9k then W cant be calculated manually
Without graphing can you write out all solutions to x2 = 2x ?
This is not entirely JEE stuff, but you can learn an interesting concept here
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20 Answers
341I presume you are asking for a polynomial expansion. Just wiki for lambert W function and read it up. There is an article in Wolfram also
11Then without olympiad level math, how will u know the other solution? [7]
62celestine W is a known function...
This is like the log formula.. it is like asking what how do we know log2 is whatever it is !!!
341Yes, I came across this technique very recently
There is a function called Lambert's W function.
Suppose xex= y (x,y>0), then x = W(y) in other words it is the inverse of the function f(x) = xex
Now we know that a negative root exists. Let us put x = -y and rewrite the equation as y2 = 2-y
So, now, we can take logarithm on both sides and obtain
2 ln y = -y ln 2 or 1/y ln(1/y) = ln2/2
If we put ln(1/y) = z, we get the equation zez = ln2/2
which means z = ln(1/y) = W(ln2/2) and hence y = e-W(ln2/2)
This is a standard function in many math software and plugging this number in we obtain x = -y = -0.7666
You can use this to solve an equation like xx = 2. Try it expressing the solution in terms of the W function.
11There r three solutions......x=2 is one. I don't know abt the other two. [4]
1Prophet please tell me , I am curious to know .
@ Priyam I knew about that , there was nothing new in it .
Remark : I can provide far better graphs . :D
341The solutions that are easy to spot are x=2 and x = 4
let f(x) = 2x-x2
f(0) = 1, and as x→∞, f(x)→-∞
The continuity of f(x) means there is a root in the interval (-∞,0)
How to find that root is pretty interesting
33Diptosh
See:
http://targetiit.com/iit_jee_forum/posts/how_many_times_63_838.html
1But isn't graphs the easiest way to do such problems ?
Obviously it has two , how can it have 3 solns ?
prophet tell us about it .
341Something to go on:
We already have an idea that there are two solutions for x>0 and one for x<0
The JEE related part is finding the +ve roots and proving that there are no more than two. x = 2 is one. That should make finding the other root easy
The tough part is finding the negative root (no, not Newton Raphson's :D)