OCR (non-MEI) Further Pure 2: 25th June 2018

I just checked, and 1 was the answer to the Weierstrass substitution question, not that one. If I remember correctly, there was only one integral that evaluated to 1. Maybe you're mixing up the questions.

But I do remember a definite integral involving sinh^2(x) - it might have been sinh^2(3x), or we might be remembering the 3 from the polar integration.

You're right, that wasn't the one that evaluated to 1, I believe I got 1/2 + (1/6)sinh(6) for this question. It wanted it in the form A + BsinhC

Reply 21

peterw55

11

Original post by L_stMemories

I believe it was the integral between 0 and 1 for (cosh(3x))^2, which gave an answer of 1/2 + (1/12)sinh(6), I hope?

Yes! Thank you. I think I got that too (and checked with my calculator, if I remember correctly).

Good luck tomorrow if you have S3 or M4. Otherwise, enjoy the break!

Reply 22

L_stMemories

OP

10

(>'-'

> UNOFFICIAL MARKSCHEME <('-'< :wink:

Below are the topics each question tests on, let me know if something doesn't look right!

Q1 - Graphs and Functions
Q2 - Hyperbolic Functions: Prove Identity and Integration
Q3 - Hyperbolic Functions: Differentiation
Q4 - Reduction Formulae (with a cheeky series/integrals question in part(iii))
Q5 - Integration with t=tan(0.5x) Substitution. (Weierstrass for extra nerdy human beings. :3)
Q6 - Maclaurin Series
Q7 - Newton-Raphson Method
Q8 - Numerical Iterative Method
Q9 - Polar Coordinates

It is entirely possible that I have mixed up the orders of Q4-Q6. Let me know if I did!

1)(i) The asymptotes of the graph are x=1 and y=1. [2]
1)(ii) y has range y≧0.5 (greater than or equal to 0.5). The only stationary point of the graph is at (-1, 0.5). [6]
1)(iii) Graph sketching! Some features: nothing below the x-axis; everything above y=0.5 (as found in part(ii)); as x tends to -ve infinity, the graph approaches y=1 from below and as x tends to +ve infinity, the graph approaches y=1 from above; asymptote at x=1 where both curves shoot upwards at both sides of the asymptote. [2]
2)(i) Show that cosh2x = 2(coshx)^2-1 (write coshx in terms of exponentials and expand the entire expression 2(coshx)^2-1). [3]
2)(ii) Integral of (cosh3x)^2 between 0 and 1 = (1/2) + (1/12)sinh(6) (in the form A+BsinhC, as required). [3]
3) Show that [hyperbolic curve] has no stationary points. (I personally found dy/dx first, then used an identity to convert it into a quadratic in sinhx, and b^2-4ac was negative and hence there are no roots of dy/dx=0, therefore no stationary points.) [5]
4) Integral In = (x^n e^(x-1)) between 0 and 1.
4)(i) Show that In=1-nI(n-1) (the n and n-1 are subscript to represent the reduction integral). [3]
4)(ii) Find the exact value of I4 = 9-24e^-1. [4] (maybe [3])
4)(iii) Given that y=x^n e^(x-1), find dy/dx and sketch the curve. Hence deduce that 0<In<1. Personally this was my shakiest ground, but I think the dy/dx proves that y is an increasing function for n≧1 (which is given). The y coordinate of the graph at x=1 is y=1, which is the unit square. Since the curve lies between the unit square, it must have an area less than 1 but (evidently) greater than 0 because...well, it has an area... [5]
5)(i) Derivative of tan^-1(3x) is 3/(1+9x^2) [2]
5)(ii) Show [something really complicated with triple derivatives involving 18 or 36]. [4]
5)(iii) Maclaurin expansion for tan^-1(3x) up to the term in x^3 is 3x-9x^3. [2] (maybe [3])
6) The integral of (1/(1+sinx)) between 0 and pi/2 = 1, using t=tan(0.5x). [5]
7)(i) x1=0.5 gives a tangent that cuts x-axis in the negative range where the curve (x^2-lnx-2) is undefined, while x1=1 tends to the other root, beta (where 1<beta<2), and so neither of these initial values give the required root of alpha (where 0<alpha<1). [2]
7)(ii) Show that the NR formula can be written as: [x(x^2+lnx+1)/(2x^2-1)]. [3]
7)(iii) Using an initial value of 0.1, find the value of alpha to [how many decimal places?]. [2]
8)(i) Cobweb diagram that shows convergence to root. [2]
8)(ii) Find x2 and x3 to 5dp given that x1=1.4. Find alpha to 4dp = 1.3488. [3]
8)(iii) Find e3/e2 and e4/e3, and comment on the gradient of the curve at the root. I'm not too sure on this one, but I wrote that the actual gradient lies between e3/e2 and e4/e3 because it is a cobweb diagram that shows fluctuations towards the root. [3]
9)(i) Show that theta=0 is a line of symmetry. Prove that f(theta)=f(2(0)-theta)=f(-theta). f(theta)=2cos(theta)-1, so f(-theta)=2cos(-theta)-1=2cos(theta)-1=f(theta), therefore theta=0 is a line of symmetry. [2]
9)(ii) Sketch the polar curve, indicating the point when theta=0 (r=1) as well as the tangents at the pole (theta=pi/3 or -pi/3). [4]
9)(iii) Area enclosed by the curve is pi-(3/2)root3. [5]

(edited 5 years ago)

Reply 23

JackMac2904

16

Nice, I got those answers. Reckon I would of lost less than 5 marks (hopefully) which should be more than enough for the A* :smile:

Reply 24

akaash13

6

Original post by L_stMemories

(>'-'

> UNOFFICIAL MARKSCHEME <('-'< :wink:

Below are the topics each question tests on, let me know if something doesn't look right!

Q1 - Graphs and Functions
Q2 - Hyperbolic Functions: Prove Identity and Integration
Q3 - Hyperbolic Functions: Differentiation
Q4 - Reduction Formulae (with a cheeky series/integrals question in part(iii))
Q5 - Integration with t=tan(0.5x) Substitution. (Weierstrass for extra nerdy human beings. :3)
Q6 - Maclaurin Series
Q7 - Newton-Raphson Method
Q8 - Numerical Iterative Method
Q9 - Polar Coordinates

It is entirely possible that I have mixed up the orders of Q4-Q6. Let me know if I did!

1)(i) The asymptotes of the graph are x=1 and y=1. [2]
1)(ii) y has range y≧0.5 (greater than or equal to 0.5). The only stationary point of the graph is at (-1, 0.5). [6]
1)(iii) Graph sketching! Some features: nothing below the x-axis; everything above y=0.5 (as found in part(ii)); as x tends to -ve infinity, the graph approaches y=1 from below and as x tends to +ve infinity, the graph approaches y=1 from above; asymptote at x=1 where both curves shoot upwards at both sides of the asymptote. [2]
2)(i) Show that cosh2x = 2(coshx)^2-1 (write coshx in terms of exponentials and expand the entire expression 2(coshx)^2-1). [3]
2)(ii) Integral of (cosh3x)^2 between 0 and 1 = (1/2) + (1/12)sinh(6) (in the form A+BsinhC, as required). [3]
3) Show that [hyperbolic curve] has no stationary points. (I personally found dy/dx first, then used an identity to convert it into a quadratic in sinhx, and b^2-4ac was negative and hence there are no roots of dy/dx=0, therefore no stationary points.) [5]
4) Integral In = (x^n e^(x-1)) between 0 and 1.
4)(i) Show that In=1-nI(n-1) (the n and n-1 are subscript to represent the reduction integral). [3]
4)(ii) Find the exact value of I4 = 9-24e^-1. [4] (maybe [3])
4)(iii) Given that y=x^n e^(x-1), find dy/dx and sketch the curve. Hence deduce that 0<In<1. Personally this was my shakiest ground, but I think the dy/dx proves that y is an increasing function for n≧1 (which is given). The y coordinate of the graph at x=1 is y=1, which is the unit square. Since the curve lies between the unit square, it must have an area less than 1 but (evidently) greater than 0 because...well, it has an area... [5]
5)(i) Derivative of tan^-1(3x) is 3/(1+9x^2) [2]
5)(ii) Show [something really complicated with triple derivatives involving 18 or 36]. [4]
5)(iii) Maclaurin expansion for tan^-1(3x) up to the term in x^3 is 3x-9x^3. [2] (maybe [3])
6) The integral of (1/(1+sinx)) between 0 and pi/2 = 1, using t=tan(0.5x). [5]
7)(i) x1=0.5 gives a tangent that cuts x-axis in the negative range where the curve (x^2-lnx-2) is undefined, while x1=1 tends to the other root, beta (where 1<beta<2), and so neither of these initial values give the required root of alpha (where 0<alpha<1). [2]
7)(ii) Show that the NR formula can be written as: [x(x^2+lnx+1)/(2x^2-1)]. [3]
7)(iii) Using an initial value of 0.1, find the value of alpha to [how many decimal places?]. [2]
8)(i) Cobweb diagram that shows convergence to root. [2]
8)(ii) Find x2 and x3 to 5dp given that x1=1.4. Find alpha to 4dp = 1.3488. [3]
8)(iii) Find e3/e2 and e4/e3, and comment on the gradient of the curve at the root. I'm not too sure on this one, but I wrote that the actual gradient lies between e3/e2 and e4/e3 because it is a cobweb diagram that shows fluctuations towards the root. [3]
9)(i) Show that theta=0 is a line of symmetry. Prove that f(theta)=f(2(0)-theta)=f(-theta). f(theta)=2cos(theta)-1, so f(-theta)=2cos(-theta)-1=2cos(theta)-1=f(theta), therefore theta=0 is a line of symmetry. [2]
9)(ii) Sketch the polar curve, indicating the point when theta=0 (r=1) as well as the tangents at the pole (theta=pi/3 or -pi/3). [4]
9)(iii) Area enclosed by the curve is pi-(3/2)root3. [5]

Cool, I got these answers too :smile:

I know this is unrelated but does anybody have predictions for 80 and 90 ums boundaries for this and other further maths papers (fp3, s2, m2 and m3 for me)?