove that (a+ib)m\n+(a-ib)m\n=2(a2+b2)m\2n .cos((m\n)(tan-1b\a))

solve (x^2+y^2-a^2)xdx + (x^2-y^2-b^2)ydy=0

  BS GREWAL 

DIFFERENTIAL EQUATION SOLUTION

   question:-solve (x^2+y^2-a^2)xdx + (x^2-y^2-b^2)ydy=0

SOLUTION:-

  another question :-solve (x^2-ay)dx =(ax-y^2)dy

Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube.

Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

solve (x^2-ay)dx =(ax-y^2)dy

 BS GREWAL 

                DIFFERENTIAL EQUATION QUESTIONS 

Question:-solve (x^2-ay)dx =(ax-y^2)dy

solution:-   

Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube.

 Bs Grewal 

maxima and minima questions

Question:- Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube.

Solution:-let the equation of a sphere of the radius is

x^2+y^2+z^2=0---------1 eqn

Let x,y,z be the discussion of the rectangular in sphere  1eqn, then, its volume,

V=xyz

Let u=v^2= (xyz)^2

u=(xy)^2(a^2-x^2-y^2)  by equation 1

u=(axy)^2-x^4y^2-x^2y^4

du/dx=2 (a^2)xy^2-4x^3y^2-2xy^4

du/dy=2 (a^2)x^2*y-2x^4*y-4x^2y^3

for maxima or minima of u,we have

du/dx=o, du/dy=0

=2x(y^2)(a^2-2x^2-y^2)=0

2(x^2)y(a^2-x^2-2y^2)=0

.i.e. (a^2-2x^2-y^2)=0  =>  a^2=2x^2+y^2 ----------2 equation

a^2-x^2-2y^2=0      => a^2=x^2+2y^2  ---------------3equation

now subtracting equation 3 from equation 2, we have

x^2-y^2=0 => x=y

putting in equation number 1, we have

x=y=z=a/√3

now r= d^2u/dx^2=2(ay)^2-12(xy)^2-2y^4

s= d^2u/dxdy=4(a^2)xy-8(x^3)y-8xy^3

t=d^2u/d^y^2=2(ax)^2 – 2x^4-12(xy)^2

at x=y=a/3,we have

r=-8a^4/9,s=-4a^4/9

t=-8a^4/9

rt-s^2=41a^8/81=+ve   i.e.  rt-s^2>0

since r<0 & rt-s^2>0 ,so the u I.e. volume of the rectangular solid (V) is maximum when x=y=z=a/3

hence, the maximum rectangular solid inscribed in a sphere is a cube ( x=y=z=a/√3)

 

 

Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

 

Reduce tan-1(cosӨ+isinӨ) to the form of a+ib. hence show that , tan-1(e^iӨ) =nπ/2+π/4-i/2* logtan(π/4-Ө/2)

      BS GREWAL

/--------------------------------------- complex number solution---------------------------------------------------/

Reduce tan-1(cosӨ+isinӨ) to the form of a+ib. hence show that, tan-1(e^iӨ) =nπ/2+π/4-i/2* log tan(π/4-Ө/2)

 

Solution :- let tan-1(cosӨ+isinӨ)=x+iy

 Then tan(x+iy)= (cosӨ+isinӨ)

  tan(x-iy)= (cosӨ-isinӨ)

now tan2x=[tan(x+iy)+ tan(x-iy)]/1- tan(x+iy) tan(x+iy)

tan2x=2cosӨ/1-(cos^2Ө+sin^2Ө)

tan2x=2cosӨ/1-1

tan2x=∞

i.e. 2x=nx+π/2

x=nx/2+π/4

tan2iy= tan(x+iy)- tan(x-iy)

= [tan(x+iy)- tan(x-iy)]/1+ tan(x+iy)*tan(x-iy)

tan2iy=2isinӨ/2

=tanh2y=sinӨ

ð  (e^2Ө-e^-2Ө)/ (e^2Ө+e^-2Ө)= sinӨ

By applying componendo and dividendo, we get

e^2Ө/ e^-2Ө=(1+ sinӨ)/ 1-sinӨ

=[(cosΘ/2+sinӨ/2)]^2/[(cosӨ/2-sinӨ/2)]^2

e^4Ө=(1+tan Ө/2)^2/(1-tan Ө/2)^2

e^2Ө=(tanπ/4+tan Ө/2)/( tanπ/4-tan Ө/2)

= tan(π/4+Ө/2)

Ө=1/2log tan(π/4+Ө/2)

e^-2Ө=(1-tan Ө/2)/(1+tan Ө/2)

=(tanπ/4-tan Ө/2)/( tanπ/4+tan Ө/2)

e^-2Ө= tan(π/-Ө/2)

-2Ө=log tan(π/-Ө/2)

 

Ө=-1/2log tan(π/4-Ө/2)

Hence , tan-1(e^iӨ) =nπ/2+π/4-i/2* log tan(π/4-Ө/2)

hence proved

 

 

 

 

 

 

 

 

 

Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

                             BS GREWAL EXERCISE QUESTION 

Question :-Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

/* --------------------------COMPLEX NUMBERS----------------------------------------------------------------*/

Solution:- let x is the nth root of unity form a series of geometric progression ,we have

x^n=1

x=(1)^1/n

x=(1+i0)1/n

x=(cos0+isin0)^1/n=(cos2rπ+isin2rπ)^1/n

x=cis2rπ/n

Putting r=0,1,2,3,3…,n-1

x=1,(cis2π/n)^r=w^r

.i.e.  x=1,w,w^2,w^3,……..,w^(n-1)

Clearly its form a geometric progression since w is one nth root of unity

Then w^n=1

w^n-1=0

(w-1)(1+w+w^2+…….+w^n-1)=0

w≠ 1

(1+w+w^2+…….+w^n-1)=0

Hence the sum of nth root of unity is zero

Again, 1*w*w^2.……..*w^n-1

w^(1+2+3+4…n-1)

w^(n(n-1)/2)

[(Cisπ)^2/n]^n(n-1)/2

(-1)^n-1

 

 hence proved😉😋

 * /--------------------------------------------------------------------------------------------------------------*/

 

If w is a complex cube root of unity ,prove that 1+w +w^2 =0

                                       Bs Grewal

Question:-if  is a complex cube root of unity, prove that 1+⍵+⍵^2  =0

 

Solution:-let x be a cube root of unity, then

x^3=1

= x=(1)^1/3 = (1+0i)^1/3

x=(cos0+isin0)^1/3

x=(cos(2nπ)+isin(2nπ))^1/3

x=(cos(2nπ/3)+isin(2nπ/3))

Putting n=0,1,2

x=1, cos2π/3+isin2π/3, cos4π/3+isin4π/3

x=1, cos2π/3+isin2π/3, (cos2π/3+isin2π/3)^2

Putting  ⍵ =  cos2π/3+isin2π/3

x=1,⍵,⍵^2

since  ⍵ is a cube root of  unity, then

⍵^3=1

⍵^3-1=0

(⍵-1)(⍵^2+⍵+1)=0

such that ⍵≠1

(⍵^2+⍵+1)=0

complex function ,separate into real and imaginary parts 1.exp(z^2),z=(x+iy) 2. exp(5+iπ/2) 3. e^(16+30i)

                                        BS GREWAL SOLUTION

   Complex functions

If for each value of the complex variable Z(x+iy) in a given region R,we have one or more values of W(-u+iv) ,then w is said to be a complex number of Z

Written as f(z) = w = u(x,y)+iv(x,y)

Exponential function of a complex variable

e^x= 1+x/1!+x^2/2!+ ---------+x^n/n!+-----------,x €R

then e^z=1+z/1!+z^2/2!+ ---------+z^n/n!+---------, z € Ө

properties:-

* Exponential form of z =r(e^iӨ)

*e^z is periodic function having imaginary period 2πi

.i.e. e^(z+2πi) =e^z* e^2πi = e^z

·         e^z is not zero for any value of z

·      e^z = e^z⎺

 

question:-separate into real and imaginary parts

1.exp(z^2),z=(x+iy)

2. exp(5+iπ/2)

3. e^(16+30i)

HERE IS THE SOLUTION

Solution   1. :-   e^(z^2)=e^(x+iy)^2

 e^(x^2-y^2+2ixy) = e^(x^2-y^2)*e^(2ixy)

(e^(x^2-y^2))*(cos2xy+isin2xy)

 real part= e^(x^2-y^2)

imaginary part= e^(cos2xy+isin2xy)

 

Solution   2.:- exp(5+iπ/2)=e^(5+iπ/2)

Let x+iy=e^5*e^ iπ/2

             =  e^5*(cosiπ/2+isin π/2)

=e^5*(i)

y= e^5*(i)

 

Solution   3.:- e^(16+30i)

X=16 and y=30

 

e^16*(cos30+isin30)

e^16*(√ 3/2+i1/2)

solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0

prove that sin3α+8sin3β+27sin3¥= 18sin(α+¥+β) and cos3α+8cos3β +27cos3¥=18cos(α+¥+β) if, sinα+2sinβ+3sin¥=0, cosα+2cosβ+3cos¥=0

                Bs Grewal

        Exercise question

  Here is the solution of a complex number

 Question:-:-if sinα+2sinβ+3sin¥=0, cosα+2cosβ+3cos¥=0 prove that sin3α+8sin3β+27sin3¥= 18sin(α+¥+β) and cos3α+8cos3β +27cos3¥=18cos(α+¥+β)

Solution:- let a= cosα+isinα  =  cisα

         b   =2( cosβ+ isinβ  )=  2cisβ

     c    = 3(cos¥ + isin¥ )=  3cis¥

      and given sinα+2sinβ+3sin¥=0,

         cosα+2cosβ+3cos¥=0

    then  a+b+c = cisα+2cisβ+3cis¥

   => a+b+c = (cosα+2cosβ+3cos¥)+i(sinα+2sinβ+3sin¥)

   => a+b+c = 0+i0=0

   => now (a+b+c)^3 = a^3+ b^3+ c^3+3(a+b+c)(ab+bc+ca)-3abc

   since,  (a+b+c)=0

   => (cisα)^3+(2cisβ)^3+(3cis¥)^3 =3 cisα cisβ cis¥

   =>Cis3α+ 8Cis3β+27 Cis3¥=18cisα cisβ cis¥

  => Cos3 α +isin3 α +8cos3 β +8isin3 β +27cos3¥  +27isin3¥ =18cis(α+ β+¥ )

  => Cos3 α+8cos3 β+27cos3¥  +i(sin3α+8sinβ+27sin3¥)=18(cos(α+β+¥)+isin(α+β+¥))

    On comparing real and imaginary part

    cos3 α+8cos3 β+27cos3¥  =18cos(α+β+¥)
   (sin3α+8sinβ+27sin3¥)= 18sin(α+β+¥)

     Hence proved

solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0

Solve the equation x^4-2x^+4x^2+6x-21=0. Given that the sum of two of its roots is zero

 

          Bs Grewal

         exercise solution

Question:- Solve the equation x^4-2x^3+4x^2+6x-21=0. Given that the sum of two of its roots is zero.

 

Solution:- let 𝛂,𝛃,Ɣ,𝛅 are the roots of the equation such that 

Also, 𝞪+𝛃+Ɣ+𝛅=2    =>Ɣ+𝛅=2

Thus the quadratic factor corresponding to α,β is of the form

X^2-0x+p and that corresponding to   is of the form of x^2-2x+q

Hence,  x^4-2x^3+4x^2+6x-21=(x^2+p)(x^2-2x+q)

 x^4-2x^3+4x^2+6x-21=x^4+(p+q)x^2-2x^3-2xp+pq

 

Equating coefficients of from On both  sides of the above equation, we get

p+q=4     and    -2p=6   => p=-3

and q=7

hence the given equation is equivalent to 

hence the roots x=±√3,1+i√3

solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0