Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

rand_althor · Jul 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$\begin{align*}\textup{Tangent at }x=a:\\y-y_1&=m(x-x_1) \\y-0&=a(a-b)(x-a) \\y&=a(a-b)(x-a) \\\textup{Tangent at }x=b: \\y-y_1&=m(x-x_1) \\y-0&=-a(a-b)(x-b) \\y&=-a(a-b)(x-b) \\\textup{Equating tangents to find M:} \\a(a-b)(x-a)&=-a(a-b)(x-b) \\x-a&=-x+b \\2x&=a+b\\x&=\frac{a+b}{2}\end{align*}$

$\textup{Therefore M has an x-coordinate of: } \frac{a+b}{2} \textup{, which is the average of the x-intercepts.}$

appleibeats · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For part c) of the same question:

Find the point V where the tangent is horizontal. Show that M is vertically above or below V and twice as far from the x-axis. Sketch the situation.

I have found Point V: (a + b /2 , -1/4 a(a-b)^2)

I know that M and V have the same x coordinates so is either vertically above or below.

Not sure how to show it is twice as far from the x - axis.

I know that point M has x coordinate a + b / 2

I think I am meant to find the y coordinate for point M but subbing x coordinate into the function will give me point V, won't it?

rand_althor · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$V is the point$ (\frac{a+b}{2},\frac{-a(a-b)^2}{4}) \\$M is the intersection of the two tangents. Hence, it is not a point on the parabola, but a point on one of the tangents.$ \\$ Subbing in $ x=\frac{a+b}{2} $ into $y=-a(a-b)(x-b): \\$

$\begin{align*}\\y&=-a(a-b)(\frac{a+b}{2}-b)\\&=-a(a-b)(\frac{a+b-2b}{2})\\&=-\frac{a}{2}(a-b)(a-b)\\&=-\frac{a(a-b)^2}{2}\end{align*}\\$Therefore M is the point $(\frac{a+b}{2},-\frac{a(a-b)^2}{2})$. As M and X share an x-coordinate, and have different y-coordinates, one is vertically above or below the other. To show that M is twice as far from the x-axis, find the ratio of their y-coordinates:$$

$\begin{align*}\frac{y_M}{y_V}&=\frac{-\frac{a(a-b)^2}{2}}{\frac{-a(a-b)^2}{4}}\\&=\frac{1}{2}\times4\\&=2\end{align*}\\$Hence, M is twice as far from the x-axis as V.$$

appleibeats · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Q4

Find i) the local maxima or minima and ii) the global maximum and minim of the funciton y = x^4 -8x^2 + 11 on each of the domains:

a) 1 less than or equal to x less than or equal to 3.

I have found the global max = 11 and global min = 4

But why is the boundary point which = local maxima ??

Are boundary points excluded??

rand_althor · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Global maximum: No global maximum as the curve approaches ∞ as x approaches ∞ and -∞.
Global minimum: f(x)=-5 when x=2 or -2.
Local maximum: f(x)=20 when x=3.
Local minimum: f(x)=-5 when x=2.

appleibeats said:
But why is the boundary point which = local maxima ??
Are boundary points excluded??

Not sure what you mean by this.

InteGrand · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Q4

Find i) the local maxima or minima and ii) the global maximum and minim of the funciton y = x^4 -8x^2 + 11 on each of the domains:

a) 1 less than or equal to x less than or equal to 3.

I have found the global max = 11 and global min = 4

But why is the boundary point which = local maxima ??

Are boundary points excluded??

rand_althor said:
Global maximum: No global maximum as the curve approaches ∞ as x approaches ∞ and -∞.
Global minimum: f(x)=-5 when x=2 or -2.
Local maximum: f(x)=20 when x=3.
Local minimum: f(x)=-5 when x=2.

Not sure what you mean by this.

They asked for the local and global extrema on the domain 1 ≤ x ≤ 3. So we only focus on the curve in this domain, in other words, sketch only the emboldened part of rand_althor's graph.

So the answers are:

Global maximum: 20 (obtained at the endpoint x = 3).
Global minimum: -5 (obtained at x = 2 )
Local maximum: none in the domain
Local minimum: -5 (obtained at x = 2).

rand_althor · Jul 15, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
They asked for the local and global extrema on the domain 1 ≤ x ≤ 3. So we only focus on the curve in this domain, in other words, sketch only the emboldened part of rand_althor's graph.

So the answers are:

Global maximum: 20 (obtained at the endpoint x = 3).
Global minimum: -5 (obtained at x = 2 )
Local maximum: none in the domain
Local minimum: -5 (obtained at x = 2).

Ah okay, apologies for the wrong answers appleibeats.

appleibeats · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Not sure how to find the horizontal asymptote of

y = x / (x^2 + 1)^1/2

VBN2470 · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

We have a function f defined by $f:=f(x)=\dfrac{x}{\sqrt{x^2+1}}=\dfrac{1}{\sqrt{1+\frac{1}{x^2}}}\rightarrow{\pm{1}}$ as $x \rightarrow \pm\infty$ . Check this graphically.

(Note that f(x) > 0 when x > 0 and f(x) < 0 when x < 0 and that f is an odd function)

appleibeats · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Are you dividing the numerator and denominator by x^2 ?? Does the square root in the denominator matter?? Or can you just ignore it, when finding the power of x to divide by.

rand_althor · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$\begin{align*} \frac{x}{\sqrt{x^2+1}}&=\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}} \\ &=\frac{x}{x\sqrt{1+\frac{1}{x^2}}} \\&=\frac{1}{\sqrt{1+\frac{1}{x^2}}} \end{align*}$

InteGrand · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

rand_althor said:
$\begin{align*} \frac{x}{\sqrt{x^2+1}}&=\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}} \\ &=\frac{x}{x\sqrt{1+\frac{1}{x^2}}} \\&=\frac{1}{\sqrt{1+\frac{1}{x^2}}} \end{align*}$

For x > 0.

rand_althor · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Is this correct for x<0?

$\begin{align*}\frac{x}{\sqrt{x^2+1}}&=\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}} \\ &=\frac{x}{|x|\sqrt{1+\frac{1}{x^2}}} \\&=\frac{x}{-x\sqrt{1+\frac{1}{x^2}}} \\&=-\frac{1}{\sqrt{1+\frac{1}{x^2}}}\end{align*}$

InteGrand · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

rand_althor said:
Is this correct for x<0?

$\begin{align*}\frac{x}{\sqrt{x^2+1}}&=\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}} \\ &=\frac{x}{|x|\sqrt{1+\frac{1}{x^2}}} \\&=\frac{x}{-x\sqrt{1+\frac{1}{x^2}}} \\&=-\frac{1}{\sqrt{1+\frac{1}{x^2}}}\end{align*}$

Yes.

(Or we can just do the analysis for positive x and then use the fact that the function is odd to reduce the asymptote for negative x).

appleibeats · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How do you find the asymptote if there was no x in the numerator, instead just the number 1.

y = 1 / (x^2 + 1)^1/2

rand_althor · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For x>0:

$\begin{align*} \lim_{x\to\infty} \frac{1}{\sqrt{x^2+1}} &= \lim_{x\to\infty} \frac{1}{\sqrt{x^2(1+\frac{1}{x^2})}} \\ &=\lim_{x\to\infty} \frac{1}{x \sqrt{1+\frac{1}{x^2}}}\\ &= \lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{x}{x} \sqrt{1+\frac{1}{x^2}}} \\ &= \lim_{x\to\infty} \frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}} \\ &=0 \textup{ as } \lim_{x\to\infty} \frac{1}{x}=0 \end{align*}$

InteGrand · Jul 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Or just note that the denominator tends to infinity as $x\to \pm \infty$ , so the overall fraction (and hence the function) approaches 0, since $\lim _{t\to \infty}\frac{1}{t}=0$ .

calebwhitaker199 · Jul 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Hey guys, I need a hand with 22 and 24 please. If you could show working that would help a lot.
The answer for 22 is a) 8x^2 +9y-72=0
B ) (0,(247/32)), y=265/32
and 24) x^2 + 4x+8y-20=0

rand_althor · Jul 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

22)

$\begin{align*}&\textup{Using the equation of a parabola with vertex }(h,k): \\(x-h)^2&=4a(y-k) \\x^2&=4a(y-8) \\&\textup{When }x=3,\, y=0: \\3^2&=4a(0-8)\\4a&=\frac{9}{-8} \\a&=\frac{9}{-32} \\\therefore x^2&=4(\frac{-9}{32})(y-8) \\8x^2&=-9(y-8) \\ 8x^2&+9y-72=0 \end{align*}$

$$Focus has coordinates $(h,k+a)\to(0,8+(-\frac{9}{32}))\to(0,\frac{247}{32}) \\$Directrix is the equation $y=k-a\to y=8-(-\frac{9}{32})\to y=\frac{265}{32} \\$

24)

$\begin{align*}&\textup{Using the vertex form of a parbola with vertex }(h,k): \\y&=a(x-h)^2+k \\y&=a(x+2)^2+3 \\&\textup{When }x=2,\, y=1 \\1&=a(2+2)^2+3 \\-2&=16a \\a&=-\frac{1}{8} \\\therefore y&=-\frac{1}{8}(x+2)^2+3 \\-8y&=x^2+4x+4-24 \\x^2&+4x+8y-20=0\end{align*}$

DatAtarLyfe · Jul 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

24/ Using the standard equation (x-h)^2 = 4a(y-k) where (h,k) is your vertex, you get (x+2)^2 = 4a(y-3)
To solve for a, sub in the point (2,1), you get (2+2)^2 = 4a(1-3)
4^2 = -8a
16 = -8a
a = -2
Sub it back into the equation and expand:
(x+2)^2 = -8(y-3)
x^2 + 4x + 4 = -8y +24
x^2 + 4x + 8y - 20 = 0
Q.E.D

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