The Student Room Group

P1 Question help

heya

i was looking through the exam papers and there were these 2 questions which i got stuck on. Even if i did one part of the question the next part i wouldn't understand. Please help
Thanx

p1 2001 January Question 4 and 6

Q4) Figure 1 shows the curve with equation y = 5 + 2x − x2 and the line with equation y = 2. The curve and the line intersect at the points A and B.

(a) Find the x-coordinates of A and B. (3 marks)

The shaded region R is bounded by the curve and the line.

(b) Find the area of R.

Q6) The points A (3, 0) and B (0, 4) are two vertices of the rectangle ABCD, as shown in Fig. 2.

(a) Write down the gradient of AB and hence the gradient of BC. (3 marks)

The point C has coordinates (8, k), where k is a positive constant.

(b) Find the length of BC in terms of k. (2 marks)

Given that the length of BC is 10 and using your answer to part (b),

(c) find the value of k, (4 marks)

(d) find the coordinates of D. (2 marks)
Reply 1
lol - what in particular are you struggling with?
Reply 2
For Question 4 b, you need to cover the integration in C2 in order to answer it.
Reply 3
do they still teach p1? i thought it was scrapped
Intersection at:

y = 5 + 2x − and y = 2.

=> 2=5+2x-x²
=> x²-2x-3=0
=> (x-3)(x+1)=0

x= 3 or -1

so A=(3,2) B=(-1,2)
Reply 5
Milli
Q6) The points A (3, 0) and B (0, 4) are two vertices of the rectangle ABCD, as shown in Fig. 2.
(a) Write down the gradient of AB and hence the gradient of BC. (3 marks)
The point C has coordinates (8, k), where k is a positive constant.
(b) Find the length of BC in terms of k. (2 marks)
Given that the length of BC is 10 and using your answer to part (b),
(c) find the value of k, (4 marks)
(d) find the coordinates of D. (2 marks)


6) (a) If A(3, 0) and B(0, 4), then m(A,B) = (4-0)/(0-3) = -4/3
....(b) length of BC = √((AB)² + (AC)²) ... work it out!
.... work this out!
(b) Find the area of R.

You need to integrate between x=-1 and x=3 then remove the rectangle below y=2.

∫(5+2x-x²)dx = [5x+x²-(1/3)x³] between x=-1,3 (3 is top limit, -1 is bottom)

=15+11/3=56/3

then the area of the rectangle between y=2 and x axis, and between x=-1,3 is:

4x2=8

So R=(56-24)/3=32/3