Question
Consider the vectors a and b such that \(a = \binom{12}{-5} and |b| = 15\)
(a) Find the possible range of values for | a + b |.
Consider the vector p such that p = a + b.
(b) Given that | a + b | is a minimum, find p.
Consider the vector q such that \(q =\binom{x}{y}\), where x , y ∈ R+ .
(c) Find q such that |q| = |b| and q is perpendicular to a.
Answer/Explanation
Ans:
Note: Award (A1)A0 for 2 and 28 seen with no indication that they are the endpoints of an interval.
(b) recognition that p or b is a negative multiple of a
Question
A scientist conducted a nine-week experiment on two plants, A and B, of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant A was given fertilizer regularly, while Plant B was not.
The scientist found that the height of Plant A, hA cm, at time t weeks can be modelled by the function hA (t) = sin(2t + 6) + 9t + 27, where 0 ≤ t ≤ 9.
The scientist found that the height of Plant B, hB cm, at time t weeks can be modelled by the function hB (t) = 8t + 32, where 0 ≤ t ≤ 9.
(a) Use the scientist’s models to find the initial height of
(i) Plant B;
(ii) Plant A correct to three significant figures.
(b) Find the values of t when hA (t) = hB (t).
(c) For t > 6, prove that Plant A was always taller than Plant B.
(d) For 0 ≤ t ≤ 9, find the total amount of time when the rate of growth of Plant B was greater than the rate of growth of Plant A.
Answer/Explanation
Ans:
(a) (i) 32 (cm)
(ii) hA (0) = sin (6) + 27
= 26.7205…
= 26.7 (cm)
(b) attempts to solve hA (t) = hB (t) for t
t = 4.0074…,4.7034…,5.88332…
t = 4.01,4.70,5.88 (weeks)
(c) hA (t) – hB (t) = sin (2t +6) +t -5
EITHER
for t > 6, t − 5 > 1
and as sin (2t +6) ≥-1 ⇒ hA (t) – hB (t) > 0
OR
the minimum value of sin (2t+6) = -1
so for t> 6, hA (t) – hB (t) = t -6 > 0
THEN
hence for t > 6, Plant A was always taller than Plant B
(d) recognises that hA‘ (t) and hB‘ (t) are required
attempts to solve hA‘ (t) = hB‘ (t) for t
t =1.18879… and 2.23598… OR 4.33038… and 5.37758… OR 7.47197… and 8.51917…
