Multiple choice help (1 Viewer)

Porogamiii · Sep 29, 2021

How to do?

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Porogamiii · Sep 29, 2021

Sorry and these two.

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CM_Tutor · Sep 29, 2021

Question 3
Strategic notes

Since $\sec\theta$ is the reciprocal of $\cos\theta$ and $-1 \leq \cos\theta \leq 1$ , $\sec\theta$ must have a magnitude of 1 or more - that is, $\sec\theta \ge 1$ or $\sec\theta \le -1$ , so answers A and D are impossible.
All the answers are negative, so we don't have to worry about signs. The more usual form of this question would give the options as -B, B, -C, and C, so you do have to think about the quadrants.
In any case, multiple choice problems like this can be solved with a calculator, which is why I wouldn't ask them (instead ask in question 11 so you have to show you can work out the answer, as shown in the methods below). Anyway, the basic idea is to use the given value of $\tan\theta$ to find $\theta$ and then $\cos\theta$ and then match the decimal you get to one of the fractions provided. NOTE, THOUGH, THAT THERE IS A TRICK... the calculator gives the angle as -0.588002... radians, which is in the fourth quadrant but you seek the angle in the second quadrant, so you need to adjust the answer to $\theta$ by adding $\pi$ ( $\theta = \pi + -0.588002... = 2.55359...\ \text{radians}$ ) and so to get $\cos\theta = -0.83205...$ . If you don't do this, you will get $\cos\theta = 0.83205...$ (with the wrong sign) and you could get the answer wrong if the options include positives and negatives.

Method one - using the related angle
The related angle, $\alpha$ , is the acute angle (in quadrant one) with

$\tan\alpha = \cfrac{2}{3}$

Using Pythagoras' Theorem, $\alpha$ is in a right angled triangle with hypotenuse $\sqrt{13}$ (because we know opposite = 2 and adjacent = 3).

$\text{So,} \quad \sec\alpha = \cfrac{1}{\cos\alpha} = \cfrac{\text{hypotenuse}}{\text{adjacent}} = \cfrac{\sqrt{13}}{3}$

And, since $\theta$ lies in quadrant 2, and so $\cos\theta < 0$ , we know that

$\sec\theta = -\sec\alpha = \cfrac{\sqrt{13}}{3}$

and so the answer is B.

Method 2 - Pythagorean identities

$\begin{align*} \text{For an angle $\theta$, we know that}\ \tan^2\theta + 1 &= \sec^2\theta \\ \left(-\cfrac{2}{3}\right)^2 + 1 &= \sec^2\theta \\ \sec^2\theta &= \cfrac{4}{9} + 1 \\ &= \cfrac{4 + 9}{9} \\ &= \cfrac{13}{9} \\ \text{Thus,} \quad \sec\theta &= \pm\cfrac{\sqrt{13}}{3} \quad \quad \textbf{Note:}\ \text{from which we can reject options A, C, and D} \\ \text{But, we know that $\cfrac{\pi}{2} < \theta < \pi$ and so}\ \sec\theta &< 0\ \text{as only $\sin\theta$ and cosec $\theta$ are positive in quadrant 2} \\ \text{Thus,} \quad \sec\theta &= -\cfrac{\sqrt{13}}{3} \quad \quad \text{and the answer must be}\ \textbf{B} \end{align*}$

Method 3 - Quadrants diagram

Similar to method 1, but drawing $\theta$ in the second quadrant, we have the "adjacent" as -3 and the opposite as 2. Pythagoras' Theorem again gives the hypotenuse (always positive) as $\sqrt{13}$ .

$\text{Thus,} \quad \sec\theta = \cfrac{1}{\cos\theta} = \cfrac{\text{hypotenuse}}{\text{adjacent}} = \cfrac{\sqrt{13}}{-3} = -\cfrac{\sqrt{13}}{3}\ \quad \text{and the answer must be}\ \textbf{B}$

CM_Tutor · Sep 29, 2021

Question 8 - the answer is B

$\begin{align*} \int_1^4 g(x)\ dx &= \int_1^4 1 - f(x)\ dx \qquad \qquad \text{as we have the definition that $g(x) = 1 - f(x)$}\\ &= \int_1^4 1\ dx - \int_1^4 f(x)\ dx \qquad \text{as integrals of sums / differences can be separated into individual terms} \\ &= \bigg[x\bigg]_1^4 - 7 \qquad \qquad \qquad \text{on integrating 1 to $x$ and using the given fact that $\int_1^4 f(x)\ dx = 7$} \\ &= \big[4 - 1\big] - 7 \\ &= 3 - 7 \\ &= -4 \end{align*}$

Porogamiii · Sep 29, 2021

Oh wow for question 3 I never thought about it as a triangle and using Pythagoras - that's really cool. And I never knew you could do that with the trig identities, but algebraically it makes sense now. Thanks a lot!

CM_Tutor · Sep 29, 2021

Question 10
Comments

Though we are not asked for the value of $k$ , it could be easily found as the total cumulative probability must be 1, and so $\int_{-k}^k \cos{2x}\ dx =1$ and the PDF can never be negative, which means that $-\cfrac{\pi}{4} \le k \le \cfrac{\pi}{4}$ . Since we also know that $k>0$ , there is only one possible value for $k$ , which is $k = \cfrac{\pi}{4}$ .
$\cos{2x}$ is an even function, and so the symmetry properties of this PDF will be the same as the standard normal distribution.
There is something wrong with this question.

Option A is true because of the even symmetry.

Option B is true for any PDF.

Option C can only be true if both of the expressions given are $\cfrac{1}{3}$ , because the given $\text{LHS}\ = P(X < -0.2)$ and adding this gives the whole distribution with a total value of 1, and so it occurs at only one value... the question becomes, is that value 0.2?

$\begin{align*} \text{Let $a > 0$ be the value where}\ P(X > a) &=\cfrac{1}{3} \\ \int_a^{\infty} f(x)\ dx &= \cfrac{1}{3} \\ \int_a^{\frac{\pi}{4}} \cos{2x}\ dx + \int_{\frac{\pi}{4}}^{\infty} 0\ dx &=\cfrac{1}{3} \\ \left[\cfrac{1}{2}\sin{2x}\right]_a^{\frac{\pi}{4}} + 0 &= \cfrac{1}{3} \\ \cfrac{1}{2}\left[\sin{2\left(\cfrac{\pi}{4}\right)} - \sin{2a}\right] &= \cfrac{1}{3} \\ \sin{\left(\cfrac{\pi}{2}\right)} - \sin{2a} &= \cfrac{2}{3} \\ 1 - \sin{2a} &= \cfrac{2}{3} \\ \sin{2a} &= \cfrac{1}{3} \\ a &= 0.169918... \end{align*}$

So, option C is false as it is actually $P(X > 0.1699) = P(-0.1699 < X < 0.1699)$

Option D can only be true if both of the expressions given are $\cfrac{1}{4}$ , because the LHS + RHS covers half the distribution, and so they must add to $\cfrac{1}{2}$ , which also occurs at only one value... and so the question again becomes, is that value 0.2?

$\begin{align*} \text{Let $a > 0$ be the value where}\ P(X > a) &=\cfrac{1}{4} \\ \int_a^{\infty} f(x)\ dx &= \cfrac{1}{4} \\ \int_a^{\frac{\pi}{4}} \cos{2x}\ dx + \int_{\frac{\pi}{4}}^{\infty} 0\ dx &=\cfrac{1}{4} \\ \left[\cfrac{1}{2}\sin{2x}\right]_a^{\frac{\pi}{4}} + 0 &= \cfrac{1}{4} \\ \cfrac{1}{2}\left[\sin{2\left(\cfrac{\pi}{4}\right)} - \sin{2a}\right] &= \cfrac{1}{4} \\ \sin{\left(\cfrac{\pi}{2}\right)} - \sin{2a} &= \cfrac{1}{2} \\ 1 - \sin{2a} &= \cfrac{1}{2} \\ \sin{2a} &= \cfrac{1}{2} \\ 2a &= \cfrac{\pi}{6} \\ a &= \cfrac{\pi}{12} \approx 0.261799... \end{align*}$

So, option D is false as it is actually $P(X > 0.2618) = P(0 < X < 0.2618)$

I have two problems with this question... the first is that there are two correct answers, unless I have made a mistake in looking at A and B. The second is that actually finding the boundaries for where expressions like those in C and D become true is a lot to ask in time and effort if there was not an obvious correct answer in A or B, starting with finding the only possible value for $k$ .

Porogamiii · Sep 29, 2021

Yeah this was from our school trial and the whole exam was kind of rushed anyways, so I guess for question 10 they just assumed we'd start with A (which was the answer on the marks sheet). Thanks for your really in depth explanations though -- you're a natural teacher!

CM_Tutor · Sep 29, 2021

Maybe you could upload the paper and the answers to the resources here at BoS, so others can use them?

Porogamiii · Sep 29, 2021

2021 2U School Maths Trial

math

www.boredofstudies.org

2021 2U School Maths Trial (sols)

math

www.boredofstudies.org

Yup -- linked above if anyone wants them

Multiple choice help (1 Viewer)

Porogamiii

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2021 2U School Maths Trial

2021 2U School Maths Trial (sols)

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