Question

Show that for real x that {[cos x + 2 sin x + 1]/[cos x + sin x] } cannot have a value between 1 and 2.

Let y = [(cos x+2 sin x + 1)/(cos x + sin x) ]

y(cos x + sin x) = (cos x + 2 sin x + 1)
sin x(y-2) + cos x(y-1)=1 , I just feel that this isn't the way to do this..

Any other way to do this?

Answer 1

cosx+2sinx+1 = (cosx+sinx) + (sinx+1)

So, your quotient can be written
1+(1+sinx)/(cosx+sinx)

Now, since cosx < 1, cosx+sinx < 1+sinx
So, (1+sinx)/(cosx+sinx) > 1

and the rest follows.

Answer 2

Hope you accept the fact that the max of cosx + sinx is √2
and its minimum is -√2
These occur at x = 45° and x = 225° (π/4 and 5π/4)
see: http://www.wolframalpha.com/input/?i=y+%3D+sinx%2Bcosx

we can write it as
y = (cosx + sinx + sinx + 1)/(cosx + sinx)
= 1 + (sinx + 1)/(sinx + cosx)

the graph of y = (cos x+2 sin x + 1)/(cos x + sin x)
suggests that the statement is true
see: http://www.wolframalpha.com/input/?i=plot+y+%3D+(cos+x%2B2+sin+x+%2B+1)%2F(cos+x+%2B+sin+x)

we have critical values at x = -45°, 135°, 315° , causing asymptotes for those values. The denominator would be zero.
Look at the first graph

consider the intervals from
a) -45° to 135° , we know that cosx+sinx has a max of √2 at 45° and is positive for that interval
and y = 1 + (sinx + 1)/(sinx + cosx)
= 1 + (positive)
y>1

at 135°, the function is undefined, or y is infinitely large ----> y > 1

for x between 135 and 315, cosx + sinx has a min of -√2 and is negative throughout the interval.

let's take some samples for y = 1 + (sinx + 1)/(sinx + cosx)
x = 140 , y = -12.3 , which gives us y < 1
x = 180, y = -2 , which gives us y < 1
x = 200, y = .4886 , which gives us y < 1
x = 225, y = .79.. , y < 1
x = 310, y = -.89.. , y < 1

according to my reasoning, at x = 320, we should get y > 1
x = 320 , y = 1 + (sin320 + 1)/(cos320+sin320) = 3.898.. , true enough

Answer 3

Looks like mine was overkill

Answer 4

Many thanks to both of you!