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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Feb 2008 12:07:42 IST
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 HI FRIENDS,
  I WOULD TRY MY BEST TO GET THE BEST MATHS QUESTIONS THAT ARE OF THE SAME STANDARD AS THAT OF THE PRESENT IIT-JEE LEVEL AND WE SHALL HAVE RELATED DISCUSSIONS ON THEM.
HEY FRIENDS,
PLEASE GIVE YOUR ANSWER WITH FULL SOLUTION SO THAT EVERY VIEWER UNDERSTANDS IT WELL........
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Feb 2008 12:14:59 IST
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QN.1 Type : Only one of the options is correct.
A sequence A1, A 2, A3 ............. , An of real numbers is such that :
A1 = 0 | A2 | = | A1 + 1 | | A3 | = | A2 + 1 | . . . | An | = | An-1 + 1 |
Then the Arithmetic Mean of all these numbers cannot be less than:
(a) - 7 / 2 (b) - 5 / 2 (c) - 3 / 2 (d) - 1 / 2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Feb 2008 12:34:17 IST
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QN.2 Type : Match the following.
(A) In = 0 /4 tan n x dx (P) I1 , I2 , I3 ........... are in A.P.
(B) In = 0 /2 cos (nx) cos n x dx (Q) I1 , I2 , I3 ........... are in G.P.
(C) In = 0 sin (2nx) / sin x dx (R) I1 , I2 , I3 ........... are in H.P.
(D) In = 0 ( sin (nx) / sin x )2 dx (S) I1 + I3 , I2 + I4 , I3 + I5 ........... are in H.P.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Feb 2008 18:16:21 IST
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No one ????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Feb 2008 18:53:48 IST
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1st one's -1/2. i will give the soln only if no one gets it.
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Time wounds all heels |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Feb 2008 00:57:32 IST
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1. -1/2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Feb 2008 16:36:13 IST
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HEY CUMMON GIVE THE SOLUTIONS NOW....
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EVEN SKY IS NOT THE LIMIT...FLY HIGHER IF YOU CAN .! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Feb 2008 19:54:05 IST
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The answer is correct for QN.1............
But, no one has got the full solution ???
What about question 2 ? Is it that difficult ?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Feb 2008 20:02:36 IST
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Bhattre! We have |a2| = |a1+1| => (a2)2 = (a1)2+2a1+1 |a3| = |a2+1| => (a3)2 = (a2)2+2a2+1 . . . |an| = |an-1+1| => (an)2 = (an-1)2+2an-1+1 |an+1| = |an+1| => (an+1)2 = (an)2+2an+1 Adding up the equations and noticing that each (ai)2 gets subtracted in the subsequent equation, we get (an+1)2 = 2(a1+a2+...+an)+n>0 Hence (a1+a2+...+an)/n>-1/2 This means the AM of the terms from 1 to n is greater than -1/2.
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Time wounds all heels |
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