IIT JEE , AIEEE Forum - Mathematics , Algebra

kishore.subramanian.b

1. Find the greatest value of |z-i| if |z+1+i|<=1.

2.If all the roots of the equation of z³+az²+bz+c=0 are of unit modulus, then

a. |a|<=3 b. |b|<=3 c. |c|=1 d. |c|>=3

3.The number of rational roots of the equation

x².3^x+1+3^|x--4|+3=x².3^|x--4|+5+3^x--1 is

a. 1 b. 2 c. 3 d. none

4. The co-efficient of x⁶¹⁵ in (x-1)(x³-3)(x⁵-5)(x⁷-7)................(x⁴⁹-49)is

a. 15 b. 25 c. 9 d.30

5. If a,b,c are in G.P. and x,y are the arithmetic means between a and b,

b and c respectly, then the value of (a/x + c/y) - (b/x + b/y) is

a. 2 b. 4 c. 0 d. -4

aysh

5)a,b,c are in G.P.
Therefore,
b²=ac

ac-b²=0.

Now,
x=(a+b)/2 & y=(b+c)/2

Thus,
(a/x+c/y)-(b/x+b/y)={2a/(a+b)+2c/(b+c)} - {2b/(a+b)+2b/(b+c)}
=(2ab+2ac+2ac+2bc-2ab-2b^2-2b^2-2bc)/
{(a+b)(b+c)}
=(4ac-4b²)/{(a+b)(b+c)}
=4(ac-b²)/{(a+b)(b+c)}
=0.

PLEAZZZZ RATE IF CORRECT...

cooldude

2)let the roots be Z1,Z2,Z3.

then Z1Z2Z3=(-c)

apply mods on both side

Iz1IIz2IIz3I=I-cI

1=IcI

correct option is c

rate if the solution is correct.

cooldude

3)assume x>4,Ix-4I=x-4

L.H.S=R.H.S

infinite solutions correct option is d

kishore.subramanian.b

2nd question is multiple answer question

nitin62225

ans 1= 1+

5

ans 2= a,b,c

ans 3= all x>=4(d)

ans 4=30

ans 5=0

soln:

1)max distanxe of z lying on the given circle from i will be along the line throgh i passing through the centre of the circle. from 2 solutions obtained the one which lies on the other side of centre as i,gives the ans.

2)let z1,z2,z3 be the roots,

so c=1,

a=Iz1+z2+z3I<=Iz1I+Iz2I+Iz3I<=3

similarly b<=3

3)posted above

4)if u check for the maximum power of x that u'll get it comes out to b 625

to obtain 615 u have to omit some powers of x while mutiplying the brackets.

u'll have 2 remove (1,9) and (3,7) powers while actually multiplying the term along them.......

this gives 1*9+3*7=30

5)posted above

plz rate.

cooldude

2)let the roots be cisP,cisQ,cisR.

b=cis(P+Q)+cis(Q+R)+cis(R+P)

on simplification IbIsquared=3+2cos(P-Q)+2cos(Q-R)+2cos(R-P)

IbImax=3,when P=Q=R

also -a=cisP+cisQ+cisR

on simplification IaIsquared=3+2cos(P-Q)+2cos(Q-R)+2cos(R-P)=IbIsquared

IaImax=3

therefore option 1,2,3 are correct.

aankurverma

ans . of 5th is 0

n for 2nd multiple ans.

let the root b z1 , z2 , z3

z1z2z3 = - c

taking mod

[z1][z2][z3] = [-c] {[z1] = [z2] = [z3] = 1}

.'. [-c] = 1

or [c] = 1

for 1st

[z+1+ i] <= 1

it represents a circle passing through (-1,-1)

or it's general eqn.

x^2 + y^2 + 2x+ 2y + 1 = 0

n

[z - i]

represents a circle passing through (0,1}

so max value is =

1+

5

coz it cut s at two points n it pass through two points

plz rate if i m right

iitkgp_bipin

Coordinate approach is good. You may also apply inequality approach :

|z₁ - z₂| >= | |z₁| - |z₂| |

Hence |z - i - (-1 - 2i) | >= | |z-i| - |-1-2i| |

| |z-i| - |1+2i| | <= |z + 1 + i| <= 1

Hence | |z-i| -

5 | <= 1

Hence maximum value of |z-i| is 1+

5.

abhijeet_0201

i coulnt get the sol of 4th ques .can somebody plz elaborate!!!!!!!!!!!!!!!

nitin62225

if u check for the maximum power of x that u'll get it comes out to b 625

to obtain 615 u have to omit some powers of x while mutiplying the brackets.

u'll have 2 remove (1,9) and (3,7) powers while actually multiplying the term along them.......

this gives 1*9+3*7=30

prathima

nitin or someone can u plz xplain it in detail

plzzzzzzzzz

abhijeet_0201

yeah,plz!!!!!!!!!!!!!

abhinavabcd

1)2 is the ans (by graphical interpretation)
2)A B C is the ans
3)none (all rationals >4 are solns)
4)D is the answer(degree is 625 to get 615 omit 3,7 &1,9 multyplying and adding 21+9=30)
5)ans is zero(easy to solve but if it is a objective Q put a=b=c=x=y)

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