Home / IBDP Maths AHL 5.12 Informal ideas of limit, continuity and convergence AA HL Paper 2- Exam Style Questions

IBDP Maths AHL 5.12 Informal ideas of limit, continuity and convergence AA HL Paper 2- Exam Style Questions

IBDP Maths AHL 5.12 Informal ideas of limit, continuity and convergence AA HL Paper 2- Exam Style Questions- New Syllabus

IB DP Maths HL Papers All Topics
Question

Consider the function \( f(x) = \frac{\sqrt{x}}{\sin x} \), \( 0 < x < \pi \).

Consider the region bounded by the curve \( y = f(x) \), the \( x \)-axis, and the lines \( x = \frac{\pi}{6} \), \( x = \frac{\pi}{3} \).

(a.i) Find the values of the constants for showing that the \( x \)-coordinate of the minimum point on the curve \( y = f(x) \) satisfies the equation \( \tan x = 2x \) [5]

(a.ii) Find the values of the constants for the values of \( x \) for which \( f(x) \) is a decreasing function [2]

(b) Find the values of the constants for sketching the graph of \( y = f(x) \) showing clearly the minimum point and any asymptotic behaviour [3]

(c) Find the values of the constants for the coordinates of the point on the graph of \( f \) where the normal to the graph is parallel to the line \( y = -x \) [4]

(d) Find the values of the constants for the volume of revolution when this region is rotated through \( 2\pi \) radians about the \( x \)-axis [3]

▶️ Answer/Explanation
Markscheme Solution

[13 marks]

(a.i) \( f'(x) = \frac{\sin x \cdot \frac{1}{2}x^{-\frac{1}{2}} – \sqrt{x} \cdot \cos x}{\sin^2 x} \) (M1, A1, A1).
Setting \( f'(x) = 0 \): \( \frac{\sin x}{2\sqrt{x}} – \sqrt{x} \cos x = 0 \) (M1).
\( \frac{\sin x}{2\sqrt{x}} = \sqrt{x} \cos x \) (A1).
\( \tan x = 2x \) (AG).
(a.ii) \( x = 1.17 \), so \( 0 < x \leq 1.17 \) (A1, A1).
(b) Concave up curve over \( 0 < x < \pi \) with one minimum point above the \( x \)-axis, approaches \( x = 0 \) and \( x = \pi \) asymptotically (A1, A1, A1).

Graph of f(x)
(c) \( f'(x) = 1 \): \( \frac{\sin x \cdot \frac{1}{2}x^{-\frac{1}{2}} – \sqrt{x} \cos x}{\sin^2 x} = 1 \) (A1).
Solving gives \( x = 1.96 \) (M1, A1).
\( y = f(1.96) = 1.51 \) (A1).
(d) \( V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx = 2.68 \) or \( 0.852\pi \) (M1, A1, A1).

Markscheme Answers:

(a.i) Quotient or product rule (M1), \( f'(x) = \frac{\sin x \cdot \frac{1}{2}x^{-\frac{1}{2}} – \sqrt{x} \cos x}{\sin^2 x} \) (A1, A1), \( f'(x) = 0 \) (M1), \( \tan x = 2x \) (A1)

(a.ii) \( x = 1.17 \) (A1), \( 0 < x \leq 1.17 \) (A1)

(b) Concave up curve with minimum (A1), asymptote at \( x = 0 \) (A1), asymptote at \( x = \pi \) (A1)

(c) \( f'(x) = 1 \) (A1), solving (M1), \( x = 1.96 \) (A1), \( y = 1.51 \) (A1)

(d) \( \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx \) (M1, A1), \( 2.68 \) or \( 0.852\pi \) (A1)

[Total 13 marks]

More DP Math AA HL Paper 2 Questions..
Scroll to Top