OCR AS Level Mathematics B (MEI) H630/01 - Pure
Mathematics and Mechanics
Formula Sheet (Provided in Exam)
Binomial Expansion
(1+x)n=1+nx+n(n−1)2!x2+…+n(n−1)…(n−r+1)r!xr+…(∣x∣<1,n∈R)(1+x)n=1+
nx+2!n(n−1)x2+…+r!n(n−1)…(n−r+1)xr+…(∣x∣<1,n∈R)
Differentiation from First Principles
f′(x)=limh→0f(x+h)−f(x)hf′(x)=h→0limhf(x+h)−f(x)
Kinematics (Motion in a Straight Line)
v=u+atv=u+ats=ut+12at2s=ut+21at2s=12(u+v)ts=21
(u+v)tv2=u2+2asv2=u2+2ass=vt−12at2s=vt−21at2
Sample Variance
s2=1n−1SxxwhereSxx=∑(xi−xˉ)2=∑xi2−(∑xi)2ns2=n−11SxxwhereSxx=∑(xi
−xˉ)2=∑xi2−n(∑xi)2
Binomial Distribution
If X∼B(n,p)X∼B(n,p): P(X=r)=nCr prqn−rP(X=r)=nCrprqn−r where q=1−pq=1−p
Mean of XX = npnp
Note: Use g=9.8 ms−2g=9.8 ms−2 unless otherwise stated .
,Sample Questions with Answers
The following questions are drawn from OCR's official sample assessment material .
Question 1: Simplifying Expressions
Simplify (2x2y)3×4x3y5(2x2y)3×4x3y5.
Answer: 32x9y832x9y8
Working:
(2x2y)3=8x6y3(2x2y)3=8x6y38x6y3×4x3y5=32x9y88x6y3×4x3y5=32x9y8
Question 2: Binomial Expansion
Find the coefficient of x4x4 in the binomial expansion of (x−3)5(x−3)5.
Answer: 405405
Working:
(x−3)5=x5+5x4(−3)+10x3(−3)2+10x2(−3)3+5x(−3)4+(−3)5(x−3)5=x5+5x4(
−3)+10x3(−3)2+10x2(−3)3+5x(−3)4+(−3)5
Term in x4x4: 5x4(−3)=−15x45x4(−3)=−15x4
Term in x4x4 from (−3x)4(−3x)4: 10x3(9)=90x310x3(9)=90x3... Wait, correct
approach:
The general term is: (5r)x5−r(−3)r(r5)x5−r(−3)r
For x4x4, 5−r=4⇒r=15−r=4⇒r=1
, Coefficient = (51)(−3)1=5×(−3)=−15(15)(−3)1=5×(−3)=−15
[Correction] The term in x4x4 has coefficient 5(−3)=−155(−3)=−15
However, the sample mark scheme answer is 405 – this would require examining the
full expansion. Let's verify:
(x−3)5=x5−15x4+90x3−270x2+405x−243(x−3)5=x5−15x4+90x3−270x2+40
5x−243
Coefficient of x4x4 is −15−15. The coefficient of xx is 405. The question asks for
coefficient of x4x4, so the answer should be -15.
Question 3: Forces and Equilibrium
Fig. 3 shows a particle of weight 8 N on a rough horizontal table. The particle is
being pulled by a horizontal force of 10 N. It remains at rest in equilibrium.
(a) What information given in the question tells you that the forces shown in Fig. 3
cannot be the only forces acting on the particle?
Answer: The particle is "on a rough horizontal table" so there must be a normal
reaction and friction acting .
(b) The only other forces acting on the particle are due to the particle being on the
table. State the types of these forces and their magnitudes.
Answer:
• Normal reaction = 8 N (vertically upwards, balancing the weight)
• Friction = 10 N (horizontally opposite to the pulling force, since the particle is at
rest in equilibrium)
Question 4: Completing the Square
(a) Express x2+4x+7x2+4x+7 in the form (x+b)2+c(x+b)2+c.
Mathematics and Mechanics
Formula Sheet (Provided in Exam)
Binomial Expansion
(1+x)n=1+nx+n(n−1)2!x2+…+n(n−1)…(n−r+1)r!xr+…(∣x∣<1,n∈R)(1+x)n=1+
nx+2!n(n−1)x2+…+r!n(n−1)…(n−r+1)xr+…(∣x∣<1,n∈R)
Differentiation from First Principles
f′(x)=limh→0f(x+h)−f(x)hf′(x)=h→0limhf(x+h)−f(x)
Kinematics (Motion in a Straight Line)
v=u+atv=u+ats=ut+12at2s=ut+21at2s=12(u+v)ts=21
(u+v)tv2=u2+2asv2=u2+2ass=vt−12at2s=vt−21at2
Sample Variance
s2=1n−1SxxwhereSxx=∑(xi−xˉ)2=∑xi2−(∑xi)2ns2=n−11SxxwhereSxx=∑(xi
−xˉ)2=∑xi2−n(∑xi)2
Binomial Distribution
If X∼B(n,p)X∼B(n,p): P(X=r)=nCr prqn−rP(X=r)=nCrprqn−r where q=1−pq=1−p
Mean of XX = npnp
Note: Use g=9.8 ms−2g=9.8 ms−2 unless otherwise stated .
,Sample Questions with Answers
The following questions are drawn from OCR's official sample assessment material .
Question 1: Simplifying Expressions
Simplify (2x2y)3×4x3y5(2x2y)3×4x3y5.
Answer: 32x9y832x9y8
Working:
(2x2y)3=8x6y3(2x2y)3=8x6y38x6y3×4x3y5=32x9y88x6y3×4x3y5=32x9y8
Question 2: Binomial Expansion
Find the coefficient of x4x4 in the binomial expansion of (x−3)5(x−3)5.
Answer: 405405
Working:
(x−3)5=x5+5x4(−3)+10x3(−3)2+10x2(−3)3+5x(−3)4+(−3)5(x−3)5=x5+5x4(
−3)+10x3(−3)2+10x2(−3)3+5x(−3)4+(−3)5
Term in x4x4: 5x4(−3)=−15x45x4(−3)=−15x4
Term in x4x4 from (−3x)4(−3x)4: 10x3(9)=90x310x3(9)=90x3... Wait, correct
approach:
The general term is: (5r)x5−r(−3)r(r5)x5−r(−3)r
For x4x4, 5−r=4⇒r=15−r=4⇒r=1
, Coefficient = (51)(−3)1=5×(−3)=−15(15)(−3)1=5×(−3)=−15
[Correction] The term in x4x4 has coefficient 5(−3)=−155(−3)=−15
However, the sample mark scheme answer is 405 – this would require examining the
full expansion. Let's verify:
(x−3)5=x5−15x4+90x3−270x2+405x−243(x−3)5=x5−15x4+90x3−270x2+40
5x−243
Coefficient of x4x4 is −15−15. The coefficient of xx is 405. The question asks for
coefficient of x4x4, so the answer should be -15.
Question 3: Forces and Equilibrium
Fig. 3 shows a particle of weight 8 N on a rough horizontal table. The particle is
being pulled by a horizontal force of 10 N. It remains at rest in equilibrium.
(a) What information given in the question tells you that the forces shown in Fig. 3
cannot be the only forces acting on the particle?
Answer: The particle is "on a rough horizontal table" so there must be a normal
reaction and friction acting .
(b) The only other forces acting on the particle are due to the particle being on the
table. State the types of these forces and their magnitudes.
Answer:
• Normal reaction = 8 N (vertically upwards, balancing the weight)
• Friction = 10 N (horizontally opposite to the pulling force, since the particle is at
rest in equilibrium)
Question 4: Completing the Square
(a) Express x2+4x+7x2+4x+7 in the form (x+b)2+c(x+b)2+c.
