Home / IBDP Maths SL 2.8 The rational function AA HL Paper 1- Exam Style Questions

IBDP Maths SL 2.8 The rational function AA HL Paper 1- Exam Style Questions

IBDP Maths SL 2.8 The rational function AA HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AA : HL Papers -1 : All Topics
Question

A function f is defined by f(x) = \(\frac{2x – 1}{x + 1}\), where x ∈ R, x ≠ -1.

(a) The graph of y = f (x) has a vertical asymptote and a horizontal asymptote.

Write down the equation of:

(i) the vertical asymptote;

(ii) the horizontal asymptote.

(b) On the set of axes below, sketch the graph of y = f (x).

On your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.

Graph axes

(c) Hence, solve the inequality \(0<\frac{2x – 1}{x + 1} <2.\)

(d) Solve the inequality \(0<\frac{2|x|-1}{|x| + 1} <2.\)

▶️ Answer/Explanation

(a)

(i) Vertical asymptote:

\[ x = -1 \]

(ii) Horizontal asymptote:

\[ y = 2 \]

[2 marks]

(b) Graph sketch:

Graph of y = f(x)

Key features:

  • Rational function shape with two branches in opposite quadrants
  • Vertical asymptote at x = -1 and horizontal asymptote at y = 2
  • x-intercept at \(x = \frac{1}{2}\)
  • y-intercept at y = -1

[3 marks]

(c) Solution to inequality:

\[ x > \frac{1}{2} \]

Alternative notation: \(\left( \frac{1}{2}, \infty \right)\) or \(]\frac{1}{2}, \infty[\)

[1 mark]

(d) Solution to absolute value inequality:

Either:

1. Sketch \(y = \frac{2|x| – 1}{|x|+1}\)

Or:

2. Solve \(2|x| – 1 = 0\) to find critical points \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\)

Final solution:

\[ x < -\frac{1}{2} \text{ and } x > \frac{1}{2} \]

Alternative notation: \(\left(-\infty, -\frac{1}{2}\right) \cup \left(\frac{1}{2}, \infty\right)\)

[2 marks]

Markscheme:

(a) Correct vertical asymptote (A1), correct horizontal asymptote (A1)

(b) Correct graph shape (A1), correct asymptotes (A1), correct intercepts (A1)

(c) Correct solution (A1)

(d) Correct method (M1), correct solution (A1)

Total: [8 marks]

Question

[Maximum mark: 7]

The function \(f\) is defined by \(f(x)=\frac{7x+7}{2x-4}\) for \(x \in \mathbb{R}, x \neq 2\).

(a) Find the zero of \(f(x)\). [2]

(b) For the graph of \(y=f(x)\), write down the equation of:

(i) the vertical asymptote;

(ii) the horizontal asymptote. [2]

(c) Find \(f^{-1}(x)\), the inverse function of \(f(x)\). [3]

▶️ Answer/Explanation

(a) Zero of f(x):

Solution:

Set \(f(x) = 0\):

\[ \frac{7x+7}{2x-4} = 0 \]

Numerator must equal zero:

\[ 7x + 7 = 0 \]

\[ x = -1 \]

[2 marks]

(b) Asymptotes:

(i) Vertical asymptote:

Set denominator equal to zero:

\[ 2x – 4 = 0 \]

\[ x = 2 \]

(ii) Horizontal asymptote:

Since degrees of numerator and denominator are equal (both 1):

\[ y = \frac{7}{2} \]

[2 marks]

(c) Inverse function:

Solution:

Start with \(y = \frac{7x+7}{2x-4}\)

Swap \(x\) and \(y\):

\[ x = \frac{7y+7}{2y-4} \]

Cross-multiply:

\[ x(2y – 4) = 7y + 7 \]

\[ 2xy – 4x = 7y + 7 \]

Gather \(y\) terms:

\[ 2xy – 7y = 4x + 7 \]

Factor out \(y\):

\[ y(2x – 7) = 4x + 7 \]

Solve for \(y\):

\[ y = \frac{4x + 7}{2x – 7} \]

Thus:

\[ f^{-1}(x) = \frac{4x + 7}{2x – 7}, \quad x \neq \frac{7}{2} \]

[3 marks]

Markscheme:

(a) Correct method (M1), correct solution \(x = -1\) (A1)

(b) Correct vertical asymptote (A1), correct horizontal asymptote (A1)

(c) Correct interchange (M1), correct working (A1), correct final answer (A1)

Total: [7 marks]

Question

Consider the function defined by \( f(x) = \frac{kx-5}{x-k} \), where \( x \in \mathbb{R} \) \ \{k\} and \( k^2 \neq 5 \).

a. State the equation of the vertical asymptote on the graph of \( y = f(x) \). [1]

b. State the equation of the horizontal asymptote on the graph of \( y = f(x) \). [1]

c. Use an algebraic method to determine whether \( f \) is a self-inverse function. [4]

Consider the case where \( k = 3 \).

d. Sketch the graph of \( y = f(x) \), stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes. [3]

e. The region bounded by the x-axis, the curve \( y = f(x) \), and the lines \( x = 5 \) and \( x = 7 \) is rotated through \( 2\pi \) about the x-axis. Find the volume of the solid generated, giving your answer in the form \( \pi(a + b \ln 2) \), where \( a, b \in \mathbb{Z} \). [6]

▶️ Answer/Explanation

a. Vertical asymptote:

\[ x = k \]

[1 mark]

b. Horizontal asymptote:

\[ y = k \]

[1 mark]

c. Self-inverse verification:

METHOD 1:

Method 1 algebraic proof

METHOD 2:

Method 2 algebraic proof

Both methods show that \( f(f(x)) = x \), proving \( f \) is self-inverse.

[4 marks]

d. Graph sketch (k=3):

Graph of y=f(x)

Key features:

  • Vertical asymptote: \( x = 3 \)
  • Horizontal asymptote: \( y = 3 \)
  • x-intercept: \( \left(\frac{5}{3}, 0\right) \)
  • y-intercept: \( \left(0, \frac{5}{3}\right) \)

[3 marks]

e. Volume of revolution:

\[ \text{Volume} = \pi \int_{5}^{7} \left(\frac{3x-5}{x-3}\right)^2 dx \]

First rewrite the integrand:

\[ \frac{3x-5}{x-3} = 3 + \frac{4}{x-3} \]

Then square:

\[ \left(3 + \frac{4}{x-3}\right)^2 = 9 + \frac{24}{x-3} + \frac{16}{(x-3)^2} \]

Integrate term by term:

\[ \pi \left[9x + 24\ln|x-3| – \frac{16}{x-3}\right]_{5}^{7} \]

Evaluate:

\[ \pi \left[(63 + 24\ln4 – 4) – (45 + 24\ln2 – 8)\right] \]

Simplify:

\[ \pi (22 + 24\ln2) \]

[6 marks]

Markscheme:

a. Correct vertical asymptote (A1)

b. Correct horizontal asymptote (A1)

c. Correct method (M1M1), complete proof (A1A1)

d. Correct graph shape (A1), correct asymptotes (A1), correct intercepts (A1)

e. Correct setup (M1), correct expansion (A1), correct integration (A1), correct evaluation (A1), correct final form (A1A1)

Total: [15 marks]

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