a = -1 and b =1OK, now that 2010 is here I am reviving the old Revision Marathon game.
Rules:
The OP posts a question, first person to get it correct asks the next question, and so on.
First Question
LaTeXhow do you get that font? that wikipedia one that allows you to do (divide by signs)?
Please read the guidelines of this exercise, you must post a question upon answering one correctly (which you did).a = -1 and b =1
u=x^2, v=ln(x)Sorry; my net got disconnected and it took ages to reconnect.
u=x^2, v=ln(x)
u'=2x, v'=1/x
y'=(vu'-uv')/v^2 (chain rule)
=[2xln(x)-x]/(lnx)^2
=x[2ln(x)-1]/(lnx)^2
Question; Solve: tanx=3cotx
ROFL! Yeah that's the one!Quotient rule you mean xD
This Q can be solved geometrically. There is also an algebraic method. If not solved by a 2010 student by tomoz, i will show bothwhere n is an integer
Find the equation of the tangent(s) to x^2 + y^2 = 4, which are parallel to y= x + 3
2U (ie. long) method:Find the equation of the tangent(s) to x^2 + y^2 = 4, which are parallel to y= x + 3
If a(x) and b(x) are even functions, a(-x)=a(x) and b(-x)=b(x)If a(x) and b(x) are even functions while c(x) and d(x) are odd functions, prove that:
(i) a(x).b(x) is an even function
(ii) a(x).c(x) is an odd function
(iii) c(x).d(x) is an even function
Really? I'm not thinking well...Wrong.
Find the eqn of the locus of the point so that it is equidistant from points 3, 2 and -1, 5