Question: 1
Two events A and B are such that P(A) = 0.5 and P(A∩B) = 0.1.
(a) Find P(A∩B’).
(b) Given that P(A’ ∩ B’) = 0.2, find:
(i) P(A ∪ B);
(ii) P(B);
(iii) P(A|B’).
Answer/Explanation
Ans:
(a) If we shade in the region A ∩ B’ on the Venn diagram below, we have
Hence, using the Venn diagram above, we deduce
P(A∩B’) = P(A) – P(A∩B)
= 0.5 – 0.1
= 0.4
(b) (i) We have
(A ∪ B)’ = A’ ∩ B’
Hence, using the complementary events formula, we get
P(A’ ∩ B’) = 1 – P(A∪B)
0.2 = 1 – P(A∪B)
P(A∪B) = 1 – 0.2
P(A∪B) = 0.8
(ii) Using the combined events formula, we obtain
P(A∪B) = P(A) + P(B) – P(A∩B)
0.8 = 0.5 + P(B) – 0.1
0.8 – 0.4 = P(B)
P(B) = 0.4
(iii) Using the complementary events formula, we know P(B’) = 1 – P(B)
P(B’) = 0.6
Hence, using the conditional probability formula, we find
\(P(A|B’)=\frac{P(A\cap B’)}{P(B’)}\)
\(= \frac{0.4}{0.6}\)
\(= \frac{2}{3}\)
Question 2:
Two functions, f and g, are defined in the following table.
| x | -4 | 0 | 2 | 5 |
| f(x) | 9 | 2 | 0 | -2 |
| g(x) | 5 | -1 | 3 | -8 |
(a) Write down the value of f(2).
(b) Find the value of (g ο f) (2).
(c) Find the value of g-1 (5) .
Answer/Explanation
Ans:
(a) f(2) = 0
(b) We have
(g ο f) (2) = g(f(2))
= g(0)
= -1
(c) We have
(g ο g-1) (5) = 5
g(g-1(5)) = 5
g-1 (5) = -4 [since g(-4) = 5]
Question 3:
The following diagram shows the curve y = a cos (k(x -d)) + c where a, k, d and c are all positive constants. The curve has a minimum point at (1.5, 2) and a maximum point at (3.5, 7).
(a) Write down the value of a and the value of c.
(b) Find the value of k.
(c) Find the smallest possible value of d, given d > 0.
Answer/Explanation
Ans:
(a) We have
\(a = \frac{y max – y min}{2}\)
\(= \frac{7 – 2}{2}\)
= 2.5
and
\(c = \frac{y max + y min}{2}\)
\(= \frac{7 + 2}{2}\)
= 4.5
(b) The period of the curve is
period = 2 (3.5 – 1.5)
= 4
Hence we get
\(k = \frac{2\pi }{period}\)
\(=\frac{2\pi }{4}\)
\(=\frac{\pi }{2}\)
Question 4:
(a) Find \(\int 6x^{2}e^{x^{3}+8}dx.\)
(b) Find f(x) given that \(f'(x) = 6x^{2}e^{x^{3}+8}\) and f(-2) = 5.
Answer/Explanation
Ans:
(a) Using integration by substitution with u = x3 + 8 and \(\frac{du}{3x^{2}}= dx\), we get
\(\int 6x^{2}e^{x^{3}+8}dx = \int 6x^{2}e^{u}\frac{du}{3x^{2}}\)
\(=\int 2e^{u}du\)
= 2eu + C
\(= 2e^{x^{3}+8}+C\)
(b) We know that f(x) = ∫ f'(x) dx and from part (a) \(f(x)= 2e^{x^{3}+8}+C\). Let’s determine C. We have
\(5= 2e^{(-2)^{3}+8}+C\)
5 = 2 + C
C = 3
We found \(f(x)= 2e^{x^{3}+8}+3\)
Question 5:
Solve the equation cos 2x – sin2x + 3 cos x, for 0 ≤ x ≤ 2π.
Answer/Explanation
Ans:
Using the Pythagorean identity and double angle identity for cosine, we get
cos 2x – sin2x = cos 2 x + 3 cos x
cos 2x – 3 cos x = cos 2 x + 3 sin2 x
cos 2x – 3 cos x = 1
(2 cos2x – 1) – 3 cos x = 1
2 cos2 x – 3 cos x – 2 = 0
(2 cos x + 1) (cos x – 2) = 0
2 cos x + 1 = 0 [ since cos x – 2 ≠ 0]
cos x = \(-\frac{1}{2}\)
\(x = \frac{2\pi }{3},\frac{4\pi }{3}\)
Question 6:
Consider f(x) = logk (8x – 2x2), for 0 < x < 4, where k > 0.
The equation f(x) = 3 has exactly one solution. Find the value of k.
Answer/Explanation
Ans: If we rewrite the equation f(x) = 3 in the form ax2 + bx + c = 0, we have
logk (8x – 2x2) = 3
8x – 2x2 = k3
0 = 2x2 – 8x + k3
Hence the quadratic function q(x) = 2x2 – 8x + k3 has two equal roots.
Therefore we get
[ discriminant of q] = 0
(-8)2 – 4(2) (k3) = 0 [Δ = b2 – 4 ac]
64 – 8k3 = 0
8 – k3 = 0
k3 = 8
k = 2
Question 7 :
Let f(x) = 3x2 + 12x + 9, for x ∈ R.
(a) For the graph of f, find:
(i) the y-intercept;
(ii) the x-intercepts.
The function f can be written in the form f(x) = a(x – h)2 + k.
(b) Find the value of a, h and k.
(c) For the graph of f, write down:
(i) the coordinates of the vertex;
(ii) the equation of the axis of symmetry.
The graph of a function g is obtained from the graph of f by a reflectionin the x-axis, followed by a translation by the vector \(\binom{0}{4}.\)
(d) Find g(x), giving your answer in the form g(x) = px2 + qx + r.
Answer/Explanation
Ans: (a) (i) Evaluating f(x) = 3x2 + 12x + 9 for x = 0, we have
f(0) = 3(0)2 + 12(0) + 9
= 9
Hence the y-intercept is A(0, 9)
(ii) Solving the equation f(x) = 0 for x, we have
3x2 + 12x + 9 = 0
x2 + 4x + 3 = 0
(x + 3) (x + 1) = 0
x = -3, -1
Hence the x-intercepts are B(-3, 0) and C(-1, 0)
(b) Using the method of completing the square, we have
f(x) = 3x2 + 12x + 9
= 3 [x2 + 4x] +9
= 3(x+ 2)2 – 3
Hence we find a = 3, h = -2 and k = -3
(c) (i) (-2, -3)
(ii) x = -2
(d) We have
g(x) = -f(x) + 4
= -(3x2 + 12x + 9) + 4
= -3x2 – 12x – 5
Question 8:
The first two terms of an infinite geometric sequence, in order, are
3 log3x, 2 log3 x, where x > 0.
(a) Find the common ratio, r.
(b) Show that the sum of the infinite sequence is 9 log3 x.
The first three terms of an arithmetic sequence, in order, are
\(log_{3}x, log_{3}\frac{x}{3}, log_{3}\frac{x}{9}, \) where x > 0.
(c) Find the common difference d, giving your answer as an integer. Let S6 be the sum of the first 6 terms of the arithmetic sequence.
(d) Show that S6 = 6 log3 x – 15.
(d) Given that S6 is equal to one third of the sum of the infinite geometric sequence, find x, giving your answer in the form ap where a, p ∈ Z.
Answer/Explanation
Ans:
(a) The common ratio is
\(r = \frac{u_{2}}{u_{1}}\)
\( = \frac{2 log_{3}x}{3 log_{3}x}\)
\(= \frac{2}{3}\)
(b) Using the sum of an infinite geometric formula \(S_{\infty }=\frac{u_{1}}{1 – r},\) we get
\(S_{\infty }=\frac{3 log_{3}x}{1 – 2/3}\)
\(=\frac{3 log_{3}x}{1/3}\)
= 9 log3 x
(c) Using the properties of logarithms, we find
d = u2 – u1
\(= log_{3}\frac{x}{3}-log_{3}x\)
= (log3 x – log3 3) – log3 x
= -1
(d) Using the sum of n terms formula \(S_{n} = \frac{n}{2}(2u_{1}+(n-1)d)\) with n = 6, we get
\(S_{6} = \frac{6}{2}(2u_{1}+(6-1)d)\)
\( = \frac{6}{2}(2{log_{3}x}+(6-1)(-1))\)
\( = \frac{6}{2}(2{log_{3}x}-5)\)
= 6 log3 x – 15
Question 9:
Consider a function f. The line L1 with equation y = 2x – 1 is a tangent to the graph of f when x = 3.
(a) (i) Write down f'(3).
(ii) Find f(3).
Let g(x) = f(x2 – 1) and P be the point on the graph of g where x = 2.
(b) Show that the graph of g has a gradient of 8 at P.
Let L2 be the tangent to the graph f g at P. The line L1 intersects L2 at the point Q.
(c) Find the y-coordinates of Q.
Answer/Explanation
Ans:
(a) (i) The slope of the tangent line L1 at x = 3 is m = 2, hence
f ‘ (3) = 2
(ii) Since L1 and f have a common point at x = 3, we get
f(3) = 2(3) – 1
= 5
(b) We have
g ‘ (x) = f ‘(x2 – 1) . 2x
g ‘(2) = f ‘(22 – 1 ) . 2(2)
= f ‘ (3) . 4
= 2 (4)
= 8