Question 1:
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 = \(\theta\), where \(\theta\) is measured in radians.
The length of arc PQR is 10cm.
(a) Find the perimeter of the shaded sector.
(b) Find \(\theta\).
(c) Find the area of the shaded sector
▶️Answer/Explanation
(a) Perimeter = arc+ 2\(\times\) radius
= 10+2\(\times\) 4
= 18 cm
(b) 10 = 4\(\theta\)
\(\theta\) = \(\frac{10}{4}\)
\(\frac{5}{2}\) = 2.5
(c) Area = \(\frac{1}{2} (\frac{10}{4}) {(4)}^2 = 1.25\times 16\)
= 20 \({cm}^2\)
Question 2:
A function f is defined by f(x) = \(1 – \frac{1}{x-2}\) , where x \(\epsilon\) R , x\(\neq\) 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
(a)(i) x=2
(ii) y=1
(b)(i) \(0,\frac{3}{2}\)
(ii) (3,0)
(c) 
Question 3:
Events A and B are such that P(A) = 0.4, P(A|B) = 0.25 and P(A\(\cup\)B) = 0.55. Find P(B).
▶️Answer/Explanation
P(A\(\cup\)B) = P(A)+P(B) – (A\(\cap\)B)
Substituting values in the above expression:
0.55 = 0.4 + P(B) – (A\(\cap\)B) ——(1)
Now , P(A|B) = \(\frac{ (A\cap B)}{P(B)}\)
0.25 = \(\frac{ (A\cap B)}{P(B)}\)
\(\Rightarrow (A\cap B)\) = 0.25 P(B)
Putting the values in in (1) :
0.55 = 0.4 + P(B) – 0.25 P(B)
Upon solving , we get :
P(B) = \(\frac{15}{75}\) = 0.2
Question 4:
The following diagram shows part of the graph of y = \(\frac{x}{x^2+2}\) for x\(\geq\) 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
\(A = \int_{0}^{c} \frac{x}{x^2+2} dx \)
Solving we get:
\(\left [ \frac{1}{2} ln (x^2+2)\right ]_{0}^{c}\)
Substituting limits , we get :
\(\frac{1}{2} ln (c^2+2) -\frac{1}{2} ln2 = ln 3\)
Solving further for c , we get :
\(\frac{c^2+2}{2} = 9\)
\(\Rightarrow c= 4\)
Question 5:
The functions f and g are defined for x \(\epsilon\) R by :
f(x) = ax+b , where a, b \(\epsilon\) 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
\({[f(x)]}^2 + f(x) + 3\)
= \({(ax+b)}^2 + ax + b + 3\)
Given : (g \(\circ\) f )(x) = \(4x^2 -14x +15\)
So , \(a^2x^2+ 2abx + b^2 + ax + b+ 3 =4x^2 -14x +15\)
Equating coefficients:
\(a^2x^2 = 4x^2 or a^2 = 4 \)
2abx + ax = -14x
2ab+a = -14
\(b^2 + b+3 = 15\)
\(a = \pm 2\)
and b = -4 ,3
f(x) = 2x+4 ( a=2 , b=-4)
f(x) = -2x+3 (a = -2 , b=3)
Question 6:
A continuous random variable X has probability density function f defined by:
\(\left\{\begin{matrix}
\frac{1}{2a} & a\leq x \leq 3a\\
0 & otherwise
\end{matrix}\right.\)
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
(a) E (X ) =2a (by symmetry)
(b) Var(X) = \(E{(X-E(X))}^2\)
Var(X) = \(\int_a^{3a} \frac{{(x-2a)}^2}{2a} dx\)
\(\left [\frac{{(x-2a)}^3}{6a} \right ]_a^{3a}\)
= \(\frac{a^3 – (-a^3)}{6a}\)
=\(\frac{a^2}{3}\)
Question 7:
Use mathematical induction to prove that \(\sum_{r=1}^{n} \frac{r}{(r+1)!} = 1- \frac{1}{(n+1)!}\) for all integers n \(\geq\) 1.
▶️Answer/Explanation
Let P(n) be the proposition that \(\sum_{r=1}^{n} \frac{r}{(r+1)!} = 1- \frac{1}{(n+1)!}\) for all integers , \(n\geq 1\).
Considering P (1) :
LHS = \(\frac{1}{2}\) and RHS = \(\frac{1}{2}\) and so P(1) is true.
Assume P(k) is true \(\sum_{r=1}^{k} \frac{r}{(r+1)!} = 1- \frac{1}{(k+1)!}\)
Considering P(k+1) :
\(\sum_{r=1}^{k+1} \frac{r}{(r+1)!} = \sum_{r=1}^{k} \frac{r}{(r+1)!} + \frac{k+1}{((k+1)+1)!}\)
= 1 – \(\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}\)
= 1 – \(\frac{(k+2)-(k-1)}{(k+2)!}\)
= 1- \(\frac{1}{(k+2)!}\)
= 1 – \(\frac{1}{((k+1)+1)!}\)
P (k+1) is true whenever P(k) is true and P (1) is true, so P( n) is true
(for all integers, \(n\geq 1\))
Question 8:
The functions f and g are defined by:
f(x) = cos x , \( 0 \leq x \leq \frac{\pi}{2}\)
g(x) = tan x , \( 0 \leq x \leq \frac{\pi}{2}\)
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 the tangent to the curve y = g(x) at P intersect at right angles.
(c) Find the value of sink . Give your answer in the form \(\frac{a+\sqrt{b}}{c}\) where a , c \(\epsilon\) Z , and c \(\epsilon Z^{+}\).
▶️Answer/Explanation
(a) \(cos k = \frac{sin k}{cos k}\)
\(cos^2 k = sin k \)
(b) f'(k) = -sin k and g'(x) = \(sec^2k\)
EITHER
f'(x)g'(x) = \(-\frac{sin k}{{cos}^2 k}\)
\({cos}^2 k = sin k \Rightarrow f'(x)g'(x) = -\frac{sin k}{sin k} = -1\)
OR
\(g'(x) = \frac{1}{cos^2 x}\)
\(cos^2 x = sin k \Rightarrow g'(x) = \frac{1}{sin k} = -\frac{1}{f'(k)}\)
Therefore, the two tangents intersect at right angles at P.
(c) \(1 – sin^2 k = sin k \)(from part a)
\(sin^2 k + sin k -1 =0\)
Solving for sin k :
\(sin k = \frac{-1 \pm \sqrt{1^2 – 4(1)(-1)}}{2}\)
For 0 < k < \(\frac{\pi}{2}\) , sin k >0
So, sin k = \(\frac{-1+\sqrt{5}}{2}\)
(a = -1 , b= 5 , c = 2)
Question 9:
The following diagram shows parallelogram OABC with \(\underset{OA}{\rightarrow}\) = a , \(\underset{OC}{\rightarrow}\) = c and |c| = 2|a| . where |a| \(\neq\) 0.
The angle between \(\underset{OA}{\rightarrow}\) and \(\underset{OC}{\rightarrow}\) is \(\theta\) , where \( 0 < \theta < \pi\).
Point M is on [AB] such that \(\underset{AM}{\rightarrow}\) = k \(\underset{AB}{\rightarrow}\) , where \( 0 \leq k \leq 1)\). and \(\underset{OM}{\rightarrow}\) \(\cdot\) \(\underset{MC}{\rightarrow}\) = 0.
(a) Express \(\underset{OM}{\rightarrow}\) and \(\underset{MC}{\rightarrow}\) in terms of a and c.
(b) Hence, use a vector method to show that \({|a|}^2 (1-2k)(2cos\theta – (1-2k)) = 0\).
(c) Find the range of values for \(\theta\) such that there are two possible positions for M.
▶️Answer/Explanation
(a) \(\underset{OM}{\rightarrow}\) = a + kc
\(\underset{MC}{\rightarrow}\) = (1-k)c – a
(b) Expanding \(\underset{OM}{\rightarrow}\) \(\cdot\) \(\underset{MC}{\rightarrow}\) = (a+kc) \(\cdot\) ((1-k)c-a)
= (1-2k)(a\(\cdot\)c) – \({|a|}^2\) + k(1-k)\(|c|^2\)
Uses: |c| = 2|a| and (a\(\cdot\)c) = 2\({|a|}^{2}\) cos \(\theta\)
= 2(1-2k) \({|a|}^2\) cos\(\theta\) – \({|a|}^2\) +4k(1-k) \({|a|}^2\)
= 2(1-2k) \({|a|}^2\) cos\(\theta\) -\({(1-2k)}^2\)\({|a|}^2\)
= \({|a|}^2\) (1-2k)(2cos\(\theta\) – (1-2k)) = 0
(c) Solving \({|a|}^2\) (1-2k)(2cos\(\theta\) – (1-2k)) = 0 for k :
k = \(\frac{1}{2}\) or k = \(\frac{1}{2} – cos\theta\) (\({|a|}^2\) > 0)
As \(0 \leq k \leq 1 , 0 \leq \frac{1}{2} – cos\theta \leq 1\)
\(-\frac{1}{2} \leq cos\theta \leq \frac{1}{2}\)
\(\frac{\pi}{3} \leq \theta \leq \frac{2\pi}{3} , \theta \neq \frac{\pi}{2}\)
\((\theta = \frac{\pi}{2}\) corresponds to only one possible position for M when k = \(\frac{1}{2}\))
Question 10:
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) Hence or otherwise, find the y-coordinate of R such that A is a maximum.
▶️Answer/Explanation
(a) \(y^2 = 9-x^2\) OR \(y = \pm \sqrt{9-x^2}\)
(Since , y> 0 )\(\Rightarrow y = \sqrt{9-x^2}\)
(b) b=2y = 2\(\sqrt{9-x^2}\) or h = x+3
Substituting the values into \(A = \frac{1}{2}bh\)
\(A = \sqrt{9-x^2} (x+3)\)
= \(\frac{2(x+3)\sqrt{9-x^2}}{2}\)
= \(x\sqrt{9-x^2} +3\sqrt{9-x^2}\)
(c) Using chain rule:
\(\frac{d}{dx} \sqrt{9-x^2}\) =
\(\frac{dA}{dx} = \sqrt{9-x^2} +(3+x) (\frac{1}{2}) {(9-x^2)}^{\frac{1}{2}} (-2x)\)
\(= \sqrt{9-x^2} – \frac{x^2+3x}{\sqrt{9-x^2}}\)
\(\frac{dA}{dx} = \frac{9-x^2}{\sqrt{9-x^2}}-\frac{x^2+3x}{\sqrt{9-x^2}}\)
\(\frac{dA}{dx} = \frac{9-3x-2x^2}{\sqrt{9-x^2}}\)
(d) \(\frac{dA}{dx} = \frac{9-3x-2x^2}{\sqrt{9-x^2}} = 0\)
Solving for x:
\( 9 – 3x- 2x^2 =0\)
-(2x-3)(x+3) = 0
x = \(\frac{3}{2}\)
Substituting the value of x into y:
y = \(\sqrt{9 – {(\frac{3}{2})}^2}\)
= \(-\sqrt{6.75}\)
Question 11:
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\(\epsilon\)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 \(1+5w+10w^2+12w^3 = 0\) where w\(\epsilon\)C.
(d) Consider the equation \(z^2 = 2z^{*}\) where z\(\epsilon\)C, z\(\neq\) 0.
By expressing z in the form a + bi, find the roots of the equation.
▶️Answer/Explanation
(a) |u| = \(\sqrt{{(-1)}^2+{(\sqrt{3})}^2} \)
= 2
Reference angle = \(\frac{\pi}{3}\) OR arg u = \(\pi – tan^{-1} (\sqrt{3})\)
Upon solving :
= \(\pi – \frac{\pi}{3}\)
= \(\frac{2\pi}{3}\) and so \(u = 2e^{i\frac{2\pi}{3}}\)
(b)(i) \(u^n\epsilon R \Rightarrow \frac{2n\pi}{3} = k\pi (k\epsilon z)\)
n=3
(ii) Substituting the value(must be a multiple of 3) of n in \(u^n\)
\(u^3 = 2^3 cos2\pi\)
= 8
(c)(i) Using sum of roots :
Let z=c be the real root
\(((-1+\sqrt{3}i) + (-1-\sqrt{3}i))c = 5\)
-2+c=-5
c=-3(so z=-3 is a root)
(ii) Factorizing into a product of linear factor and a quadratic factor:
\(1+5w+10w^2+12w^3 = (3w+1)(4w^2+2w+1)\)
\(w = -\frac{1}{3} , \frac{1}{-1\pm\sqrt{3}i} \)
(d) \({(a+bi)}^2 = 2(a-ib)\)
Expanding and equating real and imaginary parts:
\(a^2-b^2+2abi = 2a – 2bi\)
\(a^2-b^2 =2a\) and \(2ab=-2b\)
Finding the values of a and b:
2b(a+1) = 0
\(b = 0 \Rightarrow a^2 = 2a \Rightarrow a=2\) (Real root)
\(a=-1 \Rightarrow 1-b^2 = -2 \Rightarrow b = \pm\sqrt{3}\) (complex roots \(-1\pm\sqrt{3}i)\)
Question 12:
(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 \(x\geq 0\).
The curve intersects the x-axis at \(x_1,x_2,x_3……\)
The nth x- intercept of the curve \(x_n\) is given by \(x_n = \frac{{(2n-1)}^2 {\pi}^2}{4}\) where \(n \epsilon Z^{+}\).
(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 \(k n \pi\) where \(k \epsilon Z^{+}\).
(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
(a) Let \(t = \sqrt{x}\)
\(t^2 = x \Rightarrow 2t dt = dx\)
So, \(\int cos \sqrt{x} dx = 2\int t cos tdt\)
Integrating by parts:
u= 2t , dv cos t dt , du =2dt , v= sin t
\(2\int t cos t dt = 2t sin t – 2\int sin t dt \)
= 2t sin t + cos t +C
Substituting \(t =\sqrt{x} \)
= \(\int cos \sqrt{x} dx = 2\sqrt{x} sin \sqrt{x} +2cos\sqrt{x} +C\)
(b) \(x_{n+1} = \frac{{(2(n+1)-1)}^2 {\pi}^2}{4} = \frac{{(2n+1)}^2 {\pi}^2}{4}\)
(c) Area of \(R_n = |\int_{x_n}^{x_{n+1}} cos \sqrt{x} dx | \)
= \(|[2\sqrt{x} sin\sqrt{x} + 2cos \sqrt{x}]_{\frac{{(2n-1)}^2 {\pi}^2}{4}}^{\frac{{(2n+1)}^2 {\pi}^2}{4}}| \)
Substituting limits into integrated expression :
\(2| \frac{(2n+1)\pi}{2} \times sin \frac{(2n+1)\pi}{2}+ cos \frac{(2n+1)\pi}{2} – (\frac{(2n-1)\pi}{2} \times sin \frac{(2n-1)\pi}{2} + cos\frac{(2n-1)\pi}{2})|\)
Upon solving , we get:
= \(4n\pi\)
(d) d = \(R_{n+1}\) – \(R_n\)
= \(4(n+1) \pi – 4n\pi\)
= \(4\pi\)
OR
An arithmetic sequence is of the form \(u_n = dn+c\)
Comparing \(u_n = dn+c\) and \(R_n = 4n\pi\)
\(d = 4\pi , c = 0\)
So the areas of the regions form an arithmetic sequence.