Weird 2U Question in Cranbrook Trial (1 Viewer)

[studybuddy09]

Got this question on Friday's 2009 test and I've been arguing with my friends over the answer

The question was find p so that 0<(=) x <(=) pi

when sinx=px

i wrote that p>0 because i thought about the quadrants and for x to satisfy domain it must be in the first and second quadrant, while another mate of mine (scardizzle) reckons its all real p


I dont really know

Pls Help

Studybuddy09

 

[Timothy.Siu]

you are right.
for sin x=px to have a solution 0<=x<=pi

p>=0

 

[Trebla]

I'm not exactly sure what the question is asking for...is that the full question?

It's asking for values of p which satisfy 0 ≤ x ≤ π given that sin x = px. My interpretation is that it's looking for possible values of p for which the solution to sin x = px lies within 0 to π with end points inclusive.

If that's the correct interpretation then the answer would be all real p, because the solution x = 0 lies within 0 to π inclusive and also satisfies sin x = px regardless of what value p actually is...

 

[Enigmanation]

If the values of p are to be such that y=px will intercept with y=sinx (0<= x <= pi) then p>=0. p cannot be all real numbers as this would include p<0, and y=-x or any straight line with a negative gradient will not intercept with y=sinx between 0 and pi.

 

[Trebla]

The key thing is that the interval 0 ≤ x ≤ π INCLUDES x = 0. This means that say y = - x works because x = 0 satisifies sin x = - x (in other words they intersect at the origin)...

 

[Enigmanation]

Oh my bad, cant believe i didn't realise that, yes your completely spot on. All real p.

 

[scardizzle]

The key thing is that the interval 0 ≤ x ≤ π INCLUDES x = 0. This means that say y = - x works because x = 0 satisifies sin x = - x (in other words they intersect at the origin)...

yeh that's what i was thinking, it always intersects at x = 0 hence there will always be a solution reqardless of p, i think study buddy misinterpreted the question.

 

[cutemouse]

How many marks was it?

Knowing who wrote the trial, I'd say it's been pilfered from somewhere. heh

 

[phoebeee]

Omg, I don't even understand the question

 

[scardizzle]

How many marks was it?

Knowing who wrote the trial, I'd say it's been pilfered from somewhere. heh

it was worth 2 marks

Omg, I don't even understand the question

it's pretty much asking what can p be for there to be a solution for sinx = px or alternativley sinx - px = 0

when x= 0

0 - 0 = 0 regardless of p

well atleast that's what i reasoned we'll probably find out the proper solution tomorrow

 

[cutemouse]

it was worth 2 marks

Hmm, then it should be something more sophisticated.

well atleast that's what i reasoned we'll probably find out the proper solution tomorrow

Good old SKB XD