IBDP Maths SL 2.5 Composite functions fog AA HL Paper 1- Exam Style Questions- New Syllabus

(a) Write in vertex form:

\( f(x) = 5(x + 1)(x + 3) \)
Expanding gives \( f(x) = 5(x^2 + 4x + 3) = 5x^2 + 20x + 15 \).
Completing the square: \( f(x) = 5[x^2 + 4x + 4 – 4 + 3] = 5[(x + 2)^2 – 1] \).
Result: \( f(x) = 5(x + 2)^2 – 5 \), where \( a = 5, h = -2, k = -5 \).

(b) Sketch:

• Vertex: \( (-2, -5) \).
• x-intercepts: Solve \( 5(x + 1)(x + 3) = 0 \Rightarrow x = -1, -3 \).
• y-intercept: \( f(0) = 15 \).

(c) Solve \( f(x) \leq 40 \):

\( 5(x^2 + 4x + 3) \leq 40 \Rightarrow x^2 + 4x + 3 \leq 8 \).
\( x^2 + 4x – 5 \leq 0 \Rightarrow (x + 5)(x – 1) \leq 0 \).
Solution: \( -5 \leq x \leq 1 \).

(d)(i) Composite function:

\( (f \circ g)(x) = f(\ln x) = 5(\ln x + 1)(\ln x + 3) \).

(d)(ii) Solve \( (f \circ g)(x) \leq 40 \):

Using part (c), replace \( x \) with \( \ln x \): \( -5 \leq \ln x \leq 1 \).
Exponentiate: \( e^{-5} \leq x \leq e \).
Solution: \( e^{-5} \leq x \leq e \).

(e) Domain of \( g \circ f \):

\( (g \circ f)(x) = \ln(f(x)) \). We require \( f(x) > 0 \).
\( 5(x + 1)(x + 3) > 0 \).
Since the parabola opens upwards, it is positive outside the roots.
Domain: \( x < -3 \) or \( x > -1 \).