Home / IBDP Maths SL 5.4 Tangents and normals at a given point, and their equations AA HL Paper 2- Exam Style Questions

IBDP Maths SL 5.4 Tangents and normals at a given point, and their equations AA HL Paper 2- Exam Style Questions

IBDP Maths SL 5.4 Tangents and normals at a given point, and their equations AA HL Paper 2- Exam Style Questions- New Syllabus

IBDP Maths AA :HL Papers-2: All Topics
Question

Consider the function \( f(x) = 90e^{-0.5x} \) for \( x \in \mathbb{R}^+ \). The graph of \( f \) and the line \( y = x \) intersect at point \( P \). The line \( L \) has a gradient of \(-1\) and is a tangent to the graph of \( f \) at the point \( Q \).

(a) Find the \( x \)-coordinate of \( P \). [2]

(b) Find the exact coordinates of \( Q \). [4]

(c) Show that the equation of \( L \) is \( y = -x + 2 \ln 45 + 2 \). [2]

Diagram showing the shaded region A enclosed by the graph of f, the line y=x, and the line L

(d)
(i) Find the \( x \)-coordinate of the point where \( L \) intersects the line \( y = x \). [1]
(ii) Hence, find the area of the shaded region \( A \) enclosed by the graph of \( f \), the line \( y = x \), and the line \( L \). [4]

(e) The line \( L \) is tangent to the graphs of both \( f \) and the inverse function \( f^{-1} \). Find the shaded area enclosed by the graphs of \( f \) and \( f^{-1} \) and the line \( L \). [2]

▶️ Answer/Explanation
Markscheme Solution

(a) [2 marks]

Intersection at \( P \): \( 90e^{-0.5x} = x \) (M1).
Solve numerically: \( x \approx 5.56619 \approx 5.57 \) (A1).

(b) [4 marks]

\( f'(x) = 90 \cdot (-0.5)e^{-0.5x} = -45e^{-0.5x} \) (A1).
Tangent gradient \(-1\): \( -45e^{-0.5x} = -1 \implies e^{-0.5x} = \frac{1}{45} \implies -0.5x = \ln \frac{1}{45} \implies x = 2 \ln 45 \) (M1, A1).
At \( x = 2 \ln 45 \), \( y = f(2 \ln 45) = 90e^{-0.5 \cdot 2 \ln 45} = 90e^{-\ln 45} = \frac{90}{45} = 2 \) (A1).
Coordinates of \( Q \): \( (2 \ln 45, 2) \).

(c) [2 marks]

Tangent at \( Q(2 \ln 45, 2) \), slope \(-1\): \( y – 2 = -1 (x – 2 \ln 45) \) (M1).
Simplify: \( y = -x + 2 \ln 45 + 2 \) (A1).
Diagram showing the shaded region A enclosed by the graph of f, the line y=x, and the line L

(d) [5 marks]

(i) Intersection of \( L: y = -x + 2 \ln 45 + 2 \) and \( y = x \): \( x = -x + 2 \ln 45 + 2 \implies 2x = 2 \ln 45 + 2 \implies x = \ln 45 + 1 \approx 4.806 \) (A1).
(ii) Area of \( A \): From the diagram, area = \( \int_{4.806\ldots}^{5.566\ldots} \left(x – (-x + 2 \ln 45 + 2)\right) \, dx + \int_{5.566\ldots}^{7.613\ldots} \left(90e^{-0.5x} – (-x + 2 \ln 45 + 2)\right) \, dx \) (M1).
Diagram showing integral setup for the area of region A: ∫(4.806...)^(5.566...) (x - (-x + 2 ln 45 + 2)) dx + ∫(5.566...)^(7.613...) (90e^(-0.5x) - (-x + 2 ln 45 + 2)) dx
Simplify integrands: First integral = \( \int_{4.806\ldots}^{5.566\ldots} (2x – 2 \ln 45 – 2) \, dx \), second = \( \int_{5.566\ldots}^{7.613\ldots} (90e^{-0.5x} + x – 2 \ln 45 – 2) \, dx \) (M1).
Numerical integration: \( \approx 1.51965 \approx 1.52 \) (A1, A1).

(e) [2 marks]

Area enclosed by \( f \), \( f^{-1} \), and \( L \): By symmetry about \( y = x \), total area = \( 2 \times 1.52 = 3.04 \) (M1, A1).

Markscheme Answers:

(a) \( x \approx 5.57 \) (M1, A1)

(b) \( f'(x) = -45e^{-0.5x} \), \( Q: (2 \ln 45, 2) \) (A1, M1, A1, A1)

(c) \( y = -x + 2 \ln 45 + 2 \) (M1, A1)

(d)
(i) \( x = \ln 45 + 1 \approx 4.806 \) (A1)
(ii) Area \( \approx 1.52 \) (M1, M1, A1, A1)

(e) Area \( \approx 3.04 \) (M1, A1)

Total [15 marks]

Question

A family of cubic functions is defined as \( f_k(x) = k^2 x^3 – k x^2 + x \), \( k \in \mathbb{Z}^+ \).

(a)
(i) Express in terms of \( k \) the first and second derivatives \( f’_k(x) \) and \( f”_k(x) \). [2]
(ii) Find the coordinates of the points of inflexion \( P_k \) on the graphs of \( f_k \). [4]

(b) Show that all \( P_k \) lie on a straight line and state its equation. [2]

(c) Show that for all values of \( k \), the tangents to the graphs of \( f_k \) at \( P_k \) are parallel, and find the equation of the tangent lines. [5]

▶️ Answer/Explanation
Markscheme Solution

(a) [6 marks]

(i) \( f’_k(x) = 3k^2 x^2 – 2kx + 1 \) (A1).
\( f”_k(x) = 6k^2 x – 2k \) (A1).

(ii) Set \( f”_k(x) = 0 \): \( 6k^2 x – 2k = 0 \implies x = \frac{1}{3k} \) (M1, A1).
Substitute into \( f_k(x) \): \( f_k\left(\frac{1}{3k}\right) = k^2 \left(\frac{1}{3k}\right)^3 – k \left(\frac{1}{3k}\right)^2 + \frac{1}{3k} \) (M1).
Simplify: \( = \frac{k^2}{27k^3} – \frac{k}{9k^2} + \frac{1}{3k} = \frac{1}{27k} – \frac{1}{9k} + \frac{1}{3k} = \frac{7}{27k} \) (A1).
Coordinates of \( P_k \): \( \left(\frac{1}{3k}, \frac{7}{27k}\right) \).

(b) [2 marks]

Line through \( P_k \): \( y – \frac{7}{27k} = \frac{7}{9} \left(x – \frac{1}{3k}\right) \) (M1).
Simplify: \( y = \frac{7}{9}x \) (A1).
Since the equation is independent of \( k \), all \( P_k \) lie on \( y = \frac{7}{9}x \) (R1).

(c) [5 marks]

Gradient at \( P_k \): \( f’_k\left(\frac{1}{3k}\right) = 3k^2 \left(\frac{1}{3k}\right)^2 – 2k \left(\frac{1}{3k}\right) + 1 = \frac{3k^2}{9k^2} – \frac{2k}{3k} + 1 = \frac{1}{3} – \frac{2}{3} + 1 = \frac{2}{3} \) (M1, A1).
Since the gradient \( \frac{2}{3} \) is independent of \( k \), the tangents are parallel (R1).
Equation: \( y – \frac{7}{27k} = \frac{2}{3} \left(x – \frac{1}{3k}\right) \) (M1).
Simplify: \( y = \frac{2}{3}x + \frac{1}{27k} \) (A1).

Markscheme Answers:

(a)
(i) \( f’_k(x) = 3k^2 x^2 – 2kx + 1 \) (A1)
\( f”_k(x) = 6k^2 x – 2k \) (A1)
(ii) \( x = \frac{1}{3k} \) (M1, A1), \( y = \frac{7}{27k} \) (M1, A1), \( P_k \left(\frac{1}{3k}, \frac{7}{27k}\right) \)

(b) \( y = \frac{7}{9}x \) (A1), all \( P_k \) lie on this line (R1)

(c) Gradient \( \frac{2}{3} \) (M1, A1), tangents parallel (R1), equation \( y = \frac{2}{3}x + \frac{1}{27k} \) (M1, A1)

Total [13 marks]

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