Question
The following diagram shows a circle with centre O and radius 4cm.
The points P, Q and R lie on the circumference of the circle and PÔR = θ, where θ is measured in radians.
The length of arc PQR is 10cm.
(a) Find the perimeter of the shaded sector.
(b) Find θ.
(c) Find the area of the shaded sector.
Answer/Explanation
Answer:
(a) attempts to find perimeter
arc+2 × radius OR 10 + 4 + 4
= 18 (cm)
(b) 10 = 4θ
θ = \(\frac{10}{4}\) \(=\frac{5}{2}\), 2.5)
(c) area = \(\frac{1}{2}(\frac{10}{4})(4^2)\) (=1.25 × 16)
= 20 (\(cm^2\))
Question
A function f is defined by f(x) = 1 – \(\frac{1}{x-2}\), where \(x\epsilon \mathbb{R}\), x ≠ 2.
(a) The graph of y = f (x) has a vertical asymptote and a horizontal asymptote.
Write down the equation of
(i) the vertical asymptote;
(ii) the horizontal asymptote.
(b) Find the coordinates of the point where the graph of y = f (x) intersects
(i) the y-axis;
(ii) the x-axis.
(c) On the following set of axes, sketch the graph of y = f (x), showing all the features found in parts (a) and (b).
Answer/Explanation
Answer:
(a) (i) x = 2
(ii) y = 1
(b) (i) (0, \(\frac{3}{2}\))
(ii) (3, 0)
(c) 
two correct branches with correct asymptotic behaviour and intercepts clearly shown
Question
Events A and B are such that P(A) = 0.4, P(A|B) = 0.25 and P(A∪B) = 0.55. Find P(B).
Answer/Explanation
Answer:
substitutes into P(A∪B) = P(A) + P(B) – P(A∩B) to form
0.55 = 0.4 + P(B) – P(A∩B) (or equivalent)
substitutes into P(A|B) = \(\frac{P(A∩b)}{P(B)}\) to form 0.25 =\(\frac{P(A∩B)}{P(B)}\) (or equivalent)
attempts to combine their two probability equations to form an equation in P(B)
correct equation in P(B)
0.55 = 0.4 + P(B) – 0.25P(B) OR \(\frac{P(B) – 0.15}{P(B)}\) = 0.25 OR P(B) – 0.15 = 0.25P(B)
(or equivalent)
P(B) = \(\frac{15}{75}\) (=\(\frac{1}{5}=0.2\))
Question
The following diagram shows part of the graph of y = \(\frac{x}{x^2 + 2}\) for ≥ 0.
The shaded region R is bounded by the curve, the x-axis and the line x = c .
The area of R is ln3.
Find the value of c .
Answer/Explanation
Answer:
\(A = \int_{0}^{c} \frac{x}{x^2 + 2}dx\)
EITHER
attempts to integrate by inspection or substitution using \(u = x^2 + 2\) or \(u = x^2\)
OR
attempts to integrate by inspection
THEN
correctly substitutes their limits into their integrated expression
correctly applies at least one log law to their expression 
\(\frac{c^2+2}{2} = 9\) OR \(\sqrt{\frac{c^2+2}{2}} = 3\)
\(c^2 = 16\)
c = 4
Question
The functions f and g are defined for x ∈ \(\mathbb{R}\) by
f (x) = ax + b, where a, b ∈ \(\mathbb{Z}\)
g(x) = \(x^2 + x +3\).
Find the two possible functions f such that \((g\circ f)(x)=4x^2 – 14x + 15\).
Answer/Explanation
Answer:
attempts to form (g \(\circ \) f) (x)
equates their corresponding terms to form at least one equation
a = ±2 (seen anywhere)
attempt to use 2ab + a = -14 to pair the correct values (seen anywhere)
f(x) = 2x – 4 (accept a = 2 with b = -4) f(x) = -2x + 3 (accept a = -2 with b = 3)
Question
A continuous random variable X has probability density function f defined by
where a is a positive real number.
(a) State E(X) in terms of a.
(b) Use integration to find Var(X) in terms of a.
Answer/Explanation
Answer:
(a) E(X) = 2a (by symmetry)
(b) METHOD 1
METHOD 2
Question
Use mathematical induction to prove that \(\sum_{r=1}^{n}\frac{r}{(r+1)!}=1-\frac{1}{(n+1)!}\) for all integers n ≥ 1.
Answer/Explanation
Answer:
let P(n) be the proposition that \(\sum_{r=1}^{n}\frac{r}{(r+1)!}= 1-\frac{1}{(n+1)!}\) for all integers, n ≥ 1
considering P(1):
LHS = \(\frac{1}{2}\) and RHS = \(\frac{1}{2}\) and so P(1) is true
assume P(k) is true ie, \(\sum_{r=1}^{k}\frac{r}{(r+1)!}= 1-\frac{1}{(k+1)!}\)
considering P(k+1):
P (k + 1) is true whenever P(k) is true and P (1) is true, so P(n) is true
(for all integers, n ≥1)
Question
The functions f and g are defined by
f(x) = cos x, 0 ≤ x ≤ \(\frac{π}{2}\)
g(x) = tan x, 0 ≤ x ≤ \(\frac{π}{2}\)
The curves y = f(x) and y = g(x) intersect at a point P whose x – coordinate is k, where 0 < k < \(\frac{\pi}{2}\).
(a) Show that \(cos^2\) k = sin k.
(b) Hence, show that the tangent to the curve y = f(x) at P and tangent to the curve y = g(x) at P intersect at right angles.
(c) Find the value of sin k. Give your answer in the form \(\frac{a+\sqrt{b}}{c}\), where a, \(c\epsilon \mathbb{Z}\) and \(b\epsilon \mathbb{Z}^+\).
Answer/Explanation
Answer:
(a) cos k = \(\frac{sin k}{cos k}\)
\(cos^2 k = sin k\)
(b) f'(k) = -sin k and g'(k) = \(sec^2 k\)
EITHER
f'(k) g'(k) = -\(\frac{sin k}{cos^2 k}\)
\(cos^2 k = sin k ⇒ g'(k) = \frac{1}{sin k} = -\frac{1}{f'{k}}\)
(c) \(1 – sin^2 k = sin k\) (from part (a))
\(sin^2 k + sin k – 1 = 0\)
attempts to solve for sin k
sin k = \(\frac{-1 ± \sqrt{1^2 – 4(1)(-1)}}{2}\)
(for 0 < k < \(\frac{\pi}{2}\), sin k > 0) ⇒ sin k = \(\frac{-1 + \sqrt{5}}{2}\)
(a = -1, b = 5, c = 2)
Question
The following diagram shows parallelogram OABC with \(\vec{OA}=a\), \(\vec{OC}=c\) and |c| = 2|a|, where |a| \(\neq \) 0.
The angle between \(\vec{OA}\) and \(\vec{OC}\) is θ, where 0 < θ < \(\pi\).
Point M is on [AB] such that \(\vec{AM}\) = k \(\vec{AB}\), where 0 ≤ k ≤ 1 and \(\vec{OM} . \vec{MC}\) = 0.
(a) Express \(\vec{OM}\) and \(\vec{MC}\) in terms of a and c.
(b) Hence, use a vector method to show that\(|a|^2\) (1 – 2k) (2 cos θ – (1 – 2k)) = 0.
(c) Find the range of values for \(\theta \) such that there are two possible for M.
Answer/Explanation
Answer:
(a) \(\vec{OM} = a + kc\)
\(\vec{MC} = (1 – k)c – a\)
(b) attempts to expand their dot product \(\vec{OM}.\vec{MC}\) = (a + kc).((1 – k)c – a)
(c) attempts to solve \(|a|^2\) (1 – 2k)(2cos \(\theta\) – (1 – 2k)) = 0 for k
k = \(\frac{1}{2}\) or k = \(\frac{1}{2} – cos \theta \) (\(|a|^2\) > 0)
as 0 ≤ k ≤ 1, 0 ≤ \(\frac{1}{2} – cos \theta ≤ 1\)
\(- \frac{1}{2}\leq cos \theta \leq \frac{1}{22}\)
\(\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, \theta \neq \frac{\pi}{2}\)
(\(\theta = \frac{\pi}{2}\) corresponds to only one possible position for M when k = \(\frac{1}{2}\))
Question
A circle with equation \(x^2 + y^2 = 9\) has centre (0 , 0) and radius 3.
A triangle, PQR, is inscribed in the circle with its vertices at P(-3, 0), Q(x , y) and R(x , -y), where Q and R are variable points in the first and fourth quadrants respectively. This is shown in the following diagram.
(a) For point Q, show that y = \(\sqrt{9-x^2}\).
(b) Hence, find an expression for A, the area of triangle PQR, in terms of x.
(c) Show that \(\frac{dA}{dx} = \frac{9 – 3x – 2x^2}{\sqrt{9-x^2}}\).
(d)H⇒ence or otherwise, find the y – coordinate of R such that A is a maximum.
Answer/Explanation
Answer:
(a) \(y^2 = 9 – x^2\) OR y = ±\(\sqrt{9-x^2}\)
(since y > 0) ⇒ y = \(\sqrt{9-x^2}\)
(b) b = 2y (=2\(\sqrt{9-x^2}\) or h = x + 3
attempts to substitute their base expression and height expression into A = \(\frac{1}{2}bh\)
A = \(\sqrt{9-x^2} (x + 3)\) (or equivalent) (=\(\frac{2(x+3)\sqrt{9-x^2}}{2} = x\sqrt{9-x^2} + 3 \sqrt{9-x^2}\))
(c) METHOD 1
attempt to use the product rule to find \(\frac{dA}{dx}\)
attempts to use the chain rule to find \(\frac{d}{dx} \sqrt{9-x^2}\)
METHOD 2
(d) \(\frac{dA}{dx}=0 (\frac{9-3x-2x^2}{\sqrt{9-x^2}}=0)\)
attempts to solve \(9-3x-2x^2 = 0\) (or equivalent)
-(2x – 3)(x + 3)(=) OR \(x = \frac{3\pm \sqrt{(-3)^2-4(-2)(9)}}{2(-2)}\) (or equivalent)
\(x = \frac{3}{2}\)
substitutes their value of x into either \(y = \sqrt{9-x^2}\) or \(y = -\sqrt{9-x^2}\)
Question
Consider the complex number u = -1 + \(\sqrt{3}i\).
(a) By finding the modulus and argument of u, show that \(u = 2e^{i \frac{2 \pi }{3}}\).
(b) (i) Find the smallest positive integer n such that \(u^n\) is a real number.
(ii) Find the value of \(u^n\) when n takes the value found in part (b)(i).
(c) Consider the equation \(z^3 + 5z^2 + 10z + 12 = 0\), where \(z\epsilon \mathbb{C}\)
(i) Given that u is a root of \(z^3 + 5z^2 + 10z + 12 = 0\), find the other roots.
(ii) By using a suitable transformation from z to w, or otherwise, find the roots of the equation 1 + 5w + 10\(w^2\) + 12\(w^3\) = 0, where \(z\epsilon \mathbb{C}\), z \(\neq \) 0.
By expressing z in the form a + bi, find the roots of the equation.
Answer/Explanation
Answer:
(a) METHOD
\(|u| = \sqrt{(-1)^2+(\sqrt{3})^2}(=\sqrt{1+3})\)
= 2
reference angel = \(\frac{\pi}{3}\) OR arg u = \(\pi – tan^{-1} (\sqrt{3})\) OR arg u = \(\pi + tan^{-1}(-\sqrt{3})\)
= \(\pi – \frac{\pi}{3}\)
= \(\frac{2 \pi}{3}\) and so \(u = 2e^{i \frac{2 \pi }{3}}\)
METHOD 2
reference angle = \(\frac{\pi}{3}\) OR arg u = \(\pi – tan^{-1} (\sqrt{3})\) OR arg u = \(\pi + tan^{-1} (- \sqrt{3})\)
= \(\pi – \frac{\pi }{3}\)
= \(\frac{2 \pi}{3}\)
(b) (i) \(u^n \epsilon \mathbb{R} \Rightarrow \frac{2n \pi}{3} = k \pi (k\epsilon \mathbb{Z})\)
n = 3
(ii) substitutes their value (must be a multiple of 3) for n into \(u^n}\)
\(u^3 = 2^3 cos 2 \pi\)
= 8
(c) (i) -1 – \(\sqrt{3} i\) is a root (by the conjugate root theorem)
let z = c be the real root
EITHER
uses sum of roots (equated to \(\pm\) 5)
\(((-1+\sqrt{3}i)+(-1-\sqrt{3}i) + c) = -5\)
-2 + c = -5
OR
uses product of roots (equated to \(\pm\) 12)
\((-1 + \sqrt{3}i)(-1-\sqrt{3}i)=-12\)
4c = -12
OR
\((z-(-1 + \sqrt{3}i))(z-(-1-\sqrt{3}i))=z^2 + 2z + 4\)
compares coefficients eg
\((z-c)(z^2 + 2z + 4) = z^3 + 5z^2 + 10z + 12\)
-4c = 12
THEN
c = -3 (and so z = -3 is a root)
(ii) METHOD 1
compare \(z^3 + 5z^2 + 10z +12 = 0\) and \(1 + 5w + 10w^2 + 12w^2 = 0\)
z = \(\frac{1}{w} \Rightarrow w = \frac{1}{z}\)
\(w = -\frac{1}{3}, \frac{1}{-1 \pm \sqrt{3}}i (=\frac{-1 \pm \sqrt{3}i }{4})\)
METHOD 2
attempts to factorize into a product of a linear factor and a quadratic factor
\(1 + 5w + 10w^2 + 12w^3 = (3w + 1)(4w^2 + 2w + 1)\)
\(\(w = -\frac{1}{3}, \frac{1}{-1 \pm \sqrt{3}}i (=\frac{-1 \pm \sqrt{3}i }{4})\)
(d) \(a + bi)^2 = 2(a – bi)\)
attempts to expand and equate real and imaginary parts:
\(a^2 – b^2 + 2abi = 2a – 2bi\)
\(a^2 – b^2 = 2a\) and 2ab = -2b
attempts to find the value of a or b
2b(a + 1) = 0
b = 0 \(\Rightarrow a^2\) = 2a \(\Rightarrow \) a = 2 (real root)
a = -1 \(\Rightarrow 1 – b^2 = -2 \Rightarrow = \pm \sqrt{3}\) (complex roots -1 \(\pm \sqrt{3i}\))
Question
(a) By using an appropriate substitution, show that \(\int cos \sqrt{x} dx = 2 \sqrt{x} sin \sqrt{x} + 2cos \sqrt{x} + C\).
The following diagram shows part of the curve y = cos \(\sqrt{x}\) for x ≥ 0.
The curve intersects the x-axis at \(x_1, x_2, x_3, x_4\), ….
The nth x-intercept of the curve, \(x_n\), is given by \(x_n\) = \(\frac{(2n-1)^2 \pi ^2}{4}\), where \(n \epsilon \mathbb{Z}^+\).
(b) Write down a similar expression for \(x_{n+1}\).
The regions bounded by the curve and the x-axis are denoted by \(R_1 , R_2 , R_3\) , …, as shown on the above diagram.
(c) Calculate the area of region \(R_n\) .
Give your answer in the form k nπ, where \(k\epsilon \mathbb{Z}^+\)
(d) Hence, show that the areas of the regions bounded by the curve and the x-axis, \(R_1 , R_2 , R_3\) , …, form an arithmetic sequence.
Answer/Explanation
Answer:
(a) let t = \(\sqrt{x}\)
\(t^2 = x \Rightarrow 2t dt = dx\) (or equivalent)
so \(\int cos \sqrt{x} dx = 2 \int t cos t dt\)
attempts integration by parts
u = 2t, dv = cos t dt, du = 2 dt, v = sin t
\(2\int\) t cost + 2t sin t – \(2\int \) sin t dt
= 2t sin t + 2 cos t + C
substitution of t = \(\sqrt{x}\) \(\Rightarrow \int cos \sqrt{x} dx = 2\sqrt{x}sin \sqrt{x} + 2 cos \sqrt{x} + C\)
(b) \(x_{n+1} = \frac{(2(n+1)-1)^2\pi^2}{4} (=\frac{(2n+1)^2\pi^2}{4})\)
(c) area of \(R_n\) is \(\left | \int_{x_n}^{x_{n+1}}cos \sqrt{x}dx \right |\)
attempts to substitute their limits into their integrated expression
(d) EITHER
attempts to find (d=) \(R_{n+1} – R_n\)
(d=)4(n+1) \(\pi \) – 4n\(\pi \)
= \(4 \pi\)
which is a constant (common difference is \(4 \pi\))
OR
an arithmetic sequence is of the form \(u_n = dn + c (u_n = dn + u_1 – d\))
attempts to compare \(u_n = dn + c (u_n = dn + u_1 – d)\) and \(R_n = 4n\pi\)
d = \(4 \pi\) and c = 0 (\(u_1 – d = 0\))
THEN
so the ares of the regions form an arithmetic sequence