thepositiveclub Posted May 12, 2015 Report Share Posted May 12, 2015 (edited) Hello!Could someone please explain (a) (ii)? I've attached a screenshot of the question as an image. Thanks in advance! Edited May 12, 2015 by thepositiveclub 1 Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 12, 2015 Report Share Posted May 12, 2015 Just curious, which exam paper is this from? Reply Link to post Share on other sites More sharing options...
thepositiveclub Posted May 12, 2015 Author Report Share Posted May 12, 2015 Just curious, which exam paper is this from? May 2010 Paper 2 TZ2 Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 12, 2015 Report Share Posted May 12, 2015 Wouldn't the values of k be the y-coordinates of P and Q? Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 12, 2015 Report Share Posted May 12, 2015 Okay I have something to clarify while we're at this: when they say "exactly two solutions" what do they mean? Reply Link to post Share on other sites More sharing options...
thepositiveclub Posted May 12, 2015 Author Report Share Posted May 12, 2015 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 = −1.15, x =1.15k = −1.13, k =1.13 Could you please explain how you got it? Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 12, 2015 Report Share Posted May 12, 2015 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 = −1.15, x =1.15k = −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 reasons1) 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 Reply Link to post Share on other sites More sharing options...
dorianb Posted May 12, 2015 Report Share Posted May 12, 2015 (edited) 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 May 12, 2015 by dorianb 1 Reply Link to post Share on other sites More sharing options...
thepositiveclub Posted May 12, 2015 Author Report Share Posted May 12, 2015 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 = −1.15, x =1.15k = −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 reasons1) 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! Reply Link to post Share on other sites More sharing options...
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