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Hi All,
Even Abhujaba is wrong

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The Mistake
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If ab=ac then b=c. this is wrong.
If ab=ac, then either b=c or a=0.
So, when u ended up with 2f '(0) =2f '(0) .K
we have, either f '(0)=0 or K=0
So, 2 cases arise
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End of mistake
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A better solution is as follows:
f(x+y) +f(x-y) =2f(x).f(y)
put x=0 and y=0
so we get, K+K=2*K^2
There fore, either K=0 or K=1 ..................................( 1 )

Now put x=0 in the given equation. U get,
f(y) +f(-y) = 2 K f(y)

If K=0, then the function is odd.
If K=0, then the funstion is even.
So, the answer is A and B

Tarin is right.
But instead of putting in numbers, it is always advisable to put in terms of points. I mean, instead of writing

6, write

AQP.
That always keeps the solution clean.

Very simple question,
The equation of a Hyperbolic Cylinder is (x2/a2)-(y2/b2)=1, where x2 denotes x^2.
here x=0 and y=0 are called principle axes
Refer to http://en.wikipedia.org/wiki/Quadric
The given equation can be written as,
( x+y-z )^2- (y-z)^2 = 1
Hence the given equation is a Hyperbolic Cylinder with x+y-z=0 and y-z=0 as principle axes.

Dear Aditya,

First take, y =

x, then x=y^2, so, dx=2y dy
Hence, given integral =

(2*y*tan(y))dy
Now for product rule,

UdV = UV-

VdU, take, U=y and dV=tan y
use

tan x = ln |cos x|

Now, u will end up with an expression, where the only term which is unknown is

ln |cos y|dy

ln |cos y|dy=

ln |cos y|*sin y * cosec y * dy

Now take, U=cosec y, and dV = ln |cos y|*sin y
Now u will end up with an expression, where the only unknown is

^{sec y dy
Now}

sec y dy = ln |sec y + tan y|
thats alll!!!

Yes.. Priyesh is right.

Hello Eswar,
As stated by Siddhartha, the actula statement for Wilson's theorem is:

(p-1)!+1=0(mod p)
Since, divisibility of 23 is asked for, we have to relate (18!+1) to (23-1)!+1.
So lets start with (23-1)!+1. let us say a=(23-1)!+1 and b=(18!+1)
By Wilson's theorem, a is divisible by 23. So, let us say, a=23c
23c = a = (23-1)!+1 = (18!)*(19*20*21*22)+1
= (18!+1)*(19*20*21*22)-1*(19*20*21*22)+1
= b*(19*20*21*22)-175560+1
= b*(19*20*21*22)-175559
= b*(19*20*21*22)-23*7633
Therefore,
b*(19*20*21*22) = 23*(c+7633) ...........................Eqn 1
The RHS of Eqn 1 is divisible by 23, so must be the case with LHS.
Therefore, LHS is divisible by 23. But, (19*20*21*22) is not divisible by 23. Hence b must be divisible by 23.
Hence, (18!+1)is divisible by 23
Hence proved!!!