2023 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

To find the values of \( a \) and \( b \), we use the two pieces of information given:

1. The curve passes through the point \( (2, 9) \). Substituting these coordinates into the curve's equation:
\[ 9 = a(2)^2 + \frac{b}{2} \]
\[ 9 = 4a + \frac{b}{2} \]
Multiplying the entire equation by 2 to clear the fraction:
\[ 8a + b = 18 \quad \text{--- (Equation 1)} \]

2. The point \( (2, 9) \) is a stationary point, which means the derivative \( \frac{dy}{dx} = 0 \) at \( x = 2 \).
First, find the derivative \( \frac{dy}{dx} \):
\[ y = ax^2 + bx^{-1} \]
\[ \frac{dy}{dx} = 2ax - bx^{-2} = 2ax - \frac{b}{x^2} \]

Since \( \frac{dy}{dx} = 0 \) when \( x = 2 \):
\[ 0 = 2a(2) - \frac{b}{2^2} \]
\[ 0 = 4a - \frac{b}{4} \]
\[ 16a - b = 0 \implies b = 16a \quad \text{--- (Equation 2)} \]

Now substitute Equation 2 into Equation 1:
\[ 8a + 16a = 18 \]
\[ 24a = 18 \implies a = \frac{18}{24} = \frac{3}{4} \]

Substitute \( a = \frac{3}{4} \) back into Equation 2:
\[ b = 16 \left(\frac{3}{4}\right) = 12 \]

Thus, the values are \( a = \frac{3}{4} \) and \( b = 12 \).

M1: Substitute \( (2, 9) \) into the equation of the curve to obtain a linear relation in \( a \) and \( b \) (e.g., \( 8a + b = 18 \)).
M1: Differentiate \( y \) with respect to \( x \) to find \( \frac{dy}{dx} \) (with at least one term differentiated correctly).
A1: Obtain correct derivative \( \frac{dy}{dx} = 2ax - \frac{b}{x^2} \) and set to 0 when \( x = 2 \) to find a second equation (e.g., \( b = 16a \)).
A1: Solve the simultaneous equations to find both correct values: \( a = \frac{3}{4} \) (or \( 0.75 \)) and \( b = 12 \).

Let the terms of the arithmetic progression (AP) be represented by \( u_n = a + (n-1)d \).

The 2nd term is:
\[ u_2 = a + d \]

The 5th term is:
\[ u_5 = a + 4d \]

The 14th term is:
\[ u_{14} = a + 13d \]

These three terms form a geometric progression (GP), so the ratio between successive terms is constant:
\[ \frac{u_5}{u_2} = \frac{u_{14}}{u_5} = r \]

This gives the equation:
\[ (a + 4d)^2 = (a + d)(a + 13d) \]

Expanding both sides:
\[ a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \]

Subtract \( a^2 \) from both sides and collect all terms on one side:
\[ 16d^2 - 13d^2 + 8ad - 14ad = 0 \]
\[ 3d^2 - 6ad = 0 \]
\[ 3d(d - 2a) = 0 \]

Since \( r \neq 1 \), the common difference \( d \) cannot be \( 0 \). Therefore:
\[ d = 2a \]

Substitute \( d = 2a \) back into the expressions for the terms:
- The first term of the GP (the 2nd term of the AP) is:
\[ u_2 = a + (2a) = 3a \]
- The second term of the GP (the 5th term of the AP) is:
\[ u_5 = a + 4(2a) = 9a \]

The common ratio \( r \) is:
\[ r = \frac{u_5}{u_2} = \frac{9a}{3a} = 3 \]

M1: Express the three terms of the AP correctly as \( a + d \), \( a + 4d \), and \( a + 13d \).
M1: Establish a correct relation for geometric progression, e.g., \( (a + 4d)^2 = (a + d)(a + 13d) \).
A1: Simplify and solve the equation to find the correct relation between \( d \) and \( a \), namely \( d = 2a \) (or equivalent, rejecting \( d = 0 \)).
A1: Substitute \( d = 2a \) to find the correct common ratio \( r = 3 \).

We are given the equation:
\[ 2\sin^2\theta - 3\cos\theta - 3 = 0 \]

Use the trigonometric identity \( \sin^2\theta = 1 - \cos^2\theta \) to rewrite the equation in terms of \( \cos\theta \) only:
\[ 2(1 - \cos^2\theta) - 3\cos\theta - 3 = 0 \]
\[ 2 - 2\cos^2\theta - 3\cos\theta - 3 = 0 \]
\[ -2\cos^2\theta - 3\cos\theta - 1 = 0 \]

Multiply the entire equation by \(-1\):
\[ 2\cos^2\theta + 3\cos\theta + 1 = 0 \]

This is a quadratic equation in terms of \( \cos\theta \). Let's factorize it:
\[ (2\cos\theta + 1)(\cos\theta + 1) = 0 \]

This yields two possible cases:
1. \( 2\cos\theta + 1 = 0 \implies \cos\theta = -\frac{1}{2} \)
2. \( \cos\theta + 1 = 0 \implies \cos\theta = -1 \)

Now, solve each for the interval \( 0 \le \theta \le 2\pi \):

- For \( \cos\theta = -\frac{1}{2} \):
The basic angle is \( \alpha = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \).
Since cosine is negative in the second and third quadrants:
\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
\[ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \]

- For \( \cos\theta = -1 \):
Within the interval \( 0 \le \theta \le 2\pi \), this occurs at:
\[ \theta = \pi \]

Combining all solutions, we get:
\[ \theta = \frac{2\pi}{3}, \pi, \frac{4\pi}{3} \]

M1: Substitute \( \sin^2\theta = 1 - \cos^2\theta \) into the equation to obtain an expression in \( \cos\theta \) only.
A1: Form the correct quadratic equation \( 2\cos^2\theta + 3\cos\theta + 1 = 0 \) (or equivalent).
M1: Factorize or solve the quadratic equation to obtain \( \cos\theta = -\frac{1}{2} \) and \( \cos\theta = -1 \).
A1: Find all three correct solutions: \( \theta = \frac{2\pi}{3} \), \( \theta = \pi \), and \( \theta = \frac{4\pi}{3} \). (Deduct 1 mark for any extra incorrect solutions within the range).

(i) The 3rd, 15th, and 47th terms of the arithmetic progression are \( T_3 = a + 2d \), \( T_{15} = a + 14d \), and \( T_{47} = a + 46d \) respectively. Since these form a geometric progression, \( \frac{a + 14d}{a + 2d} = \frac{a + 46d}{a + 14d} \). Cross-multiplying: \( (a + 14d)^2 = (a + 2d)(a + 46d) \), which expands to \( a^2 + 28ad + 196d^2 = a^2 + 48ad + 92d^2 \). Simplifying this gives \( 104d^2 = 20ad \). Since \( d \neq 0 \), dividing by \( 4d \) yields \( 26d = 5a \). To find the common ratio \( r \), substitute \( a = 5.2d \): \( r = \frac{5.2d + 14d}{5.2d + 2d} = \frac{19.2d}{7.2d} = \frac{8}{3} \).

(ii) The sum of the first 10 terms of the arithmetic progression is given by \( S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) = 485 \). This simplifies to \( 2a + 9d = 97 \). Using \( d = \frac{5}{26}a \), we get: \( 2a + 9\left(\frac{5}{26}a\right) = 97 \implies \frac{52a + 45a}{26} = 97 \implies \frac{97a}{26} = 97 \implies a = 26 \).

(i) M1: Set up the correct ratio equation for the geometric progression. M1: Expand and simplify to obtain a linear relation between a and d. A1: Correctly show 5a = 26d. A1: Obtain r = 8/3. (ii) M1: Use the sum formula for an AP with n = 10 and equate to 485. M1: Substitute the relationship between a and d to obtain a single-variable equation. A1.75: Correctly find a = 26.

(i) Expressing \( \mathrm{f}(x) \) by completing the square: \( \mathrm{f}(x) = 2(x^2 - 6x) + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5 \). The vertex of the curve is at \( (3, -5) \). For the function to have an inverse, it must be one-to-one, so the domain must be restricted to one side of the line of symmetry \( x = 3 \). Since the domain is \( x \ge k \), the smallest value of \( k \) is \( 3 \).

(ii) Let \( y = 2(x-3)^2 - 5 \). Rearranging for \( x \): \( y + 5 = 2(x-3)^2 \implies (x-3)^2 = \frac{y+5}{2} \). Taking the positive square root since \( x \ge 3 \): \( x - 3 = \sqrt{\frac{y+5}{2}} \implies x = 3 + \sqrt{\frac{y+5}{2}} \). Therefore, \( \mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}} \). The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \). Since \( x \ge 3 \), the minimum value of \( \mathrm{f}(x) \) is \( -5 \), so the range of \( \mathrm{f} \) is \( \mathrm{f}(x) \ge -5 \). Thus, the domain of \( \mathrm{f}^{-1} \) is \( x \ge -5 \).

(i) M1: Attempt to find the axis of symmetry (by completing the square or differentiation). A1: State k = 3. (ii) M1: Rearrange y = 2(x-3)^2 - 5 to make x the subject. M1: Correctly choose the positive square root due to the domain. A1: State f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}. M1: Identify that the domain of f^{-1} is the range of f. A1.75: State the correct domain x \ge -5.

(i) Rewrite the curve equation as \( y = 16x^{-1} + x^2 \). Differentiating with respect to \( x \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = -16x^{-2} + 2x = 2x - \frac{16}{x^2} \). At the stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \): \( 2x - \frac{16}{x^2} = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2 \). Substituting \( x = 2 \) back into the curve equation: \( y = \frac{16}{2} + 2^2 = 8 + 4 = 12 \). So, the coordinates of the stationary point are \( (2, 12) \).

(ii) Differentiating again: \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(-16x^{-2} + 2x\right) = 32x^{-3} + 2 = \frac{32}{x^3} + 2 \). At \( x = 2 \): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{32}{8} + 2 = 4 + 2 = 6 \). Since \( 6 > 0 \), the stationary point is a minimum.

(i) M1: Differentiate the curve equation to find dy/dx. M1: Equate dy/dx to 0 and solve for x. A1: Find x = 2. A1: Find y = 12 and state coordinates (2, 12). (ii) M1: Correctly differentiate dy/dx to find the second derivative. A1: Substitute x = 2 to evaluate the second derivative as 6. A1.75: Conclude that the point is a minimum based on a positive second derivative.

(i) Expand the integrand first: \( \left(x^2 - \frac{3}{x^2}\right)^2 = (x^2)^2 - 2(x^2)\left(\frac{3}{x^2}\right) + \left(\frac{3}{x^2}\right)^2 = x^4 - 6 + 9x^{-4} \). Now integrate term-by-term: \( \int (x^4 - 6 + 9x^{-4}) \mathrm{d}x = \frac{x^5}{5} - 6x + \frac{9x^{-3}}{-3} + C = \frac{x^5}{5} - 6x - \frac{3}{x^3} + C \).

(ii) Using the integrated expression from part (i): \( \left[ \frac{x^5}{5} - 6x - \frac{3}{x^3} \right]_1^2 = \left( \frac{32}{5} - 12 - \frac{3}{8} \right) - \left( \frac{1}{5} - 6 - 3 \right) = \left( 6.4 - 12 - 0.375 \right) - (0.2 - 9) = -5.975 - (-8.8) = 2.825 \) or \( \frac{113}{40} \).

(i) M1: Attempt to expand the bracket with at least two correct terms. A1: Obtain expanded form x^4 - 6 + 9x^{-4}. M1: Integrate term-by-term (power increased by 1 and divided by new power). A1.75: Correct integration, including constant of integration C. (ii) M1: Substitute limits 2 and 1 into their integrated expression. M1: Perform calculation, handling signs carefully. A1: Obtain 113/40 or 2.825.

(i) The perimeter of the sector is given by \( P = 2r + r\theta = 20 \). Rearranging this to express \( \theta \) in terms of \( r \) gives \( r\theta = 20 - 2r \implies \theta = \frac{20 - 2r}{r} \). The area of the sector is \( A = \frac{1}{2} r^2 \theta \). Substituting the expression for \( \theta \): \( A = \frac{1}{2} r^2 \left(\frac{20 - 2r}{r}\right) = \frac{1}{2} r(20 - 2r) = 10r - r^2 \).

(ii) To find the maximum area, differentiate \( A \) with respect to \( r \): \( \frac{\mathrm{d}A}{\mathrm{d}r} = 10 - 2r \). Setting \( \frac{\mathrm{d}A}{\mathrm{d}r} = 0 \) gives \( r = 5 \). The second derivative is \( \frac{\mathrm{d}^2A}{\mathrm{d}r^2} = -2 < 0 \), confirming that this is a maximum. The maximum area is \( A = 10(5) - (5)^2 = 25 \) cm\(^2\). The corresponding value of \( \theta \) is \( \theta = \frac{20 - 2(5)}{5} = 2 \) radians.

(i) M1: Write down perimeter equation 2r + r\theta = 20 and rearrange for \theta. M1: Substitute this expression into area formula A = 0.5r^2\theta. A1.75: Obtain A = 10r - r^2 with clear algebraic steps. (ii) M1: Differentiate A and set dA/dr = 0 to find r. A1: Find r = 5 and maximum area = 25. M1: Substitute r = 5 back into perimeter relation to solve for \theta. A1: Obtain \theta = 2.

(i) Combine the fractions over a common denominator: \( \text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)} = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)} \). Since \( \sin^2 \theta + \cos^2 \theta = 1 \), this simplifies to: \( \frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = \text{RHS} \).

(ii) Using the identity from part (i), the equation becomes \( \frac{2}{\sin \theta} = 3 \implies \sin \theta = \frac{2}{3} \). The basic angle is \( \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ \). Since the sine value is positive, solutions lie in the first and second quadrants: \( \theta = 41.8^\circ \) and \( \theta = 180^\circ - 41.81^\circ = 138.2^\circ \) (both answers rounded to 1 decimal place).

(i) M1: Combine fractions with a correct common denominator. M1: Correctly expand the numerator (1 + cos \theta)^2. M1: Apply the fundamental identity sin^2 \theta + cos^2 \theta = 1. A1.75: Factorise numerator and cancel common factor to reach 2/sin \theta. (ii) M1: Set 2/sin \theta = 3 and rearrange to find sin \theta = 2/3. A1: Obtain first angle 41.8^\circ. A1: Obtain second angle 138.2^\circ (accept 138.2^\circ and 41.8^\circ).

(i) Equating the curve and line equations to find intersection points: \( 2x^2 + kx + 5 = 3x + 3 \implies 2x^2 + (k-3)x + 2 = 0 \). For no intersection, the quadratic equation must have no real roots, so the discriminant must be negative: \( b^2 - 4ac < 0 \implies (k-3)^2 - 4(2)(2) < 0 \implies (k-3)^2 - 16 < 0 \implies (k-3)^2 < 16 \). This yields \( -4 < k-3 < 4 \), and solving for \( k \) gives \( -1 < k < 7 \).

(ii) For \( k = 7 \), substituting into the equation yields \( 2x^2 + (7-3)x + 2 = 0 \implies 2x^2 + 4x + 2 = 0 \implies x^2 + 2x + 1 = 0 \). Factoring gives \( (x+1)^2 = 0 \implies x = -1 \). Substituting \( x = -1 \) into the line equation: \( y = 3(-1) + 3 = 0 \). Thus, the coordinates of the point of intersection are \( (-1, 0) \).

(i) M1: Equate curve and line equations. M1: Rearrange into standard quadratic form ax^2 + bx + c = 0. M1: Apply discriminant b^2 - 4ac < 0. A1.75: Correctly solve inequality to get -1 < k < 7. (ii) M1: Substitute k = 7 and solve the resulting quadratic equation. A1: Find x = -1. A1: Find y = 0 and state point (-1, 0).

(i) The midpoint of \( AB \) is \( M = \left( \frac{2+6}{2}, \frac{5+13}{2} \right) = (4, 9) \). The gradient of \( AB \) is \( m = \frac{13-5}{6-2} = \frac{8}{4} = 2 \). The gradient of the perpendicular bisector is the negative reciprocal: \( m_{\perp} = -\frac{1}{2} \). Using the midpoint, the equation of the perpendicular bisector is: \( y - 9 = -\frac{1}{2}(x - 4) \implies y - 9 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 11 \) (or \( x + 2y = 22 \)).

(ii) The line meets the \( y \)-axis where \( x = 0 \): \( y = -\frac{1}{2}(0) + 11 = 11 \), so \( C = (0, 11) \). The exact length of \( AC \) is: \( AC = \sqrt{(0-2)^2 + (11-5)^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \).

(i) M1: Find the midpoint of AB. M1: Find the gradient of AB. M1: Find the perpendicular gradient. A1.75: Correct equation of the perpendicular bisector in any valid form. (ii) M1: Set x = 0 in their equation to find coordinates of C. A1: State C is (0, 11). A1: Correctly calculate the exact length of AC as 2\sqrt{10} or \sqrt{40}.