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further pure 1

that the equation f(x) = 1-2sinx has one root which lies in the interval [0.5,0.8] .
I tried and got 0.98 for f(0.5) which is wrong, plz help
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TeeEm
19
Original post by alesha98
that the equation f(x) = 1-2sinx has one root which lies in the interval [0.5,0.8] .
I tried and got 0.98 for f(0.5) which is wrong, plz help


make sure you are using radians
Try turning your calculator into Radians Mode!
Reply 3
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alesha98
OP
4
OK thanks . I got x=pie/6. What is the next step?
Original post by alesha98
OK thanks . I got x=pie/6. What is the next step?


So that answer is positive. Now do the same for the other one, you should get a negative value out if you've done everything right :smile:

Therefore because it's a continuous function and there is a change in sign therefore the root must lie between that range of values :smile:
Reply 5
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alesha98
OP
4
Thanks I got the answer now. May you explain why I need to turn calculator into radian mode?
Original post by alesha98
Thanks I got the answer now. May you explain why I need to turn calculator into radian mode?


Normally there should be some kind of hint in the question. Can you see any radian values in the question? In these type of questions you're always looking for a change in sign. If the question dosen't tell you - try it both ways :smile:

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