IB Mathematics SL 2.9 The function ax and its graph. AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
(i) Multiply by \(-1\) and factor:
\(x^2 + 4x – 5 = 0\)
\((x + 5)(x – 1) = 0\)
\(x = -5\) or \(x = 1\)
(ii) The quadratic \(5 – 4x – x^2\) represents a downward-opening parabola.
It is positive between the roots.
\(-5 < x < 1\)
(b)
The logarithmic function is undefined when the argument is \(\le 0\). Vertical asymptotes occur at the boundaries of the domain found in (a)(ii).
(i) \(a = -5\)
(ii) \(b = 1\)
(c)
Set \(f(x) = 0\):
\(\log_k(5 – 4x – x^2) = 0\)
\(5 – 4x – x^2 = k^0 = 1\)
\(x^2 + 4x – 4 = 0\)
Using the quadratic formula:
\(x = \frac{-4 \pm \sqrt{16 – 4(1)(-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2}\)
\(x = \frac{-4 \pm 4\sqrt{2}}{2}\)
\(x = -2 \pm 2\sqrt{2}\)
(d)
The maximum value of the function \(f(x)\) occurs when the argument \(5 – 4x – x^2\) is at its maximum.
The vertex of the parabola \(y = -x^2 – 4x + 5\) is at \(x = \frac{-b}{2a} = \frac{-(-4)}{2(-1)} = -2\).
Maximum argument value: \(5 – 4(-2) – (-2)^2 = 5 + 8 – 4 = 9\).
Given the maximum value of \(f(x)\) is \(2\):
\(\log_k(9) = 2\)
\(k^2 = 9\)
Since \(k > 1\), \(k = 3\).