Thanks for this exam! So tough

For 5b), how do you know p>0?

Can you explain the whole of Q8? I had a look at the ans but I don't understand what's going on

Same goes with Q10a)

thanks for the feedback!

well basically for 5b) there's not enough information to actually solve for p since it just ends up cancelling but the reason i state it's a positive real constant is because it's probability so pr(a) has to be 0 or greater, but if p=0 things start to break down because for pr(b|a) pr(a) would be 0 and you can't divide by 0 so that's why.

for 8a) you basically work with the integral so integral of sin is -cos and sub in the bounds. then pull the -1 out the front cuz it's just a constant multiple and then multiplying by 1/h is the same as dividing by h so it ends up being first principles in disguise. so remember f'(x) = limit h->0 (f(x+h) - f(x)) / h so like f'(pi/4) = lim h->0 (f(pi/4+h) - f(pi/4)) / h where f(x) is cos(x) so you derive cos(x) and sub in pi/4 to get -(-sin(pi/4)) which is sqrt(2)/2

for 8b) you derive cos(x) and sub in x=pi/4 to get a gradient of -sqrt(2)/2. so you just sub in pi/4 into cos(x) so it becomes cos(pi/4) = sqrt(2)/2 then use the tangent line formula

for 8c) basically since it's a left endpoint estimate, you just take y(0), y(1/2) and y(1) since that will cover your interval with rectangles of 1/2 units. so then you have to add up the rectangles (the width is always 1/2 at the bottom and the length is the value of the function) so 1/2 * y(0) + 1/2 * y(1/2) + 1/2 * y(1). and then use your exact values to calculate it.

for 8d) good thing you mentioned it it's actually a mistake haha... it should just be the integral from 0 to 3/2 of y=sin(pi/3 x) and yeah because the function is increasing, the left endpoint estimate will always under estimate the area as it leaves out parts.

hope i made some sense and was able to help