(𝑯𝒐) 𝟑 - brillantmont mathematics= 2i + 2j + k and = i + 4j + 3k. the line l has vector...

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(𝑯𝒐)𝟑


1. With respect to the origin O, the points A and B have position vectors given by

= 2i + 2j + k and = i + 4j + 3k.

The line l has vector equation r = 4i – 2j + 2k + s(i + 2j + k).

Prove that the line l does not intersect the line through A and B. [5]

OA OB


2. Solve the inequality 2x > |x – 1|. [4]


3. Expand in ascending powers of x, up to and including the term in x3, simplifying

the coefficients. [4]

2

1

)41(

x


4. (i) Prove the identity

[4]

(ii) Hence solve the equation

for 0° ≤ θ ≤ 360°. [4]

.3cos82cos44cos 4

,22cos44cos


5. In a certain industrial process, a substance is being produced in a container. The mass of the

substance in the container t minutes after the start of the process is x grams. At any time, the rate

of formation of the substance is proportional to its mass. Also, throughout the process, the

substance is removed from the container at a constant rate of 25 grams per minute.

When t = 0, x = 1000 and

(i) Show that x and t satisfy the differential equation

0.1 (x – 250).

[2]

(ii) Solve this differential equation, obtaining an expression for x in terms of t. [6]

t

x

d

d

.75d

d

t

x


6. The equation x3 – x – 3 = 0 has one real root, α.

(i) Show that α lies between 1 and 2. [2]

Two iterative formulae derived from this equation are as follows:

(A)

(B)

Each formula is used with initial value x1 = 1.5.

(ii) Show that one of these formulae produces a sequence which fails to converge, and use the

other formula to calculate α correct to 2 decimal places. Give the result of each iteration

to 4 decimal places. [5]

,3–= 31+ nn xx

.)3+(= 3

1

1+ nn xx


7. By expressing 8 sin θ – 6cos θ in the form R sin (θ – α), solve the equation

8 sin θ – 6cos θ = 7,

for 0° ≤ θ ≤ 360°. [7]


8. (i) Use the substitution x = sin2 θ to show that

[4]

(ii) Hence find the exact value of

[4]

. dsin2d

1

2 θθxx

x

4

1

0.d

1x

x

x


1. State or imply a direction vector for AB is –i + 2j + 2k, or equivalent B1

State equation of AB is r = 2i + 2j + k +t(–i + 2j + 2k), or equivalent B1√

Equate at least two pairs of components of AB and l and solve for

s or for t M1

Obtain correct answer for s or for t, e.g. s = 0 or t = −2; s = or

t = or s = 5 or t = 3 A1

Verify that all three pairs of equations are not satisfied and that the

lines fail to intersect A1

2. EITHER:

State or imply non-modular inequality (2x)2 > (x – 1)

2, or

corresponding equation B1

Expand and make a reasonable solution attempt at a 2- or 3- term

quadratic M1

Obtain critical value x = A1

State answer x > only A1

OR:

State the relevant critical linear equation, i.e. 2x = 1 – x B1

Obtain critical value x = B1

State answer x > B1

State or imply by omission that no other answer exists B1

OR:

Obtain the critical value x = from a graphical method, or by

inspection, or by solving a linear inequality B2

State answer x > B1

State or imply by omission that no other answer exists B1 [4]

3

5

3

1

31

31

31

31

31

31


3. EITHER:

Obtain correct unsimplified version of the x or x2 or x

3 term M1

State correct first two terms 1 – 2x A1

Obtain next two terms 6x2 – 20x

3 A1 + A1

[The M mark is not earned by versions with unexpanded

binomial coefficients, e.g. .]

OR:

Differentiate expression and evaluate f(0) and f’(0),

where f ′(x) = k(1 + 4x M1

State correct first two terms 1 – 2x A1

Obtain next two terms 6x2 – 20x

3 A1 + A14

[4]

4. (i) EITHER:

Express cos 4θ in terms of cos 2θ and/or sin 2θ B1

Use double angle formulae to express LHS in terms of cos θ

(and maybe sin θ) M1

Obtain any correct expression in terms of cos θ alone A1

Reduce correctly to the given form A1

OR:

Use double angle formula to express RHS in terms of cos 2 θ M1

Express cos22 θ in terms of cos 4 θ. B1

Obtain any correct expression in terms of cos 4 θ and cos2 θ A1

Reduce correctly to the given form A1 4

(ii) Using the identity, carry out method for calculating one root M1

Obtain answer 27.2° (or 0.475 radians) or 27.3° (or 0.476 radians) A1

Obtain a second answer, e.g. 332.8° (or 5.81 radians) A1√

Obtain remaining answers, e.g. 152.8° and 207.2°

(or 2.67 and 3.62 radians) and no others in range A1√4 [8]

2

21

2

3

)


5. (i) State or imply that = kx –25 B1

Show that k = 0.1 and justify the given statement B12

(ii) Separate variables and attempt integration M1

Obtain ln(x – 250), or equivalent A1

Obtain O. 1t, or equivalent A1

Evaluate a constant or use limits t = 0, x = 1000 with a solution

containing terms aln(x – 250) and bt M1

Obtain any correct form of solution, e.g. ln(x – 250) = 0. 1t + ln 750 A1

Rearrange and obtain x = 250 (3e0.1t

+ 1), or equivalent A16 [8]

6. (i) Consider sign of x3 – x – 3, or equivalent M1

Justify the given statement A12

(ii) Apply an iterative formula correctly at least once, with initial value

x1 = 1.5 M1

Show that (A) fails to converge A1

Show that (B) converges A1

Obtain final answer 1.67 A1

Show sufficient iterations to justify its accuracy to 2 d.p., or show there

is a sign change in the interval (1.665, 1.675) A15 [7]

7. State or imply that R = 10 or R = −10 B1

Use trig formula to find α M1

Obtain α = 36.9° if R = 10 or α = 216.9° if R = −10, with no errors seen A1

Carry out evaluation of sin –1

( )(≈ 44.427...°) M1

Obtain answer 81.3° A1

Carry out correct method for second answer M1

Obtain answer 172.4° and no others in the range A1

[Ignore answers outside the given range.] [7]

tx

dd

10

7


8. (i) State = 2sin θ cos θ, or dx = 2sin θ cos θ dθ B1

Substitute for x and dx throughout M1

Obtain any correct form in terms of θ A1

Reduce to the given form correctly A14

(ii) Use cos 2A formula, replacing integrand by a + bcos 2θ,

where ab ≠ 0 M1*

Integrate and obtain θ – sin2 θ A1√

Use limits θ = 0 and θ = M1(dep *)

Obtain exact answer – , or equivalent A14

[8]

d

dx

2

1

6

1

6

13

4

1

(𝑯𝒐) 𝟑 - brillantmont mathematics= 2i + 2j + k and = i + 4j + 3k. the line l has vector...

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