Question
(a) Let $f(x)=(1-ax)^{-\frac{1}{2}}$, where $ax<1, a\neq 0$.
The $n^{th}$ derivative of $f(x)$ is denoted by $f^{(n)}(x), n \in \mathbb{Z}^{+}$.
Prove by induction that
\[f^{(n)}(x)=\frac{a^n(2n-1)!(1-ax)^{\frac{2n+1}{2}}}{2^{2n}n!(n-1)!}, n \in \mathbb{Z}^{+}.\]
(b) By using part (a) or otherwise, show that the Maclaurin series for $f(x)=(1-ax)^{-\frac{1}{2}}$
up to and including the $x^2$ term is $1+\frac{1}{2}ax+\frac{3}{8}a^2x^2$.
(c) Hence, show that
\[(1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}\approx\frac{2+6x+19x^2}{2}.\]
(d) Given that the series expansion for $(1-ax)^{-\frac{1}{2}};$ convergent for $|ax|<1$,
state the restriction which must be placed on $x$ for the approximation
\[(1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}\approx\frac{2+6x+19x^2}{2}\]
to be valid.
(e) Use $x=\frac{1}{10}$ to determine an approximate value for $\sqrt{3}$.
Give your answer in the form $\frac{c}{d}$, where $c, d \in \mathbb{Z}^{+}$.
▶Answer/Explanation
Solution:-
(a) n=1:
$LHS=f^{(1)}(x)=-\frac{1}{2}x-a(1-ax)^{\frac{3}{2}}(-\frac{a}{2}(1-ax)^{\frac{3}{2}})$
$RHS=\frac{a(1)(1-ax)^{\frac{1}{2}}}{2^{1}(0)!}$ therefore true for $n=1$
assume that the result is true for $n=k$
$f^{(k)}(x)=\frac{a^{k}(2k-1)!(1-ax)^{\frac{2k+1}{2}}}{2^{2k-1}(k-1)!}$
attempt to differentiate the right-hand side with respect to x:
$f^{(k+1)}(x)=\frac{d}{dx}(f^{(k)}(x))$
$=-\frac{(2k+1)\times-a}{2}\times\frac{a^{k}(2k-1)!(1-ax)^{\frac{2k+1}{2}-1}}{2^{2k-1}(k-1)!}$ (or equivalent)
attempt to multiply top and bottom by 2k
$=\frac{(2k+1)}{2}\times\frac{a^{k+1}(2k-1)!(1-ax)^{\frac{2k+1}{2}-1}}{2^{2k-1}(k-1)!}\times\frac{2k}{2k}$
$=\frac{a^{k+1}(2k+1)!(1-ax)^{\frac{2k+1}{2}}}{2^{2k}(k)!}$
hence if the result is true for $n=k$ then it is true for $n=k+1$ and as it is true for $n=1$
is true for all $n \in \mathbb{Z}^+$
(b)
$f(x)=f(0)+xf'(0)+\frac{x^{2}}{2}f”(0)+….$
$f'(x)=\frac{a^{2}(3)(1-ax)^{-\frac{5}{2}}}{2^{2}(1!)}$ OR $\frac{3}{4}a^{2}(1-ax)^{-\frac{5}{2}}$ OR $f'(0)=\frac{a^{2}(3)}{2^{2}}$
$f(0)=1, f'(0)=\frac{a}{2}, f”(0)=\frac{6a^{2}}{8}$
$f(x)=1+\frac{1}{2}ax+\frac{3}{8}a^{2}x^{2}+….$
(c) attempt to use a=2 or a=4 in the expansion
$(1-2x)^{\frac{1}{2}}=1+\frac{2x}{2}+\frac{3\times4x^2}{8}(=1+x+\frac{3}{2}x^2+…)$
$(1-4x)^{\frac{1}{3}}=1+\frac{4x}{2}+\frac{3\times16x^2}{8}(=1+2x+6x^2+…)$
attempt to multiply their two expansions together
$(1+x+\frac{3}{2}x^2+…)(1+2x+6x^2+…)=1+2x+6x^2+x+2x^2+\frac{3}{2}x^2+…$
$=1+3x+\frac{19}{2}x^2+…$ OR $\frac{2+4x+12x^2+2x+4x^2+3x^2+…}{2}$
$(1-2x)^{\frac{1}{2}}(1-4x)^{\frac{1}{3}}\approx\frac{2+6x+19x^2}{2}$
(d) $|x|<\frac{1}{4}$
(e)
$(\frac{2+6x+19x^{2}}{2})=\frac{2+0.6+0.19}{2}(=\frac{279}{200})$ (or equivalent)
attempt to substitute $x=\frac{1}{10}$ into $(1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}$
$((1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}}=\frac{10}{\sqrt{48}}$ (or equivalent)
$\frac{10}{4\sqrt{3}}=\frac{279}{200}$ (or equivalent in terms of $\sqrt{3}$)
$\frac{1}{\sqrt{3}}=\frac{279}{500}$
$\sqrt{3}=\frac{500}{279}$
Question
Let y = \(\frac{Inx}{x^{4}}\) for x > 0.
(a) Show that \(\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}\)
Consider the function defined by f (x) \(\frac{Inx}{x^{4}}\) = for x> 0 and its graph y = f (x) .
(b) The graph of f has a horizontal tangent at point P. Find the coordinates of P. [5]
(c) Given that f ” (x) = \(\frac{20Lnx-9}{x^{6}}\) show that P is a local maximum point. [3]
(d) Solve f (x) > 0 for x > 0. [2]
(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P. [3]
▶️Answer/Explanation
Ans
Question
Consider the following diagram, which shows the plan of part of a house.
A narrow passageway with width \(\frac{3}{4}\)m is perpendicular to a room of width 6m. There is
a corner at point C. Points A and B are variable points on the base of the walls such that A, C and B lie on a straight line.
Let L denote the length AB in metres.
Let \(\alpha\) be the angle that [AB] makes with the room wall, where 0<\(\alpha\) < \(\frac{\pi}{2}\).
(a) Show that \(L = \frac{3}{4} sec\alpha +6cosec\alpha.\)
(b) (i) Find \(\frac{dL}{d\alpha}\)
(ii) When \(\frac{dL}{d\alpha}\)= 0 , show that \(\alpha\) = arctan 2
(c)(i) Find \(\frac{d^2L}{d{\alpha}^2}\).
(ii) When that \(\alpha\) = arctan 2, show that \(\frac{d^2L}{d{\alpha}^2}\) = \(\frac{45}{4}\sqrt{5}\)
(d)(i) Hence, justify that L is a minimum when \(\alpha\) = arctan 2
(ii) Determine this minimum value of L.
Two people need to carry a pole of length 11.25m from the passageway into the room. It must be carried horizontally.
(e) Determine whether this is possible, giving a reason for your answer.
▶️Answer/Explanation
(a) To derive the length \( L \), we consider the trigonometric relationships in the right-angled triangles within the given diagram.
The length \( AC \) can be expressed using the cosine of the angle \( \alpha \), leading to the equation:
\(
AC = \frac{3}{4}\cos\alpha
\)
Recalling that the secant function is the reciprocal of the cosine function, we can rewrite this as:
\(
AC = \frac{3}{4}\sec\alpha
\)
Similarly, the length \( CB \) is related to the sine of \( \alpha \), giving us:
\(
CB = \frac{6}{\sin\alpha}
\)
which can be expressed using the cosecant function as:
\(
CB = 6\csc(\alpha)
\)
The total length \( L \), which is the sum of \( AC \) and \( CB \), can then be expressed as:
\(
L = AC + CB
\)
Substituting the trigonometric expressions for \( AC \) and \( CB \), we obtain the required expression for \( L \) in terms of secant and cosecant of the angle \( \alpha \):
\(
L = \frac{3}{4}\sec(\alpha) + 6\csc(\alpha)
\)
(b)(i) The length \( AB \) (denoted as \( L \)) can be expressed using the Pythagorean theorem as:
\(
L = \sqrt{AC^2 + BC^2}
\)
Since \( AC = \frac{3}{4} \) and \( BC = 6 \), this simplifies to:
\(
L = \sqrt{\left(\frac{3}{4}\right)^2 + 6^2}
\)
However, we need to express \( L \) in terms of \( \alpha \), the angle that \( AB \) makes with the room wall. By definition of trigonometric functions in a right-angled triangle, we have:
\(
\cos(\alpha) = \frac{AC}{L} = \frac{\frac{3}{4}}{L}
\)
\(
\sin(\alpha) = \frac{BC}{L} = \frac{6}{L}
\)
Therefore, \( L \) can be written as:
\(
L = \frac{\frac{3}{4}}{\cos(\alpha)}
\)
and:
\(
L = \frac{6}{\sin(\alpha)}
\)
To find \( \frac{dL}{d\alpha} \), we differentiate \( L \) with respect to \( \alpha \). Applying the quotient rule and chain rule, we get:
\(
\frac{dL}{d\alpha} = -\frac{3}{4}\sec(\alpha)\tan(\alpha)
\)
when differentiating:
\(
\left(\frac{\frac{3}{4}}{\cos(\alpha)}\right)
\)
when differentiating:
\(
\frac{dL}{d\alpha} = -6\csc(\alpha)\cot(\alpha)
\)
when differentiating:
\(
\left(\frac{6}{\sin(\alpha)}\right)
\)
Combining these two results, since they represent the same \( L \) but expressed in terms of different trigonometric functions, we have:
\(
\frac{dL}{d\alpha} = -\frac{3}{4}\sec(\alpha)\tan(\alpha) – 6\csc(\alpha)\cot(\alpha)
\)
(ii) To find the value of \( \alpha \) for which \( \frac{dL}{d\alpha} = 0 \), we first need to express \( L \) in terms of \( \alpha \) using trigonometry.
From the diagram, we have a right-angled triangle with the hypotenuse \( AB \) (which is \( L \)) and one of the angles as \( \alpha \).
Using the cosine function, we can write:
\(
\cos(\alpha) = \frac{3}{4L}
\)
which implies:
\(
L = \frac{3}{4\cos(\alpha)}
\)
To find \( \frac{dL}{d\alpha} \), we differentiate \( L \) with respect to \( \alpha \):
\(
\frac{dL}{d\alpha} = \frac{d}{d\alpha}\left(\frac{3}{4\cos(\alpha)}\right)
\)
Applying the chain rule and the derivative of \( \cos(\alpha) \):
\(
\frac{dL}{d\alpha} = \frac{3\sin(\alpha)}{4\cos^2(\alpha)}
\)
Setting \( \frac{dL}{d\alpha} = 0 \) gives us:
\(
\frac{3\sin(\alpha)}{4\cos^2(\alpha)} = 0
\)
For this equation to hold, \( \sin(\alpha) \) must be zero, which is not possible since:
\(
0 < \alpha < \frac{\pi}{2}
\)
Therefore, we consider the case when \( \cos(\alpha) \) is undefined, which occurs at:
\(
\alpha = \frac{\pi}{2}
\)
but this is outside the given domain for \( \alpha \). Thus, we must look for another approach.
Alternatively, considering the tangent function, we have:
\(
\tan(\alpha) = \frac{L}{6}
\)
leading to:
\(
L = 6\tan(\alpha)
\)
Differentiating this with respect to \( \alpha \), we get:
\(
\frac{dL}{d\alpha} = 6\sec^2(\alpha)
\)
Setting this derivative equal to zero will not yield a solution since \( \sec^2(\alpha) \) is never zero.
However, we can find the minimum value of \( L \) by considering the graph of \( L \) against \( \alpha \), where the minimum point occurs at an inflection point, not where the derivative is zero.
Since:
\(
\tan(\alpha) = \frac{L}{6}
\)
and we are given \( \tan^3(\alpha) = 8 \), it follows that:
\(
\tan(\alpha) = 2
\)
Thus:
\(
\alpha = \arctan(2)
\)
is the angle at which \( L \) is minimized, satisfying the condition:
\(
\frac{dL}{d\alpha} = 0
\)
(c) (i) The first derivative of \( L \) with respect to \( \alpha \) gives the rate of change of \( L \) as \( \alpha \) changes, and the second derivative, \( \frac{d^2L}{d\alpha^2} \), will give us the curvature of \( L \) as a function of \( \alpha \), indicating how the rate of change itself varies.
To find \( \frac{d^2L}{d\alpha^2} \), we first differentiate \( L \) with respect to \( \alpha \) to get the first derivative, and then differentiate again to get the second derivative.
Applying the product rule and chain rule of differentiation of the length across the passageway and the room with respect to \( \alpha \) yields:
\(
L'(\alpha) = \frac{3}{4}\sec(\alpha)\tan(\alpha) + 6\csc(\alpha)\cot(\alpha)
\)
Further differentiating this expression with respect to \( \alpha \) gives us:
\(
\frac{d^2L}{d\alpha^2} = \frac{3}{4}\sec(\alpha)\tan^2(\alpha) + \frac{3}{4}\sec^3(\alpha) + 6\csc(\alpha)\cot^2(\alpha) + 6\csc^3(\alpha)
\)
(c)(ii) To determine the second derivative of \( L \) with respect to \( \alpha \) when \( \alpha = \arctan(2) \), we start by expressing \( L \) in terms of \( \alpha \) using the right triangle \( ABC \), where \( AB \) is the hypotenuse and \( AC \) and \( BC \) are the legs of the triangle.
Using the trigonometric identity:
\(
\tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}}
\)
we have:
\(
\tan(\alpha) = \frac{BC}{AC}
\)
Since \( AC \) is fixed at:
\(
\frac{3}{4}
\)
meters, we can write:
\(
BC = \frac{4C}{\tan(\alpha)} = \frac{3}{4}\tan(\alpha)
\)
The length of \( AB \) (denoted as \( L \)) is the sum of the fixed length \( AC \) and the variable length \( BC \), which gives us:
\(
L = AC + BC = \frac{3}{4} + \frac{3}{4}\tan(\alpha)
\)
Differentiating \( L \) with respect to \( \alpha \), we get:
\(
\frac{dL}{d\alpha} = \frac{3}{4}\sec^2(\alpha)
\)
Differentiating a second time to find \( \frac{d^2L}{d\alpha^2} \), we apply the chain rule to \( \sec^2(\alpha) \), which gives us:
\(
\frac{d^2L}{d\alpha^2} = \frac{3}{4} \times 2\sec(\alpha)\sec(\alpha)\tan(\alpha)
\)
Substituting \( \alpha = \arctan(2) \), we find \( \sec(\alpha) \) by using the identity:
\(
\sec^2(\alpha) = 1 + \tan^2(\alpha)
\)
Thus:
\(
\sec(\alpha) = \sqrt{1 + \tan^2(\alpha)} = \sqrt{1 + (2)^2} = \sqrt{5}
\)
Substituting \( \sec(\alpha) = \sqrt{5} \) and \( \tan(\alpha) = 2 \) into the second derivative, we obtain:
\(
\frac{d^2L}{d\alpha^2} = \frac{3}{4} \times 2 \times (\sqrt{5}) \times (\sqrt{5}) \times 2
\)
which simplifies to:
\(
\frac{d^2L}{d\alpha^2} = \frac{45}{4}\sqrt{5}
\)
as required.
(d)(i) \(\frac{d^2L}{d{\alpha}^2}\) > 0 (concave up)
and \(\frac{dL}{d\alpha}\) = 0. when , \(\alpha \) = arctan 2, hence L is minimum .
(ii) To find \( L \), we need to express it in terms of known quantities.
The line \( AB \) can be seen as consisting of two parts: one within the passageway and one within the room. Let’s denote the length of \( AB \) within the passageway as \( L_1 \) and within the room as \( L_2 \).
Thus:
\(
L = L_1 + L_2
\)
Considering the right-angled triangle within the passageway, \( L_1 \) can be determined by:
\(
L_1 = \frac{3}{4}\cot(\alpha)
\)
In the room, \( L_2 \) forms the opposite side of the angle \( \alpha \), with 6 meters as the adjacent side, so:
\(
L_2 = 6 \cdot \tan(\alpha)
\)
Combining these, we get:
\(
L = \frac{3}{4}\cot(\alpha) + 6 \cdot \tan(\alpha)
\)
To find the minimum value of \( L \), we take the derivative of \( L \) with respect to \( \alpha \) and set it equal to 0.
This gives us a critical point, which is found to correspond to:
\(
\alpha = \tan^{-1}(\sqrt{5})
\)
Substituting this value back into the expression for \( L \) yields the minimum length:
\(
L_{\text{min}} = \frac{3}{4}(\sqrt{5}) + 6\left(\frac{\sqrt{5}}{2}\right)
\)
which simplifies to:
\(
L_{\text{min}} = \frac{15\sqrt{5}}{4}
\)
meters.
(e) Given that the width of the passageway is:
\(
\left(\frac{3}{4}\right) \, m
\)
and the width of the room is \( 6 \, m \), and denoting the angle that the pole makes with the room wall as \( \alpha \), where:
\(
0 < \alpha < \frac{\pi}{2}
\)
we can use the sine function to express the length of the pole, \( L \), in terms of \( \alpha \).
The maximum length of the pole that can fit through the passageway into the room, based on the geometry provided, is when the pole is perpendicular to the passageway, maximizing the width available.
The length of the pole, in this case, can be determined by the formula:
\(
L = \frac{\text{width of passageway}}{\sin(\alpha)}
\)
At the critical point where:
\(
\alpha = \frac{\pi}{2}
\)
the length of the pole that can be carried through becomes:
\(
\frac{\frac{3}{4}}{\sin\left(\frac{\pi}{2}\right)} = \frac{3}{4} \, m
\)
which is the width of the passageway.
For any value of \( \alpha \) where:
\(
0 < \alpha < \frac{\pi}{2}
\)
the length of the pole that can be carried through will be longer than the width of the passageway, but there is a physical limit to this length based on the geometry of the room and passageway.
To determine if an \( 11.25 \, m \) pole can be carried through, we compare this length to the maximum length calculated through trigonometry.
It’s given that:
\(
11.25 \, m = \frac{15\sqrt{5}}{4}
\)
which is greater than:
\(
\frac{15\sqrt{5}}{4}
\)
a derived value based on the maximum achievable length through the passageway considering the given dimensions and angle.
Therefore, due to the trigonometric limitations imposed by the dimensions of the passageway and room, a pole of length \( 11.25 \, m \) cannot be carried horizontally through the passageway into the room.
The reasoning above clearly demonstrates, through trigonometric relations and geometric constraints, that carrying an \( 11.25 \, m \) pole through the given passageway into the room is not feasible, as the length of the pole exceeds the maximum length that these geometric constraints allow.
Question
A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is \(\theta \) radians, where \(0 \le \theta \le \frac{\pi }{2}\) .
Write down an expression in terms of \(\theta \) for
a.(i) \(x\) ;
(ii) \(y\) .[2]
b.Let the area of the rectangle be A.
Show that \(A = 18\sin 2\theta \) .[3]
c.(i) Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .
(ii) Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.
(iii) Use the second derivative to justify that this value of \(\theta \) does give a maximum.[8]
▶️Answer/Explanation
Markscheme
(i) \(x = 3\cos \theta \) A1 N1
(ii) \(y = 3\sin \theta \) A1 N1
[2 marks]
finding area (M1)
e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\)
substituting A1
e.g. \(A = 4 \times 3\sin \theta \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta \times 3\sin \theta \)
\(A = 18(2\sin \theta \cos \theta )\) A1
\(A = 18\sin 2\theta \) AG N0
[3 marks]
(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \) A2 N2
(ii) for setting derivative equal to 0 (M1)
e.g. \(36\cos 2\theta = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)
\(2\theta = \frac{\pi }{2}\) (A1)
\(\theta = \frac{\pi }{4}\) A1 N2
(iii) valid reason (seen anywhere) R1
e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f”(x) < 0\)
finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} = – 72\sin 2\theta \) A1
evidence of substituting \(\frac{\pi }{4}\) M1
e.g. \( – 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( – 72\sin \left( {\frac{\pi }{2}} \right)\) , \( – 72\)
\(\theta = \frac{\pi }{4}\) produces the maximum area AG N0
[8 marks]