Fundamentals - FTSE questions.........

62

Lokesh Verma ·2008-11-10 00:16:42

yup this is correct..

and the solution is very simple u can make use of it!

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-10-20 04:06:22

Q.2 Q.3 i solved in exam, Q.1 became too lengthy does a shortcut there Q.4 nahi bana....

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-10-20 04:07:46

Q.5 (need hint)
a^y-b^y=log_xb (given)

then find the area of rectangle..... in fig.....

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-10-20 04:34:15

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

Kumar Saurabh ·2008-10-20 06:51:55

i dont understand the question!

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-10-20 06:59:13

U have 2 find the area of the rectangle formed by y axis and three lines such that two vertices are on given graph

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

Kumar Saurabh ·2008-10-20 07:41:47

my answer i get is

x(log_ax-log_bx)

how it constant for all x?

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-10-20 07:47:30

also,
a^y-b^y=log_xb (given)

UP 0 DOWN 0
Post on FacebookPost anonymously?

33

Abhishek Priyam ·2008-11-10 00:13:44

i think it should be such that light travels as shown after reflections....

any comments??

UP 0 DOWN 0
Post on FacebookPost anonymously?

62

Lokesh Verma ·2008-10-20 04:02:16

hey priyam any success on these?

The last one seems to be a bit tricky.. i guess i will have to give it some time..

din click on the first look :)

UP 0 DOWN 0
Post on FacebookPost anonymously?

62

Lokesh Verma ·2008-11-10 00:21:52

if u dont figure out.. do let me know. it is very easy..

btw when i post the solution of

http://targetiit.com/iit_jee_forum/posts/10th_november_2008_480.html

It will become crystal clear to u..

Right now u might be using similar triangles and crap.. then u might not need anything other than a straight line!!

:)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

E=mc^2 darkangel ·2008-11-20 23:30:11

is the answer (29/7,29/7) on x=y
and (29/10,0) on x axis

i took the image point q of p taking x axis as mirror then took q's image r taking x=y as mirror then joined r and p which is a straight line and also the minimum distance as we know that minimum distance b/w two points is a straight line

UP 0 DOWN 0
Post on FacebookPost anonymously?

62

Lokesh Verma ·2008-11-20 23:40:54

I remember /7 in the denominator when i solved..

ur method is perfect..
I cant say the final answer cos i will nedd to check..

just tell the 2 points that u joined.. then we will be sure that u are absovlutely correct :)

good work coolspirit :)

gr8 that u got the method i was talking :)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

E=mc^2 darkangel ·2008-11-20 23:42:14

is the answer of 3rd one 1-1/10C5

UP 0 DOWN 0
Post on FacebookPost anonymously?

62

Lokesh Verma ·2008-11-21 06:21:22

yes i got the same answer :)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

kunalb Bhattacharya ·2009-01-08 00:49:39

Let s=(a.b'+a'.b)/|ab|---1
we know |ab|=|a||b|
we consider a=x+iy--2 & b=p+iq---3...
..On putttin 2 & 3 in 1 along with resp conjugates we get-----
s=2(px+qy)/âˆš{(x²+y²)(p²+q²)}...thus s is real always...

Hence we proceed as under.....

s²=(a.b'+a'.b)²/|a|²|b|²
s²={(a.b'-a'.b)/|ab|}²+4{(aa')(bb')/|a|²|b|²}
........THE LAST STEP ABOVE IS IMP.....

now...we know aa'=|a|²....same 4 bb'....

thus now...
s²={(a.b'-a'.b)/|ab|}²+4

now subtracting any 2 diff complex num of the above exp can giv any real num other than 0....thus greatest value of the exp in braces is zero....

s²=4
......max(s)=2.......(Ans)