tmua 2022 q15

[stark beacon]

can someone help with this one

[foggy crow]

p(x) and q(x) are both symmetrical in the y axis
notice that the width of the rectangle is equal to |2x| and the height is equal to |q(x)-p(x)|
so height = |3x^2 - 10|

hence area of the rectangle can be shown as: |6x^3 - 20x|
to get the x value when this is the largest we can take the derivative and get the relative maximum/minimum (doesnt matter since we take the absolute values of x), you would probably have to check the y values of where p(x) and q(x) cross to make sure that the relative maximum is the highest value of the area within the domain in the exam

d/dx = 18x^2 - 20 = 0
getting x as +-sqrt(10)/3

subbing this into the area of rectangle, we get 40sqrt(10)/9, so the answer should be H

i just used desmos here to get a clearer diagram, but it wouldnt be difficult to draw it in the tmua anyway

[torpid pollen]

Could have drawn the lines with desmos aswell smh

[foggy crow]

https://tenor.com/view/shazz-dies-of-death-jazz-gif-24027506

i am too inexperienced with desmos unfortunately

[torpid pollen]

|x|=1 {f(x)≤y≤g(x)}

Change 1 to whatever you want

[radiant harness]

Nice explanation

[torpid pollen]

Oh and for the y values you'd do (y-1)(y-5)=0 {|x|≤1}

[stark beacon]

wow this explanation is so clear tysm