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currysauce

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Hey there, its been awhile since i've done my maths

and now i'm doing actuary!! :(

so i am rusty... and am wondering (from u genious's) how do complete these so called simple calculus questions...

1. Deduce that a+b/2 >= root(ab) for all non0negative real numbers, a,b. When does the equality hold?

i did the deduction, now i dun understand the second part... cheers

2. Show graphically, or otherwise, that

sinx = mx

has no non-zero solution for -pi < x < pi when m<0

thanks
 
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gman03

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1)

at some point you would have say (a-b)2 >= 0. The equality holds if a-b=0 (i.e. a=b)

2)

What does -pi<0 mean?
 

speedie

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1. consider {root(a)-root(b)}^2>=0
a-2root(ab)+b>=0
a+b>=2root(ab)
[a+b]/2>=root(ab)

equality holds when a,b=0, or 1
 

speedie

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2. graph y=sinx, y=mx

you'll see that at -pi in the sinx curve the y value is 0. this means that the line mx has to have a gradient of 0 to have an y value of 0 at -pi. therefore there is no 0 solution for -pi as m cannot be zero or y will not equal to mx, instead = 0x =0.
 

speedie

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currysauce said:
sorry, see new qu 2
once again, just graph y=sinx and y=mx, where m<0

between -pi and 0, the sinx curve is <0, while y=mx is >0, as m<0.
between 0 and pi, the sinx curve is >0, while y=mx is <0, as m<0.

therefore there are no non zero solutions for -pi<x<pi for m<0
 

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