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# 2 exam questions watch

1. Some exam problems I can't figure out. Hope you can help!

1) .
If n is the number of roots of this function, then
a)
b) n = 0
c) n = 2
d) n = 1

I tried doing but I got that the result is d), but in the paper option c) is marked as correct (well it might be wrong coz this was not taken from a book).

Apart from solving the equation or drawing the graph, is there any other method to find the roots of a function?

2) is a twice differentiable function and has exactly 6 roots. If n is the number of zeros of then we can say that
a) n = 6
b) n 4
c) n = 4
d) n 4

I didn't even try solving this one coz I have absolutely no idea what to do

Thanks!
2. To be honest, I would have drawn a graph for the first question.
It's easy to see that when x = 1, f(x) is negative, but when x approaches 0, f(x) tends towards infinity, and that when x > 2, f(x) is also positive. f(x) is also positive at values less than 2, but I don't have my calculator handy and really can't be bothered. All you'd need to do (for working) would be to show that f(x) is negative at x=1, but positive at x=0 and x=2. This is the change of sign method. As the highest order is 3, there cannot be more than 2 real roots. I think, anyway.
In fact, here's this:

http://www10.wolframalpha.com/input/...+-+%284%2F3%29

For the 2nd question, when we differentiate twice, it will lower the maximum number of (I assume the question is talking about real) roots by 2.
Further than that, I can't help, because I don't know what determines whether the roots of a differentiated function are real or not.
I'm quite sure you can eliminate a) and d) anyway.

I'm answering this because I'm bored. I won't have been much help though!
3. 1) f(x) = x^3/3 - ln(x) - 4/3
...f'(x) = x^2 - 1/x
Let f'(x) = 0,
0 = x^2 - 1/x
0 = x^3 - 1, has only 1 root at x = 1 hence 1 turning point hence the original function must have 2 zeroes (c)
4. (Original post by Miss Mary)
2) is a twice differentiable function and has exactly 6 roots. If n is the number of zeros of then we can say that
a) n = 6
b) n 4
c) n = 4
d) n 4!
Let the function be f(x) then if f''(x) has 6 roots, then f''(x) has 5 turning points, f'(x) must then have 6 turning points and 7 roots and so f(x) would have 7 turning points and 8 roots hence not a)

If f''(x) has 4 roots, then f''(x) has 3 turning points, and f'(x) would have 4 turning points and 5 roots so that f(x) would have 5 turning points and 6 roots hence c) it is.

By the same reasoning, it cannot be b) or d) as you would end up with less and more zeros in f(x) respectively.
5. (Original post by Karunesh)
1) f(x) = x^3/3 - ln(x) - 4/3
...f'(x) = x^2 - 1/x
Let f'(x) = 0,
0 = x^2 - 1/x
0 = x^3 - 1, has only 1 root at x = 1 hence 1 turning point hence the original function must have 2 zeroes (c)
I think there is something wrong with your reasoning.

Here is a graph with one turning point but the original function does NOT have 2 zeroes.

6. I wrote a lengthy reply to your post steve but I seemed to have refreshed my page and lost my solution to that.

Basically, further work can be done; we know from the reasoning that f(x) is a parabola as it has 1 turning point, now we need to eliminate either b) or c) as those are the cases. That's pretty simple to do..
7. Key fact: If a (continuous, differentiable) function f(x) has N roots, then f'(x) must have at least N-1 roots, but it can have more.

(As far as I'm aware, all the posted answers to Q2 are wrong, because of people not taking the last part of the sentence above into account).
8. (Original post by Karunesh)
I wrote a lengthy reply to your post steve but I seemed to have refreshed my page and lost my solution to that.

Basically, further work can be done; we know from the reasoning that f(x) is a parabola as it has 1 turning point, now we need to eliminate either b) or c) as those are the cases. That's pretty simple to do..
f isn't a parabola - that is a very specific curve (quadratic). But 1 turning point => at most two roots, which is what I think you really meant.
9. (Original post by DFranklin)
f isn't a parabola - that is a very specific curve (quadratic). But 1 turning point => at most two roots, which is what I think you really meant.
I agree that at most f has two roots BUT this does help distinguish between three of the given options of 0,1 or 2. Since the question is multiple choice it seems to me the best option is to sketch and
10. (Original post by steve2005)
I agree that at most f has two roots BUT this does help distinguish between three of the given options of 0,1 or 2. Since the question is multiple choice it seems to me the best option is to sketch and
Well, I would go: when x is large, x^3/3 dominates so y is large and positive. When x is near zero, -ln x dominates so y is again large and positive. When x=1, y = 1/3-4/3 = - 1 < 0. So y has two sign changes (one on each side of x = 1) and therefore at least two roots.
11. I have a question: Is it correct to talk about a function having roots? I thought roots applied to equations.

has no roots but if then of course the equation does have roots.
12. (Original post by steve2005)
I have a question: Is it correct to talk about a function having roots? I thought roots applied to equations.

has no roots but if then of course the equation does have roots.
The distinction seems to be something the A-level exam boards are keen to emphasise, but one real mathematicians couldn't care less about.

My inclination was to say that technically you're right, but in practice no-one is that pedantic. But in fact Wiki explicitly disagrees with you:

(Original post by http://en.wikipedia.org/wiki/Root_%28mathematics%29)
In mathematics, a root (or a zero) of a complex-valued function ƒ is a member x of the domain of ƒ such that ƒ(x) vanishes at x, that is, x s.t. f(x) = 0
If you look at discussions of the Riemann Hypothesis or the Fundamental Theorem of Algebra, you'll see things like "the roots of the Zeta function" much more than "the roots of the equation ".

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