Question

If f(x)=signum(sin^2x - sinx-1) has exactly 4 points of discontinuity in the interval 0 to n pi then value of n=?

Answer 1

sin^2x - sinx-1 is a continuous function ... looks for values of x where sin^2x - sinx-1 = 0

Answer 2

\[\Large x=\frac{1-\sqrt5}{2}\]

Answer 3

I neglected the other one since it was >1

Answer 4

this function doesn't seem to have 4 discontinuities on 0 to pi
http://www.wolframalpha.com/input/?i=sin%5E2x+-+sinx-1%3D0

Answer 5

http://www.wolframalpha.com/input/?i=Plot+signum%28sin%5E2x+-+sinx-1%29+from+0+to+pi

Answer 6

a little modification in the question :| sorry!

Answer 7

its n pi

Answer 8

the question now makes sense^

Answer 9

make modifications as you like ... the procedure is same. equate the f(x) to zero and find the values of x. Of course arcsin(X) gives you infinite values, choose any particular branckes as you like.
http://www.wolframalpha.com/input/?i=Plot+signum%28sin%5E2x+-+sinx-1%29+from+-2Pi+to+2Pi

Answer 10

its not arcsin !! :O :O

Answer 11

\[\LARGE f(x)=\sin^2x -sinx-1\]

Answer 12

of course you will have to take arcsin at some point ...

Answer 13

\[\LARGE x=\sin^{-1} \frac{(1-\sqrt{5})}{2}\]

Answer 14

the other one is neglected as i said since it is >1

Answer 15

let x be whatever, you know sin(2pi + x) = sin(x) -- (1) also you know sin(pi-x) = sin(x) --(2)
this is negative, you will not likely encounter them in 0 to pi,
you know sin is negative on pi to 2pi, using above relation (2) you will encounter (1-sqrt(5))/2 on pi to 2pi twice.
using relation (1) you have n=4

Answer 16

cool :O thanks!