if you're just learning the basics, then it's better to start with a textbook.
Then as you have a better grasp at the topics, you should attempt the past exam questions since they're a lot more difficult in general.
Well to be fair, the exam pressure would definitely cause silly mistakes, but there weren't many "different" questions that people haven't seen before, and not to mention it doesn't have permutations and combinations.
Haha I remember my 4U teacher mentioning that, he hated english with a passion. (and he failed so unless you don't care, don't do that)
But yeah to OP, I would say try to learn from your peers :P
For the 2nd question,for c = -2, P(x) is always increasing for x>2 and P(2) is negative while P(3) is positive, so the root must exist between that interval.
It's going to keep dying until Niantic releases the pokemon battle and trading features, which was what the players really wanted to have since the start of the game.
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
For 12a
The equation of the parabola is in the form (x-3)^2 = 4a(y+1)
The tangent has the equation y = -4x+7
Sub that into the parabola to get a quadratic in terms of x and a, then use the discriminate and let that equal...
To answer your second question,(assuming this for the topic in parametrics) no you are almost always required to derive the equation of the Chord unless they give you the equation in the beginning, so there's no point memorising the formulas.
This applies for most other topics in 3U math, where...
What I did was starting with an equation y = a(x+11)(x-11) and you get that f(-7) = f(7) = 3, use that to solve for a and sub in x = 0 for the maximum height.
I can't comprehend the limit from your typing, but the integral goes like this.
\int \frac{dt}{\sqrt{1+6t-3t^2}} = \\ \\ \int \frac{1}{\sqrt{3}} \frac{dt}{\sqrt{\frac{4}{3} - (t-1)^2}} = \frac{1}{\sqrt{3}} sin^{-1}(\frac{\sqrt{3}}{2}(t-1))
if you expand the quadratic, you get
ax^2+10ax+16a
so for part a) the answer is the equation with a = 3, as the coefficient of x^2 is the term in front of it.
use this expression for the remaining parts.