The Student Room Group

A level maths - trigonometry help

Hi, , I don't quite understand how to answer Q10 c, and i'd appreciate it if someone could help.

Here's the question:



Part A, I just had to draw the graph, pretty easy stuff (it's almost like a normal sec graph, just stretched and shifted up)

For part b, i just wrote down the x coordinates of the stationary points.

But for part C, i would think the maximum value of 1/(1+2sec) would be when the denominator (1+2sec) is really big, but from the graph, I can see that the range can go all the way up to negative and positive infinity, so that doesn't help, and I'm not sure why solution bank have chosen the minimum and maximum points.
(edited 10 months ago)
Reply 1
Original post by Muhammad012
Hi, , I don't quite understand how to answer Q10 c, and i'd appreciate it if someone could help.

Here's the question:



Part A, I just had to draw the graph, pretty easy stuff (it's almost like a normal sec graph, just stretched and shifted up)

For part b, i just wrote down the x coordinates of the stationary points.

But for part C, i would think the maximum value of 1/(1+2sec) would be when the denominator (1+2sec) is really big, but from the graph, I can see that the range can go all the way up to negative and positive infinity, so that doesn't help, and I'm not sure why solution bank have chosen the minimum and maximum points.

For the maximum of 1/y say. It must be positive as y has positive values, and consider some simple cases so
y = 3,4,5,6
1/y = 1/3, 1/4, 1/5, 1/6
Fairly clearly, the largest value of 1/y corrresponds to the smallest value of y.

Another similar way to look at it is as a reciprocol so
z = 1/y
or
zy = 1
When y is large, z must be small and vice versa. Max z must correspond to min y.
(edited 10 months ago)
Reply 2
Original post by mqb2766
For the maximum of 1/y say. It must be positive as y has positive values, and consider some simple cases so
y = 3,4,5,6
1/y = 1/3, 1/4, 1/5, 1/6
Fairly clearly, the largest value of 1/y corrresponds to the smallest value of y.

Another similar way to look at it is as a reciprocol so
z = 1/y
or
zy = 1
When y is large, z must be small and vice versa. Max z must correspond to min y.

My mistake, i meant to say that the *minimum* value of y would be when the denominator is really big. but even by looking at the graph, the (1+2sec) can go all the way up to infinity
Reply 3
Original post by Muhammad012
My mistake, i meant to say that the *minimum* value of y would be when the denominator is really big. but even by looking at the graph, the (1+2sec) can go all the way up to infinity


If the function was always positive, your argument would hold. However, y and 1/y can be negative and -1 is "more minimum" than 0 so the minimum of 1/y is the largest negative value, which corresponds to the smallest negative value of y. So bascially the same argument as before, but below the x-axis.
(edited 10 months ago)

Quick Reply

Latest