IBDP Maths SL 2.6 The quadratic function AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Horizontal asymptote:
Solution:
For \( f(x) = \frac{2(x+3)}{3(x+2)} \), as \( x \to \infty \) or \( x \to -\infty \):
\[ \lim_{x\to\pm\infty} \frac{2(x+3)}{3(x+2)} = \frac{2}{3} \]
Thus, the horizontal asymptote is:
\[ y = \frac{2}{3} \]
[1 mark]
(b)(i) Number of solutions for \( m > 0 \):
Solution:
Set \( f(x) = g(x) \):
\[ \frac{2(x+3)}{3(x+2)} = mx + 1 \]
Cross-multiply:
\[ 2x + 6 = 3mx^2 + 6mx + 3x + 6 \]
Simplify:
\[ 0 = 3mx^2 + (6m + 1)x \]
Factor:
\[ x(3mx + 6m + 1) = 0 \]
Solutions:
1. \( x = 0 \)
2. \( x = -2 – \frac{1}{3m} \) (which is \( < -2 \) for \( m > 0 \))
Thus, there are 2 solutions for \( m > 0 \).
[2 marks]
(b)(ii) Value of \( m \) for one solution:
Solution:
For exactly one solution, the quadratic must have discriminant zero:
\[ (6m + 1)^2 = 0 \implies m = -\frac{1}{6} \]
Thus, the required value is:
\[ m = -\frac{1}{6} \]
[3 marks]
(b)(iii) Range of \( m \) for two solutions with \( x \geq 0 \):
Solution:
We need both solutions \( \geq 0 \):
1. \( x = 0 \) is always a solution
2. For \( x = -2 – \frac{1}{3m} \geq 0 \):
\[ -2 – \frac{1}{3m} \geq 0 \implies -\frac{1}{3m} \geq 2 \implies \frac{1}{m} \leq -6 \]
Since \( m < 0 \), this becomes:
\[ -\frac{1}{6} < m < 0 \]
Thus, the range is:
\[ -\frac{1}{6} < m < 0 \]
[4 marks]
Markscheme:
(a) \( y = \frac{2}{3} \) (A1)
(b)(i) 2 solutions (A1)
(b)(ii) Correct method (M1), correct solution \( m = -\frac{1}{6} \) (A1)
(b)(iii) Correct range \( -\frac{1}{6} < m < 0 \) (A1A1)
Total: [10 marks]
▶️ Answer/Explanation
(a) Graph Sketch:
Solution:
Step 1: Domain of function
The denominator \(x – 2 = 0\), so:
Domain: \(x \neq 2\), i.e., \(x \in (-\infty, 2) \cup (2, \infty)\)
Step 2: X and Y intercept
Y-intercept (when \(x = 0\)):
\[ f(0) = \frac{4(0) + 2}{0 – 2} = \frac{2}{-2} = -1 \]
Y-intercept: \((0, -1)\)
X-intercept (when \(f(x) = 0\)):
\[ \frac{4x + 2}{x – 2} = 0 \implies 4x + 2 = 0 \implies x = -0.5 \]
X-intercept: \((-0.5, 0)\)
Step 3: Asymptotes
The denominator is zero at \(x = 2\):
As \(x \to 2^-\) (left), \(f(x) \to -\infty\)
As \(x \to 2^+\) (right), \(f(x) \to +\infty\)
Thus, \(x = 2\) is the vertical asymptote.
For horizontal asymptote:
\[ \lim_{x \to \pm\infty} \frac{4x + 2}{x – 2} = 4 \]
Thus, \(y = 4\) is the horizontal asymptote.
Graph:
[5 marks]
(b) Range of \(f\):
Solution:
From the graph and behavior, \(f(x)\) can take all real values except \(y = 4\).
Range: \(y \neq 4\) or \(y \in (-\infty, 4) \cup (4, \infty)\)
[1 mark]
(c) Show \(p = \frac{9}{2}\):
Solution:
For \(g(x) = x^2 + bx + c\), the axis of symmetry is at \(x = 2\):
\[ x = -\frac{b}{2a} = 2 \implies -\frac{b}{2} = 2 \implies b = -4 \]
Sum of roots: \(-\frac{1}{2} + p = -\frac{b}{a} = 4\)
Thus, \(p = 4 + \frac{1}{2} = \frac{9}{2}\)
[3 marks]
(d) Values of \(b\) and \(c\):
Solution:
From (c), \(b = -4\).
Product of roots: \(-\frac{1}{2} \times \frac{9}{2} = \frac{c}{1} \implies c = -\frac{9}{4}\)
Thus, \(b = -4\) and \(c = -\frac{9}{4}\).
[3 marks]
(e) Y-coordinate of vertex:
Solution:
The vertex is at \(x = 2\) (axis of symmetry):
\[ g(2) = (2)^2 + (-4)(2) + \left(-\frac{9}{4}\right) = 4 – 8 – \frac{9}{4} = -\frac{25}{4} \]
Thus, the y-coordinate is \(-\frac{25}{4}\).
[2 marks]
(f) Product of solutions to \(f(x) = g(x)\):
Solution:
Set \(f(x) = g(x)\):
\[ \frac{4x + 2}{x – 2} = \left(x + \frac{1}{2}\right)\left(x – \frac{9}{2}\right) \text{ OR } \frac{4x + 2}{x – 2} = x^2 – 4x – \frac{9}{4} \]
Multiply through by \(x – 2\) to form a cubic equation:
\[ 4x + 2 = (x – 2)\left(x + \frac{1}{2}\right)\left(x – \frac{9}{2}\right) \] \[ \text{OR} \] \[ 4x + 2 = \left(x^2 – 4x – \frac{9}{4}\right)(x – 2) \]
Expanding either form leads to:
\[ x^3 – 6x^2 + \frac{23}{4}x – \frac{5}{2} = 0 \]
For a cubic equation \(ax^3 + bx^2 + cx + d = 0\), the product of roots is \(-\frac{d}{a}\):
\[ \text{Product} = -\left(\frac{-5/2}{1}\right) = -\frac{5}{2} \]
As shown in the working:
Thus, the product is \(-\frac{5}{2}\).
[4 marks]
Markscheme:
(a) Correct intercepts and asymptotes (A1A1A1), graph sketch (A1A1)
(b) Correct range \(y \neq 4\) (A1)
(c) Correct derivation of \(p = \frac{9}{2}\) (A1A1A1)
(d) Correct \(b = -4\) (A1), correct \(c = -\frac{9}{4}\) (A1A1)
(e) Correct y-coordinate \(-\frac{25}{4}\) (A1A1)
(f) Correct product \(-\frac{5}{2}\) (A1A1A1A1)
Total: [18 marks]
▶️ Answer/Explanation
a. Axis of symmetry:
Solution:
Since the vertex is at (4, 2), the axis of symmetry is the vertical line through the vertex.
\[ x = 4 \] (must be an equation)
[1 mark]
b. Values of h and k:
Solution:
For the vertex form \(f(x) = a{(x – h)^2} + k\), the vertex is at (h, k).
Given vertex (4, 2):
\[ h = 4 \] \[ k = 2 \]
[2 marks]
c. Finding coefficient a:
Solution:
Using the y-intercept (0, 6):
\[ f(0) = a{(0 – 4)^2} + 2 = 6 \] \[ a(16) + 2 = 6 \] \[ 16a = 4 \] \[ a = \frac{4}{16} = \frac{1}{4} \]
[3 marks]
Markscheme:
a. Correct equation \(x = 4\) (A1)
b. Correct \(h = 4\) (A1), correct \(k = 2\) (A1)
c. Correct substitution (M1), correct equation (A1), correct solution \(a = \frac{1}{4}\) (A1)
Total: [6 marks]
▶️ Answer/Explanation
a. Derivative of f(x):
Solution:
Differentiating term by term:
\[ f'(x) = \cos x + x – 2 \]
[3 marks]
b. Finding g(4):
Solution:
Since \(x = 2\) is the axis of symmetry and \(g(0) = 5\), by symmetry:
\[ g(4) = g(0) = 5 \]
Visual representation of symmetry:
[3 marks]
c. (i) Value of h:
Solution:
The axis of symmetry \(x = h = 2\)
\[ h = 2 \]
(ii) Value of a:
Solution:
Using \(g(0) = 5\) in the form \(g(x) = a{(x – 2)^2} + 3\):
\[ 5 = a{(0 – 2)^2} + 3 \] \[ 5 = 4a + 3 \] \[ a = \frac{1}{2} \]
[4 marks]
d. x-value for parallel tangents:
Solution:
First, find \(g'(x)\):
\[ g(x) = \frac{1}{2}x^2 – 2x + 5 \] \[ g'(x) = x – 2 \]
Set \(f'(x) = g'(x)\):
\[ \cos x + x – 2 = x – 2 \] \[ \cos x = 0 \]
In the interval \(0 \le x \le \pi\):
\[ x = \frac{\pi}{2} \]
[6 marks]
Markscheme:
a. Correct derivative \(\cos x + x – 2\) (A1A1A1)
b. Correct recognition of symmetry (R1M1), correct answer \(g(4) = 5\) (A1)
c. Correct \(h = 2\) (A1), correct substitution (M1), correct working (A1), correct solution \(a = \frac{1}{2}\) (A1)
d. Correct derivative of \(g\) (A1A1), correct equation setup (M1A1), correct working (A1), correct solution \(x = \frac{\pi}{2}\) (A1)
Total: [16 marks]