IBDP Maths AHL 2.15 Solving inequalities AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed Solution:
(a) Transformations:
The function \( g(x) \) is obtained from \( f(x) \) through:
- Horizontal translation: A shift of \( \frac{3\pi}{2} \) units to the right
This comes from the \( x-\frac{3\pi}{2} \) inside the sine function - Vertical translation: A shift of \( q \) units upwards
This comes from the \( +q \) term added to the function
In vector form, this can be represented as \( \begin{pmatrix} \frac{3\pi}{2} \\ q \end{pmatrix} \).
(b) Finding minimum value of r:
Given \( g(x) \geq 7 \), we first find the minimum value of \( g(x) \):
The minimum value of \( \sin\left(x-\frac{3\pi}{2}\right) \) is -1, so: \[ \text{Minimum of } g(x) = 4(-1) + 2.5 + q = -4 + 2.5 + q \] Set this \( \geq 7 \): \[ -1.5 + q \geq 7 \] \[ q \geq 8.5 \] The smallest possible \( q \) is 8.5.
Now find the y-intercept \( r \) when \( q = 8.5 \): \[ g(0) = 4\sin\left(-\frac{3\pi}{2}\right) + 2.5 + 8.5 \] \[ \sin\left(-\frac{3\pi}{2}\right) = 1 \] (since it’s equivalent to \( \sin\left(\frac{\pi}{2}\right) \)) \[ r = 4(1) + 2.5 + 8.5 = 15 \] Therefore, the smallest possible value of \( r \) is 15.
Markscheme:
(a)
• Horizontal translation \( \frac{3\pi}{2} \) right (A1)
• Vertical translation \( q \) up (A1)
Note: Accept vector form \( \begin{pmatrix} \frac{3\pi}{2} \\ q \end{pmatrix} \) for both marks.
[2 marks]
(b)
• Recognizes minimum of sine function is -1 (M1)
• Sets up inequality \( -4 + 2.5 + q \geq 7 \) (A1)
• Solves to find \( q \geq 8.5 \) (A1)
• Correctly evaluates \( g(0) \) with \( q = 8.5 \) (M1)
• Finds \( r = 15 \) (A1)
[5 marks]
Total: [7 marks]
▶️ Answer/Explanation
Detailed Solution:
(a)
(i) First find the derivative using the quotient rule:
\[ f'(x) = \frac{{x \cdot \frac{1}{x} – \ln x}}{x^2} = \frac{1 – \ln x}{x^2} \]
Set \( f'(x) = 0 \):
\[ 1 – \ln x = 0 \Rightarrow \ln x = 1 \Rightarrow x = e \]
(ii) To show this is a local maximum:
For \( x < e \), \( f'(x) > 0 \) (function increasing)
For \( x > e \), \( f'(x) < 0 \) (function decreasing)
Thus, \( x = e \) is a local maximum.
(iii) The range is all real numbers \( y \) such that:
\[ y \leq \frac{1}{e} \]
(b) Point of Inflexion:
Find the second derivative:
\[ f”(x) = \frac{d}{dx} \left( \frac{1 – \ln x}{x^2} \right) = \frac{-3 + 2\ln x}{x^3} \]
Set \( f”(x) = 0 \):
\[ -3 + 2\ln x = 0 \Rightarrow \ln x = \frac{3}{2} \Rightarrow x = e^{3/2} \]
Coordinates: \( \left( e^{3/2}, \frac{3}{2e^{3/2}} \right) \).
(c) Graph Sketch:
Key features:
- Vertical asymptote at x=0
- x-intercept at (1,0)
- Local maximum at (e,1/e)
- Point of inflexion at (e^(3/2),3/(2e^(3/2)))
(d)
(i) Graph of \( g(x) \):
(ii) Range of \( g \): All real numbers.
(iii) Solution for \( h(x) > g(x) \):
Solution: \( -e^2 < x < -1 \)