How did everybody find the exam today? I've walked out knowing I left out 6 marks worth but aside from that hopefully still an A. Also can anyone remember what they got for their objective function equation in the linear programming question?
I walked out literally almost cried. The kid in my school who got 100% in gcse and always got like 95% in add maths mocks walked out and said he left so many blank and he hated it and it was horrible and he didn't understand it at all. That was a horrific paper. The past papers seemed so much more easier than this
i found it really bad just like the rest of my class but there was this one really clever girl who found it easy and just got stuck on one question (M****, if you're reading this)
what about the question about the remainder theorem, that was pretty tough
The function f(x)=x3−4x2+ax+b is such that:
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x = 3 is a root of the equation f(x) = 0
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when f(x) is divided by (x – 1) there is a remainder of 4.
(i) Find the value of a and the value of b. (ii) Solve the equation f(x) = 0.
Part (i): So, we know that if x = 3, f(x) = 0, so let's plug it in: this leads to 27 – 36 + 3a + b = 0 3a + b = 9 We also know (due to the factor theorem) that f(1) must produce 4. 1 – 4 + a + b = 4 a + b = 7 We now have simultaneous equations: (equation 1): 3a + b = 9 (equation 2): a + b = 7 (1) – (2): 2a = 2 a = 1
1 + b = 7 b = 6
Part (ii): Now, we can fill in a and b: f(x)=x3−4x2+x+6 We know that when f(x) = 0, when x = 3. This means that (x – 3) is a factor (factor theorem) Now we can find the other factors and therefore solve for all roots. x3−4x2+x+6=(x−3)(x2–x–2) (x−3)(x2–x–2)=(x−3)(x−2)(x+1) So,x = –1, x = 2,x = 3
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.
For the objective function, I put V(x, y) = 3.5x + 3y With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.
For the objective function, I put V(x, y) = 3.5x + 3y With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32
As did most of us, I believe
The graph was a bit funny – that little kink in it just pushed the objective function to that 'corner' instead of the bottom right. It was a nice spot.
when f(x) is divided by (x – 1) there is a remainder of 4.
(i) Find the value of a and the value of b. (ii) Solve the equation f(x) = 0.
Part (i): So, we know that if x = 3, f(x) = 0, so let's plug it in: this leads to 27 – 36 + 3a + b = 0 3a + b = 9 We also know (due to the factor theorem) that f(1) must produce 4. 1 – 4 + a + b = 4 a + b = 7 We now have simultaneous equations: (equation 1): 3a + b = 9 (equation 2): a + b = 7 (1) – (2): 2a = 2 a = 1
1 + b = 7 b = 6
Part (ii): Now, we can fill in a and b: f(x)=x3−4x2+x+6 We know that when f(x) = 0, when x = 3. This means that (x – 3) is a factor (factor theorem) Now we can find the other factors and therefore solve for all roots. x3−4x2+x+6=(x−3)(x2–x–2) (x−3)(x2–x–2)=(x−3)(x−2)(x+1) So,x = –1, x = 2,x = 3
The graph was a bit funny – that little kink in it just pushed the objective function to that 'corner' instead of the bottom right. It was a nice spot.
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.
For the objective function, I put V(x, y) = 3.5x + 3y With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32