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Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)
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davidgoes4wce
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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
A) Differentiate it. Sub in P to find a f(x) equation. Sub in P's x value and find the y value (gradient). Using point-gradient formula, figure out the equation, and then the x and y intercepts
B) Use point line distance formula to find the distance, which should be 13 (same as radius) - circle geo - the perpendicular of a tangent at the point of contact is the radius
C) Knowing the x & y intercepts, use the area formula for triangles (Area = x * y * 1/2)
D) Use distance formula to find the hypotenuse. Then it would be hypotenuse + x + y value = perimeter.
A) Differentiate it. Sub in P to find a f(x) equation. Sub in P's x value and find the y value (gradient). Using point-gradient formula, figure out the equation, and then the x and y intercepts
B) Use point line distance formula to find the distance, which should be 13 (same as radius) - circle geo - the perpendicular of a tangent at the point of contact is the radius
C) Knowing the x & y intercepts, use the area formula for triangles (Area = x * y * 1/2)
D) Use distance formula to find the hypotenuse. Then it would be hypotenuse + x + y value = perimeter.
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davidgoes4wce
Well-Known Member
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
^{\frac{1}{2}} )
^\frac{-1}{2} (-2x) )
^\frac{1}{2}} )
^\frac{1}{2}} =\frac{-12}{5} )
 )
 )
.$ Let x=0, y-intercept $ (0,\frac{169}{5}). $ The standard form of the equation is $ 12x+5y-169=0 )
 Perpendicular Distance $ =\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}} )
 which will be substituted into the formula Perpendicular Distance formula, given we also know the Standard Form of the equation. $ )
-169|}{\sqrt{12^2+5^2}}= \frac{169}{\sqrt{169}}=13. $ Hence, 13 is the radius of this circle. $)
 Given we know the x-intercept is $ \frac{169}{12} , $ the y-intercept is $ \frac{169}{5} $ we can use the formula of the Triangle to solve for the Area of a Triangle, which in this case is $ A=\frac{1}{2} b h )
 The perimeter of the triangle can be found by adding the distance from the Origin to the x-intercept $ = \frac{169}{12}, $ the distance from the Origin to the y-intercept $ = \frac{169}{5} $ and the diagonal distance by using Pythagoras' thereoem $ \sqrt{\frac{169}{5})^2+(\frac{169}{12})^2}=\frac{2197}{60}. $ By adding up all 3 values we are able to obtain our perimeter=$ \frac{169}{12} +\frac{169}{5}+\frac{2197}{60}= 84.5 or \frac{169}{2})
Q13 (a)
davidgoes4wce
Well-Known Member
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
I'll try my best, I do include comments in my working out as well and formulas used to get to the answer.
davidgoes4wce
Well-Known Member
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
Below is a screenshot of an example I did with the parabola on Desmos.
What you can see is if the gradients are opposite in magnitude, in this case, gradients of 5 and -5, the tangent curves intersect at an external point, (2,-10.25) . You can also see that the normals intersect at (2,2.75). Forget about the coordinates in this case, what you should be doing is focusing on the shape of that quadrilateral. What shape does it look like? The shape is in fact a kite.
 is of the form $ A=\frac{pq}{2} )
Given that you know the x-coordinates from part (a) , the tricky part here is to use algebra in order to find the y-coordinates. You then want to work out the distances of the vertical height.
With this question, its best to visualise it on a graphing facility. It's a shame that the HSC hasn't taken the same route as VCE, WACE and IB Maths in allowing students to use such a tool as I believe the learning is enhanced immensely.
Below is a screenshot of an example I did with the parabola on Desmos.
What you can see is if the gradients are opposite in magnitude, in this case, gradients of 5 and -5, the tangent curves intersect at an external point, (2,-10.25) . You can also see that the normals intersect at (2,2.75). Forget about the coordinates in this case, what you should be doing is focusing on the shape of that quadrilateral. What shape does it look like? The shape is in fact a kite.
Given that you know the x-coordinates from part (a) , the tricky part here is to use algebra in order to find the y-coordinates. You then want to work out the distances of the vertical height.
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davidgoes4wce
Well-Known Member
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
 that $ x=\frac{h+m}{2} $ for the x-coordinate at P. Substituting $ x=\frac{h+m}{2} $ into the original equation $ y=(x-h)^2+k, $ gives a y-coordinate of $ \frac{m^2}{4}+k . $ Hence, the coordinate $ P(\frac{h+m}{2}, \frac{m^2}{4}+k) )
 )
 $ we can find the equations of the Normals and the Tangents. $ )
=\frac{-1}{m}[x-(h+\frac{m}{2})] \textcircled{1})
=\frac{1}{m}[x-(h-\frac{m}{2})] \textcircled{2} )
+(h+\frac{m}{2})] )
+(h-\frac{m}{2})] )
+(h+\frac{m}{2})=m[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})] )
 )
=m[x-(h+\frac{m}{2})] \textcircled{3} )
=-m[x-(h-\frac{m}{2})] \textcircled{4} )
+(h+\frac{m}{2})]=x )
+(h-\frac{m}{2})]=x )
+(h+\frac{m}{2})=\frac{-1}{m}[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})] )
-(k-\frac{m^2}{4})=\frac{1}{2} \times (m^2+1) )
 , $ one can work out the horizontal distance between the 2 coordinates simply by subtracting the 2 x-coordinate values. Given at $ P, x={h+\frac{m}{2}} $ and at Q $, x=h-\frac{m}{2} )
-(h-\frac{m}{2})=m . )
 $ we can substitute them into the formula $, A=\frac{pq}{2} )
] m \frac{1}{2} = \frac{1}{4}m (m^2+1) )
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davidgoes4wce
Well-Known Member
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
I also want to say that every Development or Extension questions in the exercises, that you can knock off in Cambridge, you feel a greater appreciation for the mathematics language. Think any student in Year 11 or 12 should practice as many questions from here as they are exam standard pretty much. (especially the Development questions). It's also great in the fact that key concepts are highlighted in blue. (if you know those main formuale, equations and theory and go all over it again and again. You won't go wrong). Last year I said Fitzpatrick was a better book, I was wrong. Its Cambridge>Fitzpatrick>Lee>Grove but if your a keen maths student you would look at all resources possible.
As well as having a formulae book (which I purchased) for this subject, I make the effort in making my own theory notes. Just notes and formuale that I can go over again and again when I'm trying to refresh my memory on a certain question. I think all up I have around 100-120 pages worth of formulae, notes that I rigorously go all over and over, till its stored. (From time to time you do forget something )
I also want to say that every Development or Extension questions in the exercises, that you can knock off in Cambridge, you feel a greater appreciation for the mathematics language. Think any student in Year 11 or 12 should practice as many questions from here as they are exam standard pretty much. (especially the Development questions). It's also great in the fact that key concepts are highlighted in blue. (if you know those main formuale, equations and theory and go all over it again and again. You won't go wrong). Last year I said Fitzpatrick was a better book, I was wrong. Its Cambridge>Fitzpatrick>Lee>Grove but if your a keen maths student you would look at all resources possible.
As well as having a formulae book (which I purchased) for this subject, I make the effort in making my own theory notes. Just notes and formuale that I can go over again and again when I'm trying to refresh my memory on a certain question. I think all up I have around 100-120 pages worth of formulae, notes that I rigorously go all over and over, till its stored. (From time to time you do forget something )
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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
 using the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \left(a^{mx+b}\right) = a^{mx+b}\cdot m \ln a$.$)
 \ln a}$. Let $u=(mx+b)\ln a = m\ln a\cdot x + b\ln a$. Then $\frac{\mathrm{d}u}{\mathrm{d}x} = m\ln a$. Now by the chain rule, noting $y=\mathrm{e}^{u} = a^{mx+b}$, we have$)
 \\ &= a^{mx+b}\cdot \left(m\ln a\right), \end{align*}$$)
 = \ln a \cdot a^u$. This, along with the chain rule, allows us to do all the questions above.$)
I'm having a mind blank
Could someone remind me how to do these please? I know that it says what to do in the question but I'm getting stuck