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Algebra

Blazing goIITian

 Joined: 12 May 2007 Post: 574
20 May 2007 00:38:58 IST
0 People liked this
26
1530
10th KILLER QUES. OVER
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

OK , a gud 10th one........

Consider the following equation........

Summation ( r = 1 to 120 ) [  r /  ( x - r ) ]   >= 3

The values  of  x  which  satisfy  the above  eqn . are

x belongs to ( a, b ) union ( c, d ) union ( e , f ) ........................( consider all

the  sets of x for which the above holds true  )

FIND  ::  ( b + d + f + ...........)  -   ( a + c + e + ..... )

• 1
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Blazing goIITian

Joined: 12 May 2007
Posts: 574
20 May 2007 23:17:37 IST
0 people liked this

Blazing goIITian

Joined: 12 May 2007
Posts: 574
21 May 2007 01:27:54 IST
0 people liked this

ok , i am giving a final chance.........

Blazing goIITian

Joined: 12 May 2007
Posts: 574
21 May 2007 02:57:25 IST
0 people liked this

ok
instead of 20 pts , i will give u 20 salutes................

Hot goIITian

Joined: 26 Jan 2007
Posts: 156
21 May 2007 10:23:09 IST
1 people liked this

Blazing goIITian

Joined: 12 May 2007
Posts: 574
21 May 2007 16:02:20 IST
1 people liked this

fantastic.......
now , c'mon , gimme the proof................

Hot goIITian

Joined: 26 Jan 2007
Posts: 156
21 May 2007 16:52:51 IST
2 people liked this

consider,

1/(x+1) >= 3

implies

(3x - 4)/(x-1) <=0

or x lies in (1,4/3]   (upper bound(4/3) is the root of numerator and lower bound(1) is root of denominator)

similarly for

1/(x-1)  +  2/(x-2) >=3    implies

(3x2 - 11x +9)/(x-1)(x-2)  <=0

or x lies in (1, (11 - 13)/6 )(2, (11 + 13)/6 )

notice that again lower bounds and upper bounds form roots of numerator and denominator respectively.

you will find a similar case for

1/(x-1)  +  2/(x-2) + 3/(x-3) >=3

where lower and upper bounds form the roots.

therefore , for    1/(x-1)  +  2/(x-2) + 3/(x-3) ....... 120/(x-120) >=3

the roots of numerator and denominator of the simplified expression  will give upper and lower bounds.

What is asked is the sum of the roots of the numerator - sum of roots of denominator.

which you can easily find to be 2420.

That solves the problem!
the roots of the denominator will o

Blazing goIITian

Joined: 12 May 2007
Posts: 574
21 May 2007 17:18:17 IST
0 people liked this

OK , U' LL get ur points................

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