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Trigonometry past paper Qustion

here is the Question:
https://ibb.co/rf5ZBCq
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by papie
here is the Question:
https://ibb.co/rf5ZBCq

What have you tried / what are you stuck with?
Reply 2
Avatar for papie
papie
OP
16
Original post by mqb2766
What have you tried / what are you stuck with?

i) i tried to find the range by setting x=0 in eqn so i could get the y-intercept, i get q . the graph is at max when value of x=1/4 pie so i substitude x=1/4 pie to get y coordinate. then i have to solve p sin^2 *2(1/4 pie) +q. but then i get p sin^2(1/2)+q i dont know wht to do here.

ii) a,b and c i dont get it. :frown:
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by papie
i) i tried to find the range by setting x=0 in eqn so i could get the y-intercept, i get q . the graph is at max when value of x=1/4 pie so i substitude x=1/4 pie to get y coordinate. then i have to solve p sin^2 *2(1/4 pie) +q. but then i get p sin^2(1/2)+q i dont know wht to do here.

ii) a,b and c i dont get it. :frown:


i) Probably easier to think
a) what is the range of sin(2x)
b) what is the range of sin^2(2x)
c) what is the range of p*sin^2(2x)
d) what is the range of p*sin^(2x) + q
The question says "state", so it should require little/no working.

that should help for ii) as well?
(edited 4 years ago)
Reply 4
Avatar for papie
papie
OP
16
Original post by mqb2766
i) Probably easier to think
a) what is the range of sin(2x)
b) what is the range of sin^2(2x)
c) what is the range of p*sin^2(2x)
d) what is the range of p*sin^(2x) + q
The question says "state", so it should require little/no working.

that should help for ii) as well?

ok i get the 1st part
for part b of ii) i get it as well when u set f(x)=q u get p sin^2 (2x)=0 so graph is back on x-axis so u get 3 solutions which is the answer.
but i still am confused abt part a and c.
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by papie
ok i get the 1st part
for part b of ii) i get it as well when u set f(x)=q u get p sin^2 (2x)=0 so graph is back on x-axis so u get 3 solutions which is the answer.
but i still am confused abt part a and c.


You've exactly the right idea. Part iia) is very similar to iib)?
What did you get for the range?
Reply 6
Avatar for papie
papie
OP
16
Original post by mqb2766
You've exactly the right idea. Part iia) is very similar to iib)?
What did you get for the range?

wait lol i actually got the part ii)a when u set eqn equal to p+q, u get p sin^2x -p the coordinate of highest point on graph is p+q so when u substract p+q the the peaks end up below the x-axis and u have only 2 solutions. it looks smthing like this : https://ibb.co/1GLQcgq
so there's only 2 solutions which is this answer. im still thinking abt the part c
part c is simple when u substract 1/2p graph intersects x-axis at 4 points. so thx :biggrin:
(edited 4 years ago)

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