# How do you do these questions?? (1 Viewer)

#### mmmm.

##### Active Member
Could someone help/explain q 11b, 13 (is there a proper way, i was only taught trial and error, and don't rlly understand the working from document) and question 28c

Although there are answers on the document, i don't really understand how they did it.

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#### cossine

##### New Member
Using the result of 11a) you can find sin A

cos A = 23/32.

This means 23 is the adjacent side and hypotenuse is 32 units long.

Using Pythagoras theorem you can then find the value of the opposite side.

Alternatively, you could use sin^theta + cos^2 theta = 1.

I am assuming you have done 13a)

For 13b) just consider each case.

you can write y = |ax + b| as piecewise function

y = { ax + b, if x>= 2.5
{ - ax - b, if x < 2.5

That means -ax-b >= 3 for x< 2.5 or ax+b >= 3 for x>= 2.5.
Then just solve.

For cumulative distributive function, the answer is just

cdf(8) - cdf(5)

or integral pdf with lower bound 5 and upper bound 8.

You just need to know what cdf is and how to use it to perform calculations

In this case, it looks like they use complementary probability to make the calculations easier.

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#### CM_Tutor

##### Well-Known Member
Q11b... We have shown in part (a) that

$\bg_white \cos{A}=\frac{23}{32}$

\bg_white \begin{align*} \text{Rearranging the Pythagorean identity } 1&=\sin^2{A}+\cos^2{A} \\ \text{ allows us to show that } \sin^2{A}&=1-\cos^2{A} \\ &=1-\left(\frac{23}{32}\right)^2 \\ &=\frac{32^2-23^2}{32^2} \\ &=\frac{(32-23)(32+23)}{32^2} \\ &=\frac{9\times 55}{32^2} \\ \text{And since A is an acute angle, } \sin{A}&=\frac{3\sqrt{55}}{32} \text{ as \sin{A}>0} \end{align*}

$\bg_white \text{So, Area \Delta ABC}=\frac{1}{2}bc\sin{A}=\frac{1}{2}\times10\times8\times\frac{3\sqrt{55}}{32}= \frac{120\sqrt{55}}{32}= \frac{15\sqrt{55}}{4} \text{ sq cm}$

#### CM_Tutor

##### Well-Known Member
Q13... You have the function

$\bg_white f(x) = |ax + b|$

(a) From the given graph, you know at least two points that satisfy the function, $\bg_white (0,\ 5)$ and $\bg_white (2.5,\ 0)$. Using the first point:

\bg_white \begin{align*} f(0) &= 5 \\ |a \times 0 + b| &= 5 \\ |b| &= 5 \\ b &= \pm 5 \end{align*}

and using the second point,

\bg_white \begin{align*} f(2.5) &= 0 \\ |a \times 2.5 + b| &= 0 \\ |2.5a + b| &= 0 \\ 2.5a + b &= 0 \\ 5a + 2b &= 0 \\ 5a &= -2b \\ \text{Now, since b = \pm5, } 5a &= -2 \times \pm 5 = \mp 10 \\ a &= \mp 2 \end{align*}

So, we have two possible solutions:

Possibility 1: $\bg_white a = 2, b = -5$ which leads to $\bg_white f(x) = |2x - 5|$

Possibility 2: $\bg_white a = -2, b = 5$ which leads to $\bg_white f(x) = |-2x + 5|$

But these are the same, as $\bg_white |-2x + 5| = |-1(2x - 5)| = |-1| \times |2x - 5| = 1 \times |2x - 5| = |2x - 5|$

So, both solutions are valid. The solutions on the exam are flawed, however, in that they find $\bg_white a = 2$ by starting with $\bg_white b = -5$ without first establishing the value of $\bg_white b$.

Part (b) requires you to solve $\bg_white f(x) \ge 3$. The solution will be the same whichever of the previous solutions you choose as they both result in the same $\bg_white f(x)$ and so have the same equation to solve, $\bg_white |2x - 5| \ge 3$. This can be solved by cases:

Case 1: $\bg_white x \ge 2.5$, in which case $\bg_white |2x - 5| = 2x - 5$ and so

\bg_white \begin{align*} 2x - 5 &\ge 3 \\ 2x &\ge 8 \\ x &\ge 4 \end{align*}

and this solution is valid throughout the restricted domain $\bg_white x \ge 2.5$.

Case 2: $\bg_white x < 2.5$, in which case $\bg_white |2x - 5| = -2x + 5$ as $\bg_white 2x - 5 < 0$ and so

\bg_white \begin{align*} -2x + 5 &\ge 3 \\ -2x &\ge -2 \\ x &\le 1 \end{align*}

and this solution is valid throughout the restricted domain $\bg_white x < 2.5$.

So, the complete solution of $\bg_white f(x) \ge 3$ is $\bg_white x \le 1$ or $\bg_white x \ge 4$. This can be expressed in interval notation as $\bg_white x \in (-\infty,\ 1]\cup[4,\ \infty)$.

Part (b) can also be solved graphically by adding the line $\bg_white y = 3$ to the given diagram, noting that there are intersections at $\bg_white (1,\ 3)$ and $\bg_white (4,\ 3)$, and then recognising that $\bg_white f(x) \ge 3$ when the given graph lies above the line that you have added. The solution $\bg_white x \in (-\infty,\ 1]\cup[4,\ \infty)$ immediately follows.