IIT JEE , AIEEE Forum - Mathematics , Algebra

sumi_coolguy

Q1:the no of +ve integral solutions of abc=30 are____________

Q2:if a_r is co efficient of x^r in expansion (1+x+x²)²⁰then -a₁+2a₂-- 3a₃+.......+40a₄₀=_______

Q3:the tangent at a point (1,-2) to the circle x²+y²=5 touches circle x²+y²- 8x+6y+20=0 at (p,q)then p+q=______________

tanvi586

ans for Q3 what i got is p+q=2.

respond if it is right i will give the solution

taruntanuj007

For question 1

a=1 b= 6 c =5 arrange them so 3!

a=1 b= 15 c =2 arrange them so 3! again

total 2*3! = 12 solutions where a,b,c take different values

priyesh

Ans to question 1 is 27

because when a=1

b=1,c=30

b=2,c=15

b=3,c=10

b=5,c=6

total cases for a =1 are 4 multiplied by 2(because b & c can interchange there places) = 8

similarly for a =2 we have 4 cases

for a=3 & a=5 we have 4 cases each

for a= 6 we have two cases

for a= 10 & a=15 we have 2 cases each

for a = 30 we have one case

total cases =8 + 4 + 4 + 4 + 2 + 2 + 2 +1 = 27

therefore ans is 27

SMARTY

HI FRIEND, I GOT ONE OF THE SOLUTIONS
IT IS THE THIRD ONE THE RIGHT ANSWER IS 2
TELL ME WHETHER I AM RIGHT

ankurgupta91

3. the equation of the tangent at ao point (1,-2) to the circle x^2+y^2=5 is given by
x-2y=5 (by the formula xx'+yy'=a^2)
nw this tangent and the circle x^2+y^2- 8x+6y+20=0 touch each other at (p,q)
so they should satisfy the both equatiions
then, p-2q=5 ---------------(1)
p^2+y^2-8p+6q+20=0 -------(2)
by solving these two equations , we get
p=3 and q=-1
so p+q=2
thats the answer
hope u can understand the logic.............

frenied

let (1+x+x²)²⁰=a₀+a₁x+...........

differnetiate both sides w.r.t. x to get

20(1+x+x²)¹⁹(0+1+2x)=0+a₁+2a₂x+3a₃x²+.......

put x=1 on both sides....

20(1+1+1)¹⁹(1+2)=Reqd. Sum=20.3²⁰

ankurgupta91

freined is right bt put x=-1
let (1+x+x^2)^20=a0+a1x+..............
differnetiate both sides w.r.t. x to get
20(1+x+x^2)^19(0+1+2x)=0+a1+2a2^x+3a3x^2+........
nw put x=-1 on both sides.......
20(1-1+1)^19(0+1-1)=0+a1-2a2+3a3-4a4.........
a1-2a2+3a3-4a4....................=20

frenied

hey ankur we've to calculate the sum of all +ve terms.i thnk thats wat the question asks.so thats wat ive done.or let the person who posted the question decide but i need a salute for that.

shenoy_apoorva

hi friend
the problem you have sent is tricky but i tried to solve it the answer for the third one is p+q=2
please inform me if i am right and please give me a salute for my hardwork

SMARTY

HEY FRIEND YOU HAVE FORGOTTEN TO GIVE ME A SALUTE

DON007

everybofy is discussing 3rd one & 2nd one.........

i should discuss the 1st one....isnt it...........now....

look...........

abc= 30.......

triplets can be formed is as follows

1) 6 5 1 = 3!

2)15 1 2 = 3!

3) 2 3 5 = 3!

4) 3 10 1 = 3!

add them we get................

= 24 .........(ans).

hope i m correct..............

thanku............

and all r right .....pakka..........

in third one p+q=2....

now 4 second prblm............

(1+x+x²)²⁰=a₀+a₁x+...........

differnetiate both sides w.r.t. x to get

20(1+x+x²)¹⁹(0+1+2x)=0+a₁+2a₂x+3a₃x²+.....

multiply by x we get

x *20(1+x+x²)¹⁹(0+1+2x)=0+a_1^x+2a₂x²+3a₃x³+.......

now put...........

x= - 1

we get........

-a₁+2a₂-- 3a₃+.......+40a_{40 ^{= 20.......}}

freniend and ankur has done a good job .............

but first multiply by x again............

okkkkkkkk

........ thanku..........

hope u understand.........

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