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# [A Math] - Need help with my school prelim paper

• Hi all, I would like to ask a few questions with regards to my A maths prelim paper 2...

Q1. A curve is such that y =(10 - 3x)(2x + 1)^5. Obtain the coordinates of the stationary points of the curve and determine their nature.

Is it true that is has only one stationary point? Can anyone verify it for me? When I checked my answers, there seems to be only one stationary point...

Q2. Variables N and t are related by N = ab^t. Linearise the equation in the form Y = mX + C

I took ln N = ln a + t ln b on everything... will I get the whole question wrong when I should be taking lg on everything? I only realised it after the paper 2.

Q3. A curve is such that (Integral UL 4 LL 0) f(x)dx = 5

Evaluate (Integral UL 4 LL 2)f(x) dx - (Integral UL 0 LL 2)f(x) dx

UL stands for upper limit and LL stands for lower limit. Will I get the whole thing wrong if I were to evaluate f(x) which is 5/4 since the question didn't state that f(x) should not be found? AND the examiner told me that this question has no flaws and there's something I need to find... so when I evaluate each definite integral, do I have to take modulus when it is not a 'area under graph' question? Or is there another trick to this question that I didn't notice?

• Originally posted by tanpantheon1:

Q1. A curve is such that y =(10 - 3x)(2x + 1)^5. Obtain the coordinates of the stationary points of the curve and determine their nature.

Is it true that is has only one stationary point? Can anyone verify it for me? When I checked my answers, there seems to be only one stationary point...

No, there are two stationary points.

dy/dx would give you
-3(2x+1)^5 + (10-3x)5(2x+1)^4  * 2
= (2x+1)^4 (-6x+3 + 100-30x)
= (2x+1)^4 (-36x+103)

Solving gives you x = -1/2 and 103/36

Originally posted by tanpantheon1:

Q2. Variables N and t are related by N = ab^t. Linearise the equation in the form Y = mX + C

I took ln N = ln a + t ln b on everything... will I get the whole question wrong when I should be taking lg on everything? I only realised it after the paper 2.

No nothing wrong.

Originally posted by tanpantheon1:

Q3. A curve is such that (Integral UL 4 LL 0) f(x)dx = 5

Evaluate (Integral UL 4 LL 2)f(x) dx - (Integral UL 0 LL 2)f(x) dx

UL stands for upper limit and LL stands for lower limit. Will I get the whole thing wrong if I were to evaluate f(x) which is 5/4 since the question didn't state that f(x) should not be found? AND the examiner told me that this question has no flaws and there's something I need to find... so when I evaluate each definite integral, do I have to take modulus when it is not a 'area under graph' question? Or is there another trick to this question that I didn't notice?

I don't understand your question. If this is only part of the full question, please post the full question.

If unsure what you need to post, please look at our forum guidelines.

_____________________________________________________________________________________

(4)
A curve is such that [(integral sign) f(x) dx] = 5

(0)

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(4)                                 (0)

Evaluate {[(integral sign) f(x) dx] - [(integral sign) f(x)dx]}

(2)                                 (2)

• Sorry I don't understand your question. Isn't it obvious that the answer is just 5?

(4)                                 (0)

Evaluate {[(integral sign) f(x) dx] - [(integral sign) f(x)dx]}

(2)                                 (2)

is same as

(4)                                 (2)

Evaluate {[(integral sign) f(x) dx] + [(integral sign) f(x)dx]}

(2)                                 (0)

which is the same as

(4)

Evaluate {[(integral sign) f(x) dx]

(0)

Why do you need to find f(x) and how is it 5/4?

You have too many sub questions with regards to your exam question, and I have no idea what your exact question is.

You could always post a picture of the actual question.