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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: "TARGET IIT - JEE 2008 || CHALLENGING PROBLEMS FROM MATHS"
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Avinash_Bhat (625)

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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........

    
Avinash_Bhat (625)

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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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Avinash_Bhat (625)

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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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Avinash_Bhat (625)

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No one ????
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hsbhatt (4475)

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1st one's -1/2. i will give the soln only if no one gets it.

Time wounds all heels
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rockey (168)

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1. -1/2
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sonalexam (19)

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HEY CUMMON GIVE THE SOLUTIONS NOW....

EVEN SKY IS NOT THE LIMIT...FLY HIGHER IF YOU CAN .!
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Avinash_Bhat (625)

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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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hsbhatt (4475)

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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(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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