Saif95
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Can you guys please help me with part iii and iv of this question please?

According to examiners report these were very hard questions and hardly anyone got full marks.
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Mohid Khan
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hi, which exam board is this?
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Mohid Khan
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i think this is how you do it:

(iii) put the x value of a and c into the gradient function to get the gradient and if you times the two gradients together you should get -1.
(iv) simply integrate then substitute the cordinates of a to find +C , Now just use the x coordinate of a and substitute into the intigrated equation, this will give you the area.

Hope this helps!
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Phichi
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(Original post by Saif95)
Can you guys please help me with part iii and iv of this question please?

According to examiners report these were very hard questions and hardly anyone got full marks.
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For iii). multiply the gradients of either tangent together. When multiplying the gradients of perpendicular lines, you should get -1.

iv). Integrate the original y= equation you're given, remembering the proof you're given in the question. You're finding the area between the lines CD and BE, bounded by the curve and the x axis. So integrate between the two x co-ordinates, relating to point E and D. Integrate from D to E.
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the bear
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it does not seem all that difficult ?
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Pride
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I think the difficulty is that you could make lots of silly mistakes here, but it's actually quite normal in terms of difficulty (compared to other A-level papers).
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Phichi
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Stick to the question, no need to share opinions on difficulty when OP didn't ask for it.
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Khallil
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(Original post by Saif95)
Can you guys please help me with part iii and iv of this question please?

According to examiners report these were very hard questions and hardly anyone got full marks.
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For the tangents to be perpendicular in part iii, the product of their respective gradients must equal -1.

Regarding part iv, since there isn't a standard integral for \ln x that we know of, I suggest letting u=\ln x

The formula for integration by parts is

\displaystyle \int u \ \dfrac{\text{d}v}{\text{d}x} \ \text{d}x = uv - \int v \ \dfrac{\text{d}u}{\text{d}x} \ \text{d}x

and is derived from the product rule of differentiation.

If you're interested
\displaystyle \text{Let } u \text{ and } v \text{ be functions of } x \\ \\ \therefore \dfrac{\text{d}}{\text{d}x} \ uv = u \ \dfrac{\text{d}v}{\text{d}x} + v \ \dfrac{\text{d}u}{\text{d}x} \\ \\ \implies \int \dfrac{\text{d}}{\text{d}x} uv \ \text{d}x = \int \left( u \ \dfrac{\text{d}v}{\text{d}x} + v \ \dfrac{\text{d}u}{\text{d}x} \right) \ \text{d}x \\ \\ \implies uv = \int u \ \dfrac{\text{d}v}{\text{d}x} \ \text{d}x + \int v \ \dfrac{\text{d}u}{\text{d}x} \ \text{d}x \\ \\ \implies \int u \ \dfrac{\text{d}v}{\text{d}x} \ \text{d}x = uv - \int v \ \dfrac{\text{d}u}{\text{d}x} \ \text{d}x
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Davelittle
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It's not that hard, think about what you've done in class and the advice given here and have a shot at it
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