Practice Paper DJS C1.3 No Calculator is allowed in C1
1 Write the terms up to x³ in the expansion of (2x – 3)⁷ [4]
2 Evaluate (1/36) –3/2 and 256 - 5/6 [4]
3 Simplify (16a⁴) –1/6 and (512 a-⁹ b18) – 2/9 [4]
4 Simplify √1225 + √12 [2]
5 Express 48 / (5 + √17) in the form a + b √17, where a & b are integers. [3]
6 Find the gradient and intercepts of the line 2x + 7y =12. [3]
7 Write the equation of the line parallel to this that goes thro’ (3,-4) [4]
8 The middle of three consecutive odd integers is 2n-5. Write down the other two.
Prove that the sum of the three integers is divisible by 3. [3]
9 Make q the subject of 1/p = 1/q – q/p [4]
10 Make r the subject of πr² + πr√(r²+h²) = A. Use completion of the square, not the quadratic formula [5]
Practice Paper DJS C1.4 No Calculator is allowed in C1
11 Write the terms up to p³ in the expansion of (p/2 – 5q)⁶ .
Use this to approximate 0.505 ⁶ (p=1, q = 1/100) to 6 d.p. [6]
12 Find the coefficient of x⁴ in the expansion of (2x – 7)⁸ [4]
13 Simplify (1/a12) –5/4 and (243 a30 b⁻⁵) - 3/5 [4]
14 Simplify √168 + √84 [2]
15 Express 25 / (6 - √11) in the form a + b √11, where a & b are integers. [3]
16 Write the equation of the normal line to 2x + 7y =12 that goes through (4,5) [5]
17 The middle of three consecutive even integers is 2n+6. Write down the other two. Find the highest common multiple of the product of the three integers. [4]
18 Differentiate (x² + 1/x)² [3]
19 Expand the expression (x – 1/x²)³ hence integrate y = (x – 1/x²)³ for 1<x<3 [5]
A section B question, or extension work….
20 The probability of passing the driving test varies with the number of times the test has been taken. Those who pass do not retake. Suggested pass rates are 1st time, 30%: 2nd time 35%: 3rd 20%: 4th 10%, 5 and more the remaining 5%.
Write (i) P(X≤3) (ii) P(X≥3).
How else might this be modelled? Suppose the probability of passing any one test is fixed at p, then the distribution can be written as
p + qp + q²p + q³p +….. = p / (1-q) =p/p = 1
(iii) Explain what q is (deduce this). Explain why the function must equal unity.
If p = 0.30, then the other terms are 0.210, 0.147, 0.103, 0.072
If p = 0.35, then the other terms are 0.228, 0.148, 0.096, 0.062
(iv) IS this consistent with your understanding of the problem?
