# Past paper question

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Hello!

Could someone please explain (a) (ii)? I've attached a screenshot of the question as an image.

Edited by thepositiveclub
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Just curious, which exam paper is this from?

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Just curious, which exam paper is this from?

May 2010 Paper 2 TZ2

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Wouldn't the values of k be the y-coordinates of P and Q?

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Okay I have something to clarify while we're at this: when they say "exactly two solutions" what do they mean?

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Okay I have something to clarify while we're at this: when they say "exactly two solutions" what do they mean?

I've got the same question, its really confusing Your answer is correct though..

This is the mark scheme: a)

(ii) recognizing that it occurs at P and Q (M1)
e.g. x = âˆ’1.15, x =1.15
k = âˆ’1.13, k =1.13

Could you please explain how you got it?

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Okay I have something to clarify while we're at this: when they say "exactly two solutions" what do they mean?

I've got the same question, its really confusing Your answer is correct though..

This is the mark scheme: a)

(ii) recognizing that it occurs at P and Q (M1)
e.g. x = âˆ’1.15, x =1.15
k = âˆ’1.13, k =1.13

Could you please explain how you got it?

Well I assumed (which is the problem here, it was an assumption) that "exactly two solutions" were the x-coordinates that were found in (i) for two reasons

1) they would never have a (ii) if it weren't related to (i)

2) And this is where I was confused...When they said tangent parallel to the x-axis, I visualized a line going through x=1.15 and x=-1.15 and it would pass through two points on the graph. So would those be "exactly two solutions" that the question refers to? <- This assumption led me to believe that the k values would simply be the y-coordinates of the x-values found previously.

I know that probably made no sense...I was just giving you what went through my head. I'll work on it unless someone else provides an answer

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After some thought I think I found the answer.

First thing I struggled with was to understand the question. Essentially it is asking you to find all values of y that that can be found with exactly two x-values. And the only y-values on the graph that have two x-values are at the max and min points.

The min is at (-1.15, -1.13). By plugging in a straight line, x= -1.13, and finding where it intersects with the graph, you will find it intersects at (1.86, -1.13). Hence one k value = -1.13.

The max is at (1.15, 1.13). Doing the same thing the intersection is at (-1.86, 1.13), owing to the symmetry of the function. Hence the other k value is = 1.13, as it has two x-values (1.15 and -1.86)

Edited by dorianb
• 1

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Okay I have something to clarify while we're at this: when they say "exactly two solutions" what do they mean?

I've got the same question, its really confusing Your answer is correct though..

This is the mark scheme: a)

(ii) recognizing that it occurs at P and Q (M1)
e.g. x = âˆ’1.15, x =1.15
k = âˆ’1.13, k =1.13

Could you please explain how you got it?

Well I assumed (which is the problem here, it was an assumption) that "exactly two solutions" were the x-coordinates that were found in (i) for two reasons

1) they would never have a (ii) if it weren't related to (i)

2) And this is where I was confused...When they said tangent parallel to the x-axis, I visualized a line going through x=1.15 and x=-1.15 and it would pass through two points on the graph. So would those be "exactly two solutions" that the question refers to? <- This assumption led me to believe that the k values would simply be the y-coordinates of the x-values found previously.

I know that probably made no sense...I was just giving you what went through my head. I'll work on it unless someone else provides an answer

After some thought I think I found the answer.

First thing I struggled with was to understand the question. Essentially it is asking you to find all values of y that that can be found with exactly two x-values. And the only y-values on the graph that have two x-values are at the max and min points.

The min is at (-1.15, -1.13). By plugging in a straight line, x= -1.13, and finding where it intersects with the graph, you will find it intersects at (1.86, -1.13). Hence one k value = -1.13.

The max is at (1.15, 1.13). Doing the same thing the intersection is at (-1.86, 1.13), owing to the symmetry of the function. Hence the other k value is = 1.13, as it has two x-values (1.15 and -1.86)

This makes more sense now, thank you both so much!

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