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Edexcel A Level Pure Mathematics 1 (Oct 2020)
Mark Scheme Legend
- M1: Method mark
- A1: Accuracy mark
- B1: Unconditional accuracy mark
- dM1: Dependent method mark
- ft: Follow through
- awrt: Answers which round to
Table of Contents
- Question 1 (Binomial Expansion)
- Question 2 (Logarithms)
- Question 3 (Vectors)
- Question 4 (Functions)
- Question 5 (Sequences)
- Question 6 (Trigonometric Modelling)
- Question 7 (Inequalities & Regions)
- Question 8 (Differential Equations)
- Question 9 (Differentiation)
- Question 10 (Integration by Substitution)
- Question 11 (Circles & Trigonometry)
- Question 12 (Trigonometric Proof & Equations)
- Question 13 (Periodic Sequences)
- Question 14 (Connected Rates of Change)
- Question 15 (Implicit Differentiation)
- Question 16 (Proof by Contradiction)
Question 1 (5 marks)
(a) Find the first four terms, in ascending powers of \( x \), of the binomial expansion of
\[ (1 + 8x)^{\frac{1}{2}} \]giving each term in simplest form.
(3)
(b) Explain how you could use \( x = \frac{1}{32} \) in the expansion to find an approximation for \( \sqrt{5} \).
There is no need to carry out the calculation.
(2)
Worked Solution
Step 1: Binomial Expansion (Part a)
Why we do this: We use the binomial expansion formula \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \) where \( n = \frac{1}{2} \) and we replace \( x \) with \( 8x \).
Step 2: Connecting to \( \sqrt{5} \) (Part b)
Why we do this: We need to show how substituting \( x = \frac{1}{32} \) into the left-hand side \( (1+8x)^{1/2} \) relates to \( \sqrt{5} \).
Substitute \( x = \frac{1}{32} \) into \( (1 + 8x)^{\frac{1}{2}} \):
\[ \left(1 + 8\left(\frac{1}{32}\right)\right)^{\frac{1}{2}} = \left(1 + \frac{1}{4}\right)^{\frac{1}{2}} = \left(\frac{5}{4}\right)^{\frac{1}{2}} = \frac{\sqrt{5}}{\sqrt{4}} = \frac{\sqrt{5}}{2} \]Step 3: Conclusion
What this tells us: Since the expansion approximates \( \frac{\sqrt{5}}{2} \), we must multiply the result of the expansion by 2 to get \( \sqrt{5} \).
Substitute \( x = \frac{1}{32} \) into the expansion \( 1 + 4x – 8x^2 + 32x^3 \), calculate the result, and multiply by 2.
✓ (A1) Explanation complete
Final Answer:
(a) \( 1 + 4x – 8x^2 + 32x^3 \)
(b) Substitute \( x = \frac{1}{32} \) into the expansion and multiply the result by 2.
Question 2 (3 marks)
By taking logarithms of both sides, solve the equation
\[ 4^{3p-1} = 5^{210} \]giving the value of \( p \) to one decimal place.
Worked Solution
Step 1: Take Logarithms
Why we do this: To solve for an exponent, we take logarithms (base 10 or natural log) of both sides. This allows us to use the power law \( \log(a^b) = b \log a \).
✓ (M1) Power law applied correctly
Step 2: Solve for \( p \)
Method: Rearrange the equation to isolate \( p \).
✓ (dM1) Valid method to isolate p
Step 3: Calculation
Using calculator:
\[ \frac{210 \log 5}{\log 4} \approx 243.832 \] \[ 3p \approx 244.832 \] \[ p \approx 81.61 \]Rounding to one decimal place:
\[ p = 81.6 \]Final Answer:
\[ p = 81.6 \]Question 3 (4 marks)
Relative to a fixed origin \( O \)
- point \( A \) has position vector \( 2\mathbf{i} + 5\mathbf{j} – 6\mathbf{k} \)
- point \( B \) has position vector \( 3\mathbf{i} – 3\mathbf{j} – 4\mathbf{k} \)
- point \( C \) has position vector \( 2\mathbf{i} – 16\mathbf{j} + 4\mathbf{k} \)
(a) Find \( \vec{AB} \)
(2)
(b) Show that quadrilateral \( OABC \) is a trapezium, giving reasons for your answer.
(2)
Worked Solution
Step 1: Calculate vector AB (Part a)
Formula: \( \vec{AB} = \mathbf{b} – \mathbf{a} \)
Step 2: Identify Trapezium Property (Part b)
Definition: A trapezium is a quadrilateral with at least one pair of parallel sides. We need to check if any pair of vectors from \( \vec{OA}, \vec{AB}, \vec{BC}, \vec{OC} \) are parallel. Parallel vectors are scalar multiples of each other.
We have \( \vec{AB} = \mathbf{i} – 8\mathbf{j} + 2\mathbf{k} \).
Let’s find \( \vec{OC} \) (position vector of C):
\[ \vec{OC} = 2\mathbf{i} – 16\mathbf{j} + 4\mathbf{k} \]Compare \( \vec{OC} \) and \( \vec{AB} \):
\[ \vec{OC} = 2(\mathbf{i} – 8\mathbf{j} + 2\mathbf{k}) \] \[ \vec{OC} = 2\vec{AB} \]Step 3: Conclusion
Since \( \vec{OC} = 2\vec{AB} \), the vector \( \vec{OC} \) is parallel to \( \vec{AB} \).
Since the quadrilateral \( OABC \) has a pair of parallel sides (\( OC \) and \( AB \)), it is a trapezium.
Final Answer:
(a) \( \mathbf{i} – 8\mathbf{j} + 2\mathbf{k} \)
(b) \( \vec{OC} = 2\vec{AB} \), therefore \( OC \) is parallel to \( AB \), so \( OABC \) is a trapezium.
Question 4 (5 marks)
The function \( f \) is defined by
\[ f(x) = \frac{3x-7}{x-2} \quad x \in \mathbb{R}, x \neq 2 \](a) Find \( f^{-1}(7) \)
(2)
(b) Show that \( ff(x) = \frac{ax+b}{x-3} \) where \( a \) and \( b \) are integers to be found.
(3)
Worked Solution
Step 1: Finding \( f^{-1}(7) \) (Part a)
Strategy: We can set \( f(x) = 7 \) and solve for \( x \), because \( f(x) = y \iff x = f^{-1}(y) \).
Step 2: Finding composite function \( ff(x) \) (Part b)
Method: Substitute \( f(x) \) into itself: replace every ‘x’ in the expression with \( \frac{3x-7}{x-2} \).
Step 3: Simplifying the fraction
Technique: Multiply numerator and denominator by \( (x-2) \) to clear the fractions.
Final Answer:
(a) \( \frac{7}{4} \)
(b) \( ff(x) = \frac{2x – 7}{x – 3} \) (so \( a=2, b=-7 \))
Question 5 (6 marks)
A car has six forward gears.
The fastest speed of the car
- in 1st gear is 28 km h\(^{-1}\)
- in 6th gear is 115 km h\(^{-1}\)
Given that the fastest speed of the car in successive gears is modelled by an arithmetic sequence,
(a) find the fastest speed of the car in 3rd gear.
(3)
Given that the fastest speed of the car in successive gears is modelled by a geometric sequence,
(b) find the fastest speed of the car in 5th gear.
(3)
Worked Solution
Step 1: Arithmetic Model (Part a)
Formula: \( n \)-th term of arithmetic sequence is \( u_n = a + (n-1)d \).
We know \( u_1 = 28 \) and \( u_6 = 115 \).
Substitute \( a = 28 \):
\[ 28 + 5d = 115 \] \[ 5d = 87 \] \[ d = 17.4 \]Step 2: Calculate 3rd term
Step 3: Geometric Model (Part b)
Formula: \( n \)-th term of geometric sequence is \( u_n = ar^{n-1} \).
We know \( u_1 = 28 \) and \( u_6 = 115 \).
Step 4: Calculate 5th term
Alternatively, using \( r^5 = \frac{115}{28} \), \( u_5 = \frac{ar^5}{r} = \frac{115}{r} \):
\[ u_5 \approx \frac{115}{1.3265} \approx 86.69 \]Final Answer:
(a) 62.8 km h\(^{-1}\)
(b) 86.7 km h\(^{-1}\) (3 s.f.)
Question 6 (7 marks)
(a) Express \( \sin x + 2\cos x \) in the form \( R\sin(x + \alpha) \) where \( R \) and \( \alpha \) are constants, \( R > 0 \) and \( 0 < \alpha < \frac{\pi}{2} \).
Give the exact value of \( R \) and give the value of \( \alpha \) in radians to 3 decimal places.
(3)
The temperature, \( \theta \)°C, inside a room on a given day is modelled by the equation
\[ \theta = 5 + \sin\left(\frac{\pi t}{12} – 3\right) + 2\cos\left(\frac{\pi t}{12} – 3\right) \quad 0 \leq t < 24 \]where \( t \) is the number of hours after midnight.
Using the equation of the model and your answer to part (a),
(b) deduce the maximum temperature of the room during this day,
(1)
(c) find the time of day when the maximum temperature occurs, giving your answer to the nearest minute.
(3)
Worked Solution
Step 1: Harmonic Form (Part a)
Method: We equate \( \sin x + 2\cos x = R\sin(x+\alpha) = R(\sin x \cos \alpha + \cos x \sin \alpha) \).
Comparing coefficients:
- \( R \cos \alpha = 1 \) (coefficient of \( \sin x \))
- \( R \sin \alpha = 2 \) (coefficient of \( \cos x \))
Step 2: Maximum Temperature (Part b)
Method: Substitute the harmonic form into the temperature equation.
Let \( X = \frac{\pi t}{12} – 3 \). The equation becomes \( \theta = 5 + (\sin X + 2\cos X) = 5 + \sqrt{5}\sin(X + 1.107) \).
The maximum value of sine is 1.
Or as a decimal: \( \approx 7.24 \)°C.
Step 3: Time of Maximum (Part c)
Method: Set the sine part to 1. The argument of sine must be \( \frac{\pi}{2} \) (or \( \frac{5\pi}{2} \), etc.).
Convert decimal hours to time:
0.23 hours = \( 0.23 \times 60 \approx 14 \) minutes.
Time is 13:14 (or 1:14 pm).
Final Answer:
(a) \( R = \sqrt{5} \), \( \alpha = 1.107 \)
(b) \( 5 + \sqrt{5} \) °C (or 7.24°C)
(c) 13:14
Question 7 (5 marks)
Figure 1 shows a sketch of a curve \( C \) with equation \( y = f(x) \) and a straight line \( l \).
The curve \( C \) meets \( l \) at the points \( (-2,13) \) and \( (0,25) \) as shown.
The shaded region \( R \) is bounded by \( C \) and \( l \) as shown in Figure 1.
Given that
- \( f(x) \) is a quadratic function in \( x \)
- \( (-2,13) \) is the minimum turning point of \( y = f(x) \)
use inequalities to define \( R \).
(5)
Worked Solution
Step 1: Equation of the Line \( l \)
Method: The line passes through \( (-2,13) \) and \( (0,25) \). Use \( y = mx + c \).
Y-intercept \( c = 25 \).
Gradient \( m = \frac{25 – 13}{0 – (-2)} = \frac{12}{2} = 6 \).
Equation of line \( l \): \( y = 6x + 25 \).
Step 2: Equation of the Curve \( C \)
Method: We are told \( f(x) \) is quadratic with a minimum at \( (-2,13) \). Using vertex form \( y = a(x-h)^2 + k \), we have \( y = a(x+2)^2 + 13 \).
Use the point \( (0,25) \) to find \( a \).
Equation of curve \( C \): \( y = 3(x+2)^2 + 13 \).
(Expanded: \( y = 3(x^2+4x+4)+13 = 3x^2 + 12x + 25 \))
Step 3: Defining Region \( R \)
Reasoning: The region \( R \) is:
- Above the curve \( C \)
- Below the line \( l \)
- Bounded by the intersection points \( x = -2 \) and \( x = 0 \) (though the inequalities for y naturally handle the bounds if we specify the x range).
Inequalities:
\[ 3(x+2)^2 + 13 < y < 6x + 25 \]Valid for \( -2 < x < 0 \).
Commonly written set notation or combined inequality:
\[ \{ (x,y) : 3(x+2)^2 + 13 < y < 6x + 25 \} \]Final Answer:
\[ 3(x+2)^2 + 13 < y < 6x + 25 \]Question 8 (2 marks)
A new smartphone was released by a company.
The company monitored the total number of phones sold, \( n \), at time \( t \) days after the phone was released.
The company observed that, during this time,
the rate of increase of \( n \) was proportional to \( n \)
Use this information to write down a suitable equation for \( n \) in terms of \( t \).
(You do not need to evaluate any unknown constants in your equation.)
Worked Solution
Step 1: Mathematical Translation
Why we do this: “Rate of increase of \( n \)” translates to the derivative \( \frac{dn}{dt} \). “Proportional to \( n \)” means \( = k \times n \).
Step 2: Solve Differential Equation
Method: This is a standard differential equation representing exponential growth.
Let \( A = e^C \).
\[ n = A e^{kt} \]Final Answer:
\[ n = A e^{kt} \](where \( A \) and \( k \) are constants)
Question 9 (9 marks)
Figure 2 shows a sketch of the curve \( C \) with equation \( y = f(x) \) where
\[ f(x) = 4(x^2 – 2)e^{-2x} \quad x \in \mathbb{R} \](a) Show that \( f'(x) = 8(2 + x – x^2)e^{-2x} \)
(3)
(b) Hence find, in simplest form, the exact coordinates of the stationary points of \( C \).
(3)
The function \( g \) and the function \( h \) are defined by
\[ g(x) = 2f(x) \quad x \in \mathbb{R} \] \[ h(x) = 2f(x) – 3 \quad x \geq 0 \](c) Find (i) the range of \( g \)
(ii) the range of \( h \)
(3)
Worked Solution
Step 1: Differentiation (Part a)
Method: Use the Product Rule with \( u = 4(x^2-2) \) and \( v = e^{-2x} \).
✓ (A1*) Shown correctly
Step 2: Stationary Points (Part b)
Method: Set \( f'(x) = 0 \). Since \( e^{-2x} \neq 0 \), we solve the quadratic part.
Find y-coordinates:
When \( x = -1 \):
\[ y = 4((-1)^2 – 2)e^{-2(-1)} = 4(-1)e^2 = -4e^2 \]When \( x = 2 \):
\[ y = 4(2^2 – 2)e^{-2(2)} = 4(2)e^{-4} = 8e^{-4} \]Coordinates: \( (-1, -4e^2) \) and \( (2, 8e^{-4}) \)
Step 3: Ranges (Part c)
(i) Range of g: \( g(x) = 2f(x) \). This scales the graph vertically by 2. The range is determined by the global minimum.
From the sketch and points, global min is at \( x = -1 \), \( y = -4e^2 \). Scaled by 2, min is \( -8e^2 \). The curve goes to \( +\infty \) on the left.
Range of g: \( g(x) \geq -8e^2 \).
(ii) Range of h: \( h(x) = 2f(x) – 3 \) for \( x \geq 0 \).
We only consider \( x \geq 0 \). In this domain, find the min and max.
At \( x=0 \), \( f(0) = 4(-2) = -8 \). So \( h(0) = 2(-8) – 3 = -19 \).
There is a stationary point (max) at \( x=2 \) in this domain. \( f(2) = 8e^{-4} \). So \( h(2) = 2(8e^{-4}) – 3 = 16e^{-4} – 3 \).
Is there a minimum? The curve asymptotes to \( y=0 \) from above, so \( h(x) \to -3 \). The lowest point in \( x \geq 0 \) is at the start \( x=0 \) because the curve goes from -8 up to the max and then down to 0.
Min value: -19. Max value: \( 16e^{-4} – 3 \).
(i) Range of g: \( \{ y \in \mathbb{R} : y \geq -8e^2 \} \)
(ii) Range of h: \( \{ y \in \mathbb{R} : -19 \leq y \leq 16e^{-4} – 3 \} \)
Final Answer:
(b) \( (-1, -4e^2) \) and \( (2, 8e^{-4}) \)
(c)(i) \( g(x) \geq -8e^2 \)
(c)(ii) \( -19 \leq h(x) \leq 16e^{-4} – 3 \)
Question 10 (10 marks)
(a) Use the substitution \( x = u^2 + 1 \) to show that
\[ \int_5^{10} \frac{3dx}{(x-1)(3 + 2\sqrt{x-1})} = \int_p^q \frac{6du}{u(3+2u)} \]where \( p \) and \( q \) are positive constants to be found.
(4)
(b) Hence, using algebraic integration, show that
\[ \int_5^{10} \frac{3dx}{(x-1)(3 + 2\sqrt{x-1})} = \ln a \]where \( a \) is a rational constant to be found.
(6)
Worked Solution
Step 1: Substitution (Part a)
Method: Differentiate \( x = u^2 + 1 \) and change the limits.
Also \( x – 1 = u^2 \implies \sqrt{x-1} = u \).
Limits:
- When \( x = 5 \), \( 5 = u^2 + 1 \implies u^2 = 4 \implies u = 2 \) (since u>0 implicitly from sqrt).
- When \( x = 10 \), \( 10 = u^2 + 1 \implies u^2 = 9 \implies u = 3 \).
So \( p=2, q=3 \).
Substitute into integral:
\[ \int \frac{3(2u \, du)}{(u^2)(3 + 2u)} = \int \frac{6u \, du}{u^2(3+2u)} = \int \frac{6 \, du}{u(3+2u)} \]Step 2: Partial Fractions (Part b)
Method: We need to integrate \( \frac{6}{u(3+2u)} \). Use partial fractions.
Set \( u=0 \): \( 6 = 3A \implies A = 2 \).
Set \( u=-1.5 \): \( 6 = -1.5B \implies B = -4 \).
\[ \frac{6}{u(3+2u)} = \frac{2}{u} – \frac{4}{3+2u} \]Step 3: Integration
(Note: divide by 2 due to chain rule on \( 2u \))
\[ = \left[ 2\ln u – 2\ln(3+2u) \right]_2^3 \]Step 4: Evaluate Limits
Upper limit (3):
\[ 2\ln 3 – 2\ln(3+6) = 2\ln 3 – 2\ln 9 = 2\ln 3 – 4\ln 3 = -2\ln 3 \]Lower limit (2):
\[ 2\ln 2 – 2\ln(3+4) = 2\ln 2 – 2\ln 7 \]Subtract:
\[ (-2\ln 3) – (2\ln 2 – 2\ln 7) \] \[ = 2\ln 7 – 2\ln 2 – 2\ln 3 \] \[ = 2(\ln 7 – \ln 2 – \ln 3) \] \[ = 2\ln\left(\frac{7}{2 \times 3}\right) = 2\ln\left(\frac{7}{6}\right) \] \[ = \ln\left(\left(\frac{7}{6}\right)^2\right) = \ln\left(\frac{49}{36}\right) \]Final Answer:
(a) \( p=2, q=3 \)
(b) \( \ln\left(\frac{49}{36}\right) \) (so \( a = \frac{49}{36} \))
Question 11 (8 marks)
Circle \( C_1 \) has equation \( x^2 + y^2 = 100 \)
Circle \( C_2 \) has equation \( (x – 15)^2 + y^2 = 40 \)
The circles meet at points \( A \) and \( B \) as shown in Figure 3.
(a) Show that angle \( AOB = 0.635 \) radians to 3 significant figures, where \( O \) is the origin.
(4)
The region shown shaded in Figure 3 is bounded by \( C_1 \) and \( C_2 \).
(b) Find the perimeter of the shaded region, giving your answer to one decimal place.
(4)
Worked Solution
Step 1: Find Intersection Points (Part a)
Method: Solve the equations simultaneously to find the x-coordinate of A and B.
\( C_1: x^2 + y^2 = 100 \implies y^2 = 100 – x^2 \)
\( C_2: (x-15)^2 + y^2 = 40 \)
Step 2: Calculate Angle AOB
Method: Use trigonometry in the circle \( C_1 \). The triangle formed by the origin, the x-axis, and point A (let’s call the projection M) is a right-angled triangle.
Hypotenuse \( OA = \text{radius} = 10 \). Adjacent \( OM = x = 9.5 \).
Let \( \alpha \) be the angle \( AOM \).
The angle \( AOB \) is \( 2\alpha \) by symmetry.
\[ \text{Angle } AOB = 2 \times 0.31756… = 0.63512… \] \[ \approx 0.635 \text{ rad (3 s.f.)} \]✓ (A1*) Shown correctly
Step 3: Perimeter – Arc Lengths (Part b)
Method: The perimeter consists of the minor arc \( AB \) of \( C_1 \) and the minor arc \( AB \) of \( C_2 \).
For \( C_1 \) arc: \( s_1 = r_1 \theta_1 \). Radius \( r_1 = 10 \). Angle \( \theta_1 \) is the reflex angle? No, looking at the diagram, the shaded region is the intersection (lens). The boundary is the chord-side arc of \( C_1 \) (minor arc?) No, wait.
Look at Figure 3. The shaded region is the intersection of the two disks.
Boundary 1: Minor arc of \( C_2 \) (closest to origin? No, farthest). Wait, center of \( C_1 \) is (0,0). Center of \( C_2 \) is (15,0). Intersection x is 9.5.
The region is to the right of the arc from \( C_1 \) and to the left of the arc from \( C_2 \)? No. Intersection means inside both.
Left boundary is arc of \( C_2 \)? No, \( C_2 \) center is at 15. The arc of \( C_2 \) bulges left towards 15-radius. \( 15 – \sqrt{40} \approx 15-6.3 = 8.7 \). Since \( 8.7 < 9.5 \), the left boundary is the arc of \( C_2 \).
Right boundary is arc of \( C_1 \). \( C_1 \) radius is 10. \( 10 > 9.5 \).
Arc 1 (from \( C_1 \)): This is the arc passing through A and B on the right side of the lens. This corresponds to the chord AB in circle C1. The angle subtended at the center O is \( AOB = 0.635 \). The arc length is \( r_1 \theta_1 = 10 \times 0.635 = 6.35 \).
Arc 2 (from \( C_2 \)): This is the arc passing through A and B on the left side of the lens. We need the angle subtended at the center of \( C_2 \) (let’s call it P, at (15,0)).
Calculate angle at center of \( C_2 \):
Center \( P(15,0) \). Radius \( r_2 = \sqrt{40} \).
x-coordinate of intersection is 9.5. Distance from center P to chord line is \( 15 – 9.5 = 5.5 \).
Let \( \beta \) be the half-angle at P.
\[ \cos \beta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5.5}{\sqrt{40}} \] \[ \beta = \cos^{-1}\left(\frac{5.5}{\sqrt{40}}\right) \approx 0.5158 \text{ rad} \]Total angle \( \theta_2 = 2\beta \approx 1.0316 \text{ rad} \).
Calculate Arc Lengths:
Arc of \( C_1 \) (Minor): \( s_1 = r_1 \theta_1 = 10 \times 0.635 = 6.35 \).
Arc of \( C_2 \) (Minor): \( s_2 = r_2 \theta_2 = \sqrt{40} \times 1.0316 \approx 6.324 \times 1.0316 \approx 6.524 \).
Wait, check the diagram shape.
The shaded region is bounded by the minor arc of C1 and the minor arc of C2? Or major?
Since intersection x=9.5 is between 0 and 10 (inside C1) and between 8.7 and 21.3 (inside C2), we are looking at the minor arcs for both.
However, we must be careful. Which arc is which?
The region is defined by \( x > \text{something} \) for \( C_1 \) part? No, \( x < 10 \) for C1.
Actually, looking at the x-values:
- Points A, B have x=9.5.
- Tip of C1 is at x=10.
- Tip of C2 (left side) is at \( 15 – \sqrt{40} \approx 8.68 \).
The region spans from x=9.5 (chord?) No.
The intersection is the area where BOTH \( x^2+y^2 < 100 \) AND \( (x-15)^2+y^2 < 40 \).
The “right” edge of the shaded region is the arc of \( C_1 \) (from 9.5 to 10). Length = \( 10 \times 0.635 \)?
Correction: The angle 0.635 is subtended by the chord at the origin. The arc corresponding to this angle is the one “cutting off” the segment on the right. Yes. So \( s_1 = 6.35 \).
The “left” edge of the shaded region is the arc of \( C_2 \). The x-values are 9.5. The center is 15. The x-distance is \( 15-9.5=5.5 \). The angle calculated \( 1.0316 \) is correct. The arc length is \( \sqrt{40} \times 1.0316 \approx 6.52 \).
However, usually “perimeter of shaded region” in these intersection problems involves the “outer” boundary or the “inner” boundary?
Looking at the SVG/Diagram provided in the source: The shaded bit is the “lens”.
Perimeter = Arc(C1) + Arc(C2).
Are we using the correct angles?
For C1: The segment is defined by the chord at x=9.5. The arc is the small cap at x=10? No.
Let’s re-read the geometry. A and B are at x=9.5.
For C1 (centered 0,0), the chord is at x=9.5. The region is the small segment to the right of the chord. Angle is correct. Arc length = \( 10 \times 0.635 \). But wait, is the shaded region to the right of the chord? No.
The intersection of two circles.
Left circle (C1) covers x up to 10.
Right circle (C2) covers x from 8.68 to 21.32.
Intersection is between x=8.68 and x=10?
Intersection points are at x=9.5.
So the region is bounded by:
1. Left boundary: The arc of C2 from A to B (the part closer to origin). This arc bulges towards the origin. It goes from A(9.5, y) to (8.68, 0) to B(9.5, -y).2. Right boundary: The arc of C1 from A to B. This arc bulges away from origin. It goes from A to (10,0) to B.
Calculation for Arc 1 (C1, right side):
Angle subtended by chord AB at origin = \( 0.635 \). Radius = 10. Arc length = \( 10 \times 0.635 = 6.35 \). (Wait, 0.635 is small. 10*0.6 = 6. Chord length is \( 2 \times 3.12 = 6.24 \). Arc > Chord. Correct).
Calculation for Arc 2 (C2, left side):
Angle subtended by chord AB at center P(15,0). Distance to chord is 5.5. Radius is \( \sqrt{40} \).
Angle \( \theta_2 = 2 \arccos(5.5/\sqrt{40}) \).
Wait, we need the major arc or minor arc?
The boundary is the arc closer to the origin. From center (15,0), the chord is at x=9.5. The arc goes through x=8.68. This is the “far” side from the center P? No, it’s the near side to the origin, but for the circle C2, it’s the side with the smaller x values. Relative to P(15,0), x=9.5 is distance 5.5. The radius is 6.32. So the chord is “in front” of the center? No, 9.5 is to the left of 15. So it’s on the left side.
The arc we want is the one that closes the region on the left. This is the minor arc subtended by the angle calculated.
Angle \( \phi = 2 \arccos(5.5/\sqrt{40}) \).
Values: \( 5.5^2 = 30.25 \). \( 40 \). \( \cos \beta = \sqrt{30.25/40} \approx 0.869 \). \( \beta \approx 0.517 \). \( 2\beta \approx 1.034 \).
Arc 2 Length = \( \sqrt{40} \times 1.034 \approx 6.324 \times 1.034 \approx 6.54 \).
Total Perimeter = \( 6.35 + 6.54 = 12.89 \).
Step 4: Final Calculation
Check mark scheme values:
Angle AOB = 0.635.
Perimeter calculation: \( 10(2\pi – 0.635) \) in MS? No, that’s for area perhaps or major arc. Wait.
Let’s check the mark scheme for Q11(b).
MS says: “Attempts \( 10 \times (2\pi – 0.635) \) … = 56.48”. Then “Attempts to find angle AXB …”. “Attempts \( 10(2\pi – 0.635) + \sqrt{40}(2\pi – 2\beta) \)”.
Result: 89.7.
CRITICAL CORRECTION: The mark scheme implies the shaded region is the UNION of the circles or the OUTER boundary? Or maybe the diagram shows the region outside the intersection but inside the figure formed by both?
Let’s look at Figure 3 in the PDF (Page 30).
Figure 3 shows two circles. The shaded region is the UNION of the two disks? No, it looks like the whole shape formed by the two circles joined together. The intersection is NOT shaded? Or the whole thing is shaded?
Let’s re-examine the crop.
Page 30 crop: The region shown shaded is the entire combined shape (like a figure-8 but merged). The intersection line is dashed. The perimeter is the Major Arc of C1 + Major Arc of C2.
My previous analysis assumed “intersection”. The MS calculation (using \( 2\pi – \theta \)) confirms it’s the Major Arcs.
Revised Calculation:
Perimeter = Major Arc \( C_1 \) + Major Arc \( C_2 \).
Angle for \( C_1 \) (reflex) = \( 2\pi – 0.635 \).
Arc 1 = \( 10(2\pi – 0.635) \approx 56.48 \).
Angle for \( C_2 \) (reflex). We need angle at center P. Chord is at distance 5.5. \( \beta = \arccos(5.5/\sqrt{40}) \). Minor angle = \( 2\beta \). Reflex = \( 2\pi – 2\beta \).
Arc 2 = \( \sqrt{40}(2\pi – 2\beta) \).
\( 2\beta \approx 1.03 \).
Arc 2 = \( 6.324 \times (6.283 – 1.03) = 6.324 \times 5.253 \approx 33.22 \).
Total = \( 56.48 + 33.22 = 89.7 \).
This matches the MS result of 89.7.
Final Answer:
(a) \( 0.635 \) radians
(b) \( 89.7 \) (1 d.p.)
Question 12 (8 marks)
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(a) Show that
\[ \text{cosec}\,\theta – \sin\theta \equiv \cos\theta \cot\theta \quad \theta \neq (180n)^\circ \quad n \in \mathbb{Z} \](3)
(b) Hence, or otherwise, solve for \( 0 < x < 180^\circ \)
\[ \text{cosec}\,x – \sin x = \cos x \cot(3x – 50^\circ) \](5)
Worked Solution
Step 1: Proof (Part a)
Strategy: Convert everything to sine and cosine.
Combine fractions:
\[ = \frac{1 – \sin^2\theta}{\sin\theta} \]Use identity \( 1 – \sin^2\theta = \cos^2\theta \):
\[ = \frac{\cos^2\theta}{\sin\theta} \] \[ = \cos\theta \times \frac{\cos\theta}{\sin\theta} \] \[ = \cos\theta \cot\theta \] \[ = \text{RHS} \]✓ (A1) Proof complete
Step 2: Solving the Equation (Part b)
Method: Replace the LHS using the identity from part (a).
Bring terms to one side:
\[ \cos x \cot x – \cos x \cot(3x – 50^\circ) = 0 \] \[ \cos x [ \cot x – \cot(3x – 50^\circ) ] = 0 \]Step 3: Solve Factors
Case 1: \( \cos x = 0 \)
For \( 0 < x < 180^\circ \), \( x = 90^\circ \).
Case 2: \( \cot x = \cot(3x – 50^\circ) \)
This implies:
\[ x = 3x – 50^\circ + 180n^\circ \](General solution for cotangent involves \( + 180n \))
Rearranging:
\[ 50 – 180n = 2x \] \[ x = 25 – 90n \]Test values for \( n \):
- If \( n=0 \), \( x = 25^\circ \). (Valid)
- If \( n=-1 \), \( x = 25 + 90 = 115^\circ \). (Valid)
- If \( n=1 \), \( x = -65^\circ \). (Invalid)
Also check periodicity directly: \( 3x – 50 = x \implies 2x = 50 \implies x = 25 \).
Next period: \( 3x – 50 = x + 180 \implies 2x = 230 \implies x = 115 \).
Final Answer:
(b) \( x = 25^\circ, 90^\circ, 115^\circ \)
Question 13 (7 marks)
A sequence of numbers \( a_1, a_2, a_3, \dots \) is defined by
\[ a_{n+1} = \frac{k(a_n + 2)}{a_n} \quad n \in \mathbb{N} \]where \( k \) is a constant.
Given that
- the sequence is a periodic sequence of order 3
- \( a_1 = 2 \)
(a) show that \( k^2 + k – 2 = 0 \)
(3)
(b) For this sequence explain why \( k \neq 1 \)
(1)
(c) Find the value of
\[ \sum_{r=1}^{80} a_r \](3)
Worked Solution
Step 1: Generate Terms (Part a)
Method: Calculate \( a_2, a_3, a_4 \) in terms of \( k \). Since the order is 3, \( a_4 \) must equal \( a_1 \).
Set \( a_4 = a_1 = 2 \):
\[ \frac{k(k+3)}{k+1} = 2 \] \[ k^2 + 3k = 2(k+1) \] \[ k^2 + 3k = 2k + 2 \] \[ k^2 + k – 2 = 0 \]✓ (A1*) Shown correctly
Step 2: Explain \( k \neq 1 \) (Part b)
Reasoning: If \( k=1 \), calculate the terms to see if the period is 3.
If \( k=1 \):
\[ a_1 = 2 \] \[ a_2 = 2(1) = 2 \] \[ a_3 = 1+1 = 2 \]The sequence would be \( 2, 2, 2, \dots \), which has order 1, not 3.
Step 3: Calculate Sum (Part c)
Method: First find \( k \) and the terms of the sequence. Then sum the repeating cycle.
Solve \( k^2 + k – 2 = 0 \):
\[ (k+2)(k-1) = 0 \]Since \( k \neq 1 \), \( k = -2 \).
Terms:
\[ a_1 = 2 \] \[ a_2 = 2(-2) = -4 \] \[ a_3 = -2 + 1 = -1 \]Cycle is \( 2, -4, -1 \). Sum of one cycle \( S_{cycle} = 2 – 4 – 1 = -3 \).
Summation is for 80 terms.
\[ 80 = 26 \times 3 + 2 \]Total sum = \( 26 \times S_{cycle} + a_1 + a_2 \)
\[ = 26(-3) + 2 + (-4) \] \[ = -78 – 2 = -80 \]Final Answer:
(c) -80
Question 14 (10 marks)
A large spherical balloon is deflating.
At time \( t \) seconds the balloon has radius \( r \) cm and volume \( V \) cm\(^3\).
The volume of the balloon is modelled as decreasing at a constant rate.
(a) Using this model, show that
\[ \frac{dr}{dt} = -\frac{k}{r^2} \]where \( k \) is a positive constant.
(3)
Given that
- the initial radius of the balloon is 40 cm
- after 5 seconds the radius of the balloon is 20 cm
- the volume of the balloon continues to decrease at a constant rate until the balloon is empty
(b) solve the differential equation to find a complete equation linking \( r \) and \( t \).
(5)
(c) Find the limitation on the values of \( t \) for which the equation in part (b) is valid.
(2)
Worked Solution
Step 1: Chain Rule (Part a)
Method: We know \( \frac{dV}{dt} = -c \) (constant rate). We need \( \frac{dr}{dt} \). Use chain rule \( \frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \).
Volume of sphere \( V = \frac{4}{3}\pi r^3 \).
Let \( k = \frac{c}{4\pi} \) (a positive constant).
\[ \frac{dr}{dt} = -\frac{k}{r^2} \]✓ (A1*) Shown correctly
Step 2: Solve Differential Equation (Part b)
Method: Separate variables and integrate.
(where \( B = 3A \))
Step 3: Find Constants
Condition 1: \( t=0, r=40 \).
\[ 40^3 = B \implies B = 64000 \]Condition 2: \( t=5, r=20 \).
\[ 20^3 = -3k(5) + 64000 \] \[ 8000 = -15k + 64000 \] \[ 15k = 56000 \] \[ 3k = 11200 \]Substitute back into equation \( r^3 = -3kt + 64000 \):
\[ r^3 = -11200t + 64000 \]Step 4: Limitation on \( t \) (Part c)
Reasoning: The radius \( r \) cannot be negative. The model stops when the balloon is empty (\( r=0 \)).
Limitation: \( 0 \leq t \leq \frac{40}{7} \) (or approx 5.71 seconds).
Final Answer:
(b) \( r^3 = 64000 – 11200t \)
(c) \( 0 \leq t \leq \frac{40}{7} \) (seconds)
Question 15 (7 marks)
The curve \( C \) has equation
\[ x^2 \tan y = 9 \quad 0 < y < \frac{\pi}{2} \](a) Show that
\[ \frac{dy}{dx} = \frac{-18x}{x^4 + 81} \](4)
(b) Prove that \( C \) has a point of inflection at \( x = \sqrt[4]{27} \)
(3)
Worked Solution
Step 1: Implicit Differentiation (Part a)
Method: Differentiate \( x^2 \tan y = 9 \) with respect to \( x \). Use product rule on LHS.
Step 2: Eliminate \( y \)
Method: Substitute \( \tan y = \frac{9}{x^2} \) and use identity \( \sec^2 y = 1 + \tan^2 y \).
Multiply numerator and denominator by \( x^2 \):
\[ = \frac{-18x}{x^4 + 81} \]✓ (A1*) Shown correctly
Step 3: Point of Inflection (Part b)
Method: A point of inflection occurs where \( \frac{d^2y}{dx^2} = 0 \) and changes sign. We need to find the second derivative using the quotient rule.
Let \( u = -18x \) and \( v = x^4 + 81 \).
\[ u’ = -18 \] \[ v’ = 4x^3 \] \[ \frac{d^2y}{dx^2} = \frac{(-18)(x^4+81) – (-18x)(4x^3)}{(x^4+81)^2} \] \[ = \frac{-18x^4 – 1458 + 72x^4}{(x^4+81)^2} \] \[ = \frac{54x^4 – 1458}{(x^4+81)^2} \]Set \( \frac{d^2y}{dx^2} = 0 \):
\[ 54x^4 – 1458 = 0 \] \[ x^4 = \frac{1458}{54} = 27 \] \[ x = \sqrt[4]{27} \]Check change of sign around \( x^4 = 27 \):
- If \( x^4 < 27 \), \( 54x^4 < 1458 \), \( y'' < 0 \).
- If \( x^4 > 27 \), \( 54x^4 > 1458 \), \( y” > 0 \).
Since the concavity changes, it is a point of inflection.
Final Answer:
(b) \( \frac{d^2y}{dx^2} = 0 \) at \( x = \sqrt[4]{27} \) and sign changes, therefore it is a point of inflection.
Question 16 (4 marks)
Prove by contradiction that there are no positive integers \( p \) and \( q \) such that
\[ 4p^2 – q^2 = 25 \](4)
Worked Solution
Step 1: Assumption
Method: Assume the statement is false. Assume there ARE positive integers \( p \) and \( q \) such that \( 4p^2 – q^2 = 25 \).
Assume \( p, q \in \mathbb{Z}^+ \).
Step 2: Factorisation
Method: Factorise the difference of two squares.
Since \( p \) and \( q \) are positive integers, \( 2p+q \) must be a positive integer factor of 25. Also \( 2p-q \) must be a factor.
Since \( p, q > 0 \), \( 2p+q > 2p-q \). Also \( 2p+q > 0 \).
Factors of 25 are 1, 5, 25.
Possible pairs \( (2p-q, 2p+q) \):
- Case A: \( 1 \times 25 \)
- Case B: \( 5 \times 5 \)
Step 3: Test Cases
Case A:
\[ 2p – q = 1 \] \[ 2p + q = 25 \]Add equations:
\[ 4p = 26 \implies p = 6.5 \]This contradicts that \( p \) is an integer.
Case B:
\[ 2p – q = 5 \] \[ 2p + q = 5 \]Add equations:
\[ 4p = 10 \implies p = 2.5 \]This contradicts that \( p \) is an integer.
(Alternatively, subtracting gives \( 2q=0 \implies q=0 \), not a positive integer).
Step 4: Conclusion
Both cases lead to non-integer solutions for \( p \) (and \( q \)).
This contradicts the assumption that there exist integer solutions.
Therefore, there are no positive integers \( p \) and \( q \) such that \( 4p^2 – q^2 = 25 \).
✓ (A1) Proof complete
Final Answer:
Statement proven by contradiction.