The Student Room Group

Alevel maths question

Hi can someone help me find the value of b, I get how to integrate the question but I’m stuck in the third line.
Thanks
Of course. Whats the question?
By inspection I suppose. Bring the e^2 and e^-4 to the other side and compare both sides like an identity
Original post by beatriceaj71
Of course. Whats the question?

To find the value of b. I’ve attached the method but I don’t know how to go from the third line to b=2
Original post by Y12_FurtherMaths
By inspection I suppose. Bring the e^2 and e^-4 to the other side and compare both sides like an identity

I’ve tried to do that but I’m still confused sorry
Original post by Jessica_rica
I’ve tried to do that but I’m still confused sorry


e^2=e^b and e^-2b =e^-4 so clearly b=2
Original post by Jessica_rica
I’ve tried to do that but I’m still confused sorry


Seems rather an odd bit of working, what was the original question?
Original post by Jessica_rica
Hi can someone help me find the value of b, I get how to integrate the question but I’m stuck in the third line.
Thanks


B....=…..2
Original post by Jessica_rica
I’ve tried to do that but I’m still confused sorry


Inspection is the simplest way to do it.



But also you can just deduce bb without any integration. You should be aware of that ex>0e^x > 0 for all xx therefore 3ex+6e2x>03e^x + 6e^{-2x} > 0 for all xx.

If you're going to be integrating between 2 and b then you're finding the area under this curve between these two values. But since our funcion is always positive, the area between any two values is going to be positive (or in other words, non-zero). The only way the area is zero is if the length of the region over which we are integrating is zero. Hence b2=0b- 2 = 0 and the result follows.


As pointed out by @ghostwalker, this workout seems odd (and rather trivial) so we're just wondering what the question was asking for in the first place.
(edited 4 years ago)

Quick Reply

Latest