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Question from C3 mock - how is it supposed to work?

My whole class just had its C3 mock exam, which didn't go well for us at all. As the teachers aren't allowed to go through the papers until February, could anyone help me and my class understand a question that none of us managed to understand?

f(t) = ke^t where k is a constant with one real solution. Find k. (4 marks)

Nobody in the entire class has a clue where to start, so any help would be appreciated.
Original post by ChrisWolff
My whole class just had its C3 mock exam, which didn't go well for us at all. As the teachers aren't allowed to go through the papers until February, could anyone help me and my class understand a question that none of us managed to understand?

f(t) = ke^t where k is a constant with one real solution. Find k. (4 marks)

Nobody in the entire class has a clue where to start, so any help would be appreciated.


There would be one real root, so if a quadratic equation can be formed b^2-4ac=0.
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ChrisWolff
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Original post by NotNotBatman
There would be one real root, so if a quadratic equation can be formed b^2-4ac=0.


Yeah, we just discussed that as a class, but no clue how it turns into a quadratic.
Original post by ChrisWolff
Yeah, we just discussed that as a class, but no clue how it turns into a quadratic.


What was the full question?
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ChrisWolff
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Original post by NotNotBatman
What was the full question?


That's all we were given.
Original post by ChrisWolff
My whole class just had its C3 mock exam, which didn't go well for us at all. As the teachers aren't allowed to go through the papers until February, could anyone help me and my class understand a question that none of us managed to understand?

f(t) = ke^t where k is a constant with one real solution. Find k. (4 marks)

Nobody in the entire class has a clue where to start, so any help would be appreciated.


Differentiate then differentiate again?
Or take logs then réarrangé??
This just seems like a question with a constant it's not over confusing?
Natural logs cant be négative And remove The exponential term to get a straight Line graph
Original post by ChrisWolff
That's all we were given.


That's not possible. I think I've found the question; in part a.) you're givenf(x)=3−2e−x f(x) = 3-2e^{-x} so f(t)=3−2e−tf(t) = 3-2e^{-t} . Now solve for one real root.
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TajwarC
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Original post by ChrisWolff
That's all we were given.


I vaguely remember seeing this question before. That's not all the information you were given, there should be an f:x somewhere in the question too. Just replace the x with t and then you will have f(t) which you can then solve for k using discriminant
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ChrisWolff
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Original post by NotNotBatman
That's not possible. I think I've found the question; in part a.) you're givenf(x)=3−2e−x f(x) = 3-2e^{-x} so f(t)=3−2e−tf(t) = 3-2e^{-t} . Now solve for one real root.


Oh, thanks, that might have been it. I don't think I got as far as the next bit. What paper was that from, if that's alright? I'd love to have another go at it for peace of mind.
Original post by ChrisWolff
Oh, thanks, that might have been it. I don't think I got as far as the next bit. What paper was that from, if that's alright? I'd love to have another go at it for peace of mind.


Jan 2014 IAL.

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