13Q. 6. A rectangle is inscribed inside a triangle of area M .
What is the maximum area of the rectangle??
I think M/2 is the ans...but not getting the proof now...
Here are the Questions i could remember .....
Q.1. What should be the values of x and y such that x2 + y2 is minimum and (x + 5)2 + (y – 12)2 = 142 ??
I got the answer as (x, y) = ( 5/13, -12/13 ) and min value as 1
Q.2. sorry i had written wrong Q here b4 ....
correct Qsn in post # 35
Q.3. If p is a prime number > 5, and its reciprocal can be written as
{the bar is over entire a1 to ar } where 
is the recurring part, prove that 10r on dividing by p leaves remainder 1.
This was the easiest of the lot.
Q.4. If a, b,c are odd integers, prove that the roots of ax2 + bx + c cannot be rational.
i proved that the Discriminant is of the form 8n + 5 .... so can't be perf. Sq That wud do?
Q. 5. There are six different paints given to you, and u have to paint all the faces of a cube with a different colour ….. In how many can this be done? {not sure but i think this was the Question}
Q. 6. A rectangle is inscribed inside a triangle of area M . What is the maximum area of the rectangle??
Is it M/2 ? Plz say yes
Q. 7. a, b, c , d are integers such that 
, b1 and b2 are integer multiples of ad – bc. Prove that the equations ax + by = b1 and cx + dy = b2 can have simultaneous solutions in integers.
Q. 8. Consider this sequence of natural numbers without the digit ZERO appearing in them : 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13,. ….. Find { an } . Compare this with a geometric series and prove that 
.
What is meant by the symbol {an} ?? [7] is it general term?
Q. 9. x, y, z are real numbers none equal to zero.
are complex numbers with 
.
If 
, prove that 
.
Q.10 in post # 35
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UP 0 DOWN 0 3 50
50 Answers
13
14)
if it would be rational, its denominator would be odd (otherwise, easy contradiction), so it would map to F2 (the field with 2 elements), but x^2+x+1 is irreducible over F2
1oh ... people know me as the ELIMINATOR ... so i would have failed in ISI .. no options .. no eliminations :(
1is the second Q subject to any other conditions(lyk a,b or c>0) becoz i can have a quadratic polynomial such dat its less dan 1 in [-1,1] but still less dan -3/2 in between!
19)
consider the triangle with vertices α,β,γ
0 is its circumcenter
0=xα+yβ+zγ means that x:y:z are the homogeneous barycentric coordinates of the circumcenter of this triangle
now we have to just show that these coordinates never sum to zero unless the triangle is degenerate
but this is clear , since the points whose barycentric coordinates sum to 0 are the points at infinity
hence proved
13Q. 5. There are six different paints given to you, and u have to paint all the faces of a cube with a different colour ….. In how many can this be done?
ans 30...once upon a time done on tiit
16)
by shear transformations you can transform the triangle into an isosceles right triangle
and that preserves maximality
the answer is obvious for a right triangle; it's essentially AM-GM
12) its stated wrong
this will be easiest if you draw it
split it up into f convex and f concave
and try to construct the extreme
once you've constructed the extreme you can prove it's correct
are you sure you've stated it correctly?
as stated it's clearly wrong
1i actually did the first Question by Calculus ...
@ platinum : yes this was how it was given.
orr ... if anybody else gave d ISI, can u pls confirm Q 2 ?
18) i am not going to solve but give some clue
how many n-digit numbers are there without digit 0
many can answer - its 9n
now ,you have a series, break it into the "1-digits" part, the "2-digits" part etc.
9in #14
wat are these barycentric and other high level stuff used
pls explain
11cele boy
can u xplain #10
if it would be rational, its denominator would be odd (otherwise, easy contradiction), so it would map to F2 (the field with 2 elements), but x^2+x+1 is irreducible over F2!!!!!!!!
11yaar meko uss ne bataya nahin !!!!!
meko laga ki yeh bahut assan hoga
tabhie poocha
jab meine poocha uss se uss ne taal diya![2]
91 Ans is 1
2 Q statement incomplete :(
3 10r/p = a1a2..ar. a1a2...ar =a1a2...ar + 1/p
so 10r = (a1a2..ar) p + 1 = pλ +1 hence proved
4 mirka yup thats the right method
5 is it 30 ???
opposite pairs for paints are chosen in 6C2 . 4C2 = 90 ways
these can be arranged in 2 unique ways so ans = 180
6 let h be one side of rectangle with base along BC
Area = h .( a - h(cotB +cotC ) ) use a= 2RsinA and complete the square
Area max = R2 sinAsinBsinC = M/2
7 y = b2a - b1c / ad - bc now b2,b1 multiples of ad -bc so y is integer
similar for x
8 Question not very clear :(
9 solving
u get
α = (Yβ + Zγ) / ( Y +Z )
ie α lies on chord joining β,γ as well as on circle
this is possible only if all 3 are same pts