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Mat q3 2010

q3 2010.pngq3 2010 sol.png
Could someone explain why the answer to part (iv) is what it is? Why is c tangential at that point? I feel I'm missing something obvious.
(edited 2 years ago)
Original post by username53983805
q3 2010.pngq3 2010 sol.png
Could someone explain why the answer to part (iv) is what it is? Why is c tangential at that point? I feel I'm missing something obvious.

Rearranging part ii to get it in terms of c, you’ll see that c is tangential to the equation and that it must be repeated at the second hump for 5 solutions, as the equation sin(x)/x is odd you’ll just need to consider the positive second hump. At least that’s how I did it
Original post by The A.G
Rearranging part ii to get it in terms of c, you’ll see that c is tangential to the equation and that it must be repeated at the second hump for 5 solutions, as the equation sin(x)/x is odd you’ll just need to consider the positive second hump. At least that’s how I did it

i don't understand, could you expand on what you mean?
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mqb2766
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Edit - the stuff below (original response) seems a bit verbose for 2 marks, so another simpler way would be to simply note that locally around the tangent point
sin(x) <= cx
so
sin(x)/x <= c
with equality occuring at the tangent point (first inequality) of sin(x) with cx. So y=c must correspond to the local maximum (second inequality) of sin(x)/x. This would seem about right for 2 marks and the derivative stuff is in the next question part.


Edit - original response
As above, but fleshing it out a bit
At that tangent point you have
sin(x) = cx
or
sin(x)/x = c
So the line y=c intersects with sin(x)/x at that point.

Also the derivatives are equal, so
cos(x) = c
or
cos(x)-c = 0
The derivative of sin(x)*x^-1 =
x^-1(cos(x) - sin(x)/x) = x^-1(cos(x) - c) = 0
at that point.

So y=c is a tangent to sin(x)/x at that point, which is the first positve maximum (ignoring zero).
(edited 2 years ago)

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