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# June 2005 OCR C3 question! watch

1. Hey guys having a bit of a problem on (ii) and (iii) of this question, I don't know how to go about it. I'll type out whole question

The function f is defined by f(x) = sq root (mx+7) - 4, where x is equal to or greater than -7/m and m is a positive constant.

(i) a sequence of transformations maps the curve y = sq root x to curve f(x). give details of transformations. <<<---- ok

(ii) Explain how you can tell that f is one-one function and find expression for f^-1 (x). <<<---- can explain why it is 1:1 function but don't know how to find expression

(iii) y=f(x) and y =f^-1 (x) do not meet. Explain how it can be deduced that neither curve meets the line y=x, and hence determine the set of possible values of m.

Thanks
2. (Original post by silly_billy)
Hey guys having a bit of a problem on (ii) and (iii) of this question, I don't know how to go about it. I'll type out whole question

The function f is defined by f(x) = sq root (mx+7) - 4, where x is equal to or greater than -7/m and m is a positive constant.

(i) a sequence of transformations maps the curve y = sq root x to curve f(x). give details of transformations. <<<---- ok

(ii) Explain how you can tell that f is one-one function and find expression for f^-1 (x). <<<---- can explain why it is 1:1 function but don't know how to find expression

(iii) y=f(x) and y =f^-1 (x) do not meet. Explain how it can be deduced that neither curve meets the line y=x, and hence determine the set of possible values of m.

Thanks
i) stretch by 1/m, shift 7 left, four down.

ii) you could argue since gradient is always increasing, it's one to one.

f(x) = √(mx+7) - 4
mx + 7 = (f(x)+4)²
f^-1x = ((x+4)² - 7)/m = (x² + 8x + 9)/m

iii) finding the inverse is essentially swapping the x and y values round. hence if f(x) ≠ f^-1(x), y≠x

second part.. give me a second.
3. (Original post by chewwy)
i) stretch by 1/m, shift 7 left, four down.

ii) you could argue since gradient is always increasing, it's one to one.

f(x) = √(mx+7) - 4
mx + 7 = (f(x)+4)²
f^-1x = ((x+4)² - 7)/m = (x² + 8x + 9)/m

iii) finding the inverse is essentially swapping the x and y values round. hence if f(x) ≠ f^-1(x), y≠x

second part.. give me a second.
thank you so much. can i take u into the exam please?
4. i remember this question from june. i didnt get part 3 then either
5. (Original post by Undry1)
i remember this question from june. i didnt get part 3 then either
damn and u do further maths
6. http://www.thestudentroom.co.uk/show...9&postcount=14

thats someone's explanation of how they did it in the exam.
7. i don't understand

ok, so let's find the range of values of m that do cause an intersection. then the other values of m won't. so:

f(x) = x = f^-1(x)
m = x+8 + 9/x

now what? surely this means (since x+8 + 9/x has a range everywhere but 0) that they'll intersect if m is anything but 0.

8. (Original post by chewwy)
i don't understand

ok, so let's find the range of values of m that do cause an intersection. then the other values of m won't. so:

f(x) = x = f^-1(x)
m = x+8 + 9/x

now what? surely this means (since x+8 + 9/x has a range everywhere but 0) that they'll intersect if m is anything but 0.

what i think the guy was saying was:
x = (x²+8x+9)/m
xm = x²+8x+9
0 = x²+(8-m)x+9

using the discriminant to find values of m which cause the equation to have no roots:

(8-m)²- 36 < 0
m² - 16m +28 < 0
(m-14)(m-2) < 0

2 < m < 14
9. Thnks you guys for all ur help - u explained it really well, will have to remember how to do them

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Updated: November 13, 2005
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