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c3 2014 past paper question 5

B0258E6C-538B-4888-8777-D6C999118AD7.jpg.jpeg
How do you arrive at that answer for part b?
Reply 1
Avatar for Notnek
Notnek
21
Original post by Rockgo288
B0258E6C-538B-4888-8777-D6C999118AD7.jpg.jpeg
How do you arrive at that answer for part b?

If you use a division method you can split the fraction into

g(x)=1+3x−2\displaystyle g(x) = 1+\frac{3}{x-2}

Does that help?
Reply 2
Avatar for JP0458B
JP0458B
16
They subbed in the smallest value that x can be (technically speaking it can't be 3 but for inspection purposes let it be so, and they then showed what g(3) is, and thus you know it must be less than that

hope that helps
Reply 3
Avatar for Rockgo288
Rockgo288
OP
11
Original post by Notnek
If you use a division method you can split the fraction into

g(x)=1+3x−2\displaystyle g(x) = 1+\frac{3}{x-2}

Does that help?


Omg!! Thanks!!! Never thought of the need to do long division!!
Reply 4
Avatar for Notnek
Notnek
21
Original post by Rockgo288
Omg!! Thanks!!! Never thought of the need to do long division!!

Yes even for a simple fraction like this, if the degree of the numerator is the same or higher than the deinoniator then division can be useful.

Alternatively you can think about what happens as x gets really big and consider that the +1 on the top and the -2 on the bottom become irrelevant so the fraction tends to x/x = 1.

If all else fails with these types of questions, plug in loads of values into it using your calculator to get an idea about what happens as x varies.

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