IBDP Maths AHL 5.13 The evaluation of limits AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) [3 marks]
Evaluate \( \lim_{x \to 0} \frac{\arctan(\cos x) – k}{x^2} \).
At \( x = 0 \): \( \cos 0 = 1 \), so \( \arctan(\cos 0) = \arctan 1 = \frac{\pi}{4} \).
Numerator: \( \arctan(\cos 0) – k = \frac{\pi}{4} – k \).
Denominator: \( x^2 = 0 \).
Form: \( \frac{\frac{\pi}{4} – k}{0} \). For a finite limit, the numerator must be zero to form \( \frac{0}{0} \) (M1).
Set: \( \frac{\pi}{4} – k = 0 \implies k = \frac{\pi}{4} \) (A1).
If \( k \neq \frac{\pi}{4} \), e.g., \( k = c \), then numerator \( \frac{\pi}{4} – c \neq 0 \), leading to \( \pm \infty \), not finite.
Thus, finite limit exists only for \( k = \frac{\pi}{4} \) (A1, AG).
(b) [6 marks]
With \( k = \frac{\pi}{4} \), evaluate \( \lim_{x \to 0} \frac{\arctan(\cos x) – \frac{\pi}{4}}{x^2} \).
At \( x = 0 \): \( \arctan(\cos 0) – \frac{\pi}{4} = \frac{\pi}{4} – \frac{\pi}{4} = 0 \), denominator \( x^2 = 0 \), form \( \frac{0}{0} \) (A1).
Apply l’Hôpital’s rule: \( \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \), where \( f(x) = \arctan(\cos x) – \frac{\pi}{4} \), \( g(x) = x^2 \) (M1).
Derivatives:
\( g'(x) = 2x \).
\( f'(x) = \frac{d}{dx} \left( \arctan(\cos x) – \frac{\pi}{4} \right) = \frac{1}{1 + (\cos x)^2} \cdot (-\sin x) = -\frac{\sin x}{1 + \cos^2 x} \) (A1).
Limit: \( \lim_{x \to 0} \frac{-\frac{\sin x}{1 + \cos^2 x}}{2x} = \lim_{x \to 0} -\frac{\sin x}{2x (1 + \cos^2 x)} \).
At \( x = 0 \): \( \sin 0 = 0 \), \( 2 \cdot 0 = 0 \), \( 1 + \cos^2 0 = 2 \), form \( \frac{0}{0} \) (A1).
Apply l’Hôpital’s rule again:
Numerator: \( p(x) = -\sin x \), \( p'(x) = -\cos x \).
Denominator: \( q(x) = 2x (1 + \cos^2 x) \), \( q'(x) = 2 (1 + \cos^2 x) + 2x \cdot (-2 \sin x \cos x) = 2 + 2 \cos^2 x – 4x \sin x \cos x \) (A1).
Limit: \( \lim_{x \to 0} \frac{-\cos x}{2 + 2 \cos^2 x – 4x \sin x \cos x} \).
At \( x = 0 \): Numerator: \( -\cos 0 = -1 \); Denominator: \( 2 + 2 \cdot 1 – 4 \cdot 0 \cdot 0 \cdot 1 = 4 \).
\( \frac{-1}{4} = -\frac{1}{4} \) (A1, AG).
Markscheme Answers:
(a) \( k = \frac{\pi}{4} \) (M1A1A1)
(b) \( -\frac{1}{4} \) (A1A1M1A1A1A1)
Total [9 marks]
▶️ Answer/Explanation
A.a. [4 marks]
Given \( f(x) = \ln (1 + \sin x) \), find the second derivative.
First derivative: Using the chain rule, \( f'(x) = \frac{1}{1 + \sin x} \cdot \cos x = \frac{\cos x}{1 + \sin x} \) (M1A1).
Second derivative: Apply the quotient rule with \( u = \cos x \), \( v = 1 + \sin x \), so \( u’ = -\sin x \), \( v’ = \cos x \).
\( f”(x) = \frac{(-\sin x)(1 + \sin x) – \cos x \cdot \cos x}{(1 + \sin x)^2} = \frac{-\sin x – \sin^2 x – \cos^2 x}{(1 + \sin x)^2} \) (M1).
Use \( \sin^2 x + \cos^2 x = 1 \): Numerator = \( -\sin x – 1 \).
Thus, \( f”(x) = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = \frac{-1}{1 + \sin x} \) (A1, AG).
A.b. [6 marks]
Maclaurin series: \( f(x) = f(0) + f'(0)x + \frac{f”(0)}{2!} x^2 + \frac{f”'(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \ldots \).
Evaluate derivatives at \( x = 0 \):
\( f(x) = \ln (1 + \sin x) \), \( f(0) = \ln (1 + 0) = 0 \).
\( f'(x) = \frac{\cos x}{1 + \sin x} \), \( f'(0) = \frac{\cos 0}{1 + 0} = 1 \).
\( f”(x) = \frac{-1}{1 + \sin x} \), \( f”(0) = \frac{-1}{1 + 0} = -1 \).
Third derivative: \( f”'(x) = \frac{d}{dx} \left( \frac{-1}{1 + \sin x} \right) = \frac{\cos x}{(1 + \sin x)^2} \) (A1).
\( f”'(0) = \frac{\cos 0}{1^2} = 1 \).
Fourth derivative: \( f^{(4)}(x) = \frac{d}{dx} \left( \frac{\cos x}{(1 + \sin x)^2} \right) \).
Quotient rule: Let \( u = \cos x \), \( v = (1 + \sin x)^2 \), so \( u’ = -\sin x \), \( v’ = 2(1 + \sin x) \cos x \).
\( f^{(4)}(x) = \frac{-\sin x (1 + \sin x)^2 – \cos x \cdot 2(1 + \sin x) \cos x}{(1 + \sin x)^4} \).
Numerator: \( -\sin x (1 + \sin x)^2 – 2 \cos^2 x (1 + \sin x) = (1 + \sin x) [-\sin x (1 + \sin x) – 2 \cos^2 x] \).
Simplify: \( -\sin x (1 + \sin x) – 2 \cos^2 x = -\sin x – \sin^2 x – 2 \cos^2 x \).
Use \( \sin^2 x + \cos^2 x = 1 \): \( -\sin x – (1 – \cos^2 x) – 2 \cos^2 x = -\sin x – 1 – \cos^2 x \).
So, \( f^{(4)}(x) = \frac{(1 + \sin x) (-\sin x – 1 – \cos^2 x)}{(1 + \sin x)^4} = \frac{-\sin x – 1 – \cos^2 x}{(1 + \sin x)^3} \).
At \( x = 0 \): Numerator = \( -0 – 1 – 1 = -2 \), denominator = \( (1 + 0)^3 = 1 \), so \( f^{(4)}(0) = -2 \) (A1).
Values: \( f(0) = 0 \), \( f'(0) = 1 \), \( f”(0) = -1 \), \( f”'(0) = 1 \), \( f^{(4)}(0) = -2 \) (A2).
Series: \( f(x) = 0 + x + \frac{-1}{2} x^2 + \frac{1}{6} x^3 + \frac{-2}{24} x^4 + \ldots = x – \frac{x^2}{2} + \frac{x^3}{6} – \frac{x^4}{12} + \ldots \) (M1A1).
A.c. [2 marks]
For \( \ln (1 – \sin x) \), use \( \ln (1 – \sin x) = \ln (1 + \sin (-x)) \).
From A.b: \( \ln (1 + \sin u) = u – \frac{u^2}{2} + \frac{u^3}{6} – \frac{u^4}{12} + \ldots \).
Let \( u = -x \): \( \ln (1 + \sin (-x)) = (-x) – \frac{(-x)^2}{2} + \frac{(-x)^3}{6} – \frac{(-x)^4}{12} + \ldots = -x – \frac{x^2}{2} – \frac{x^3}{6} – \frac{x^4}{12} + \ldots \) (M1A1).
A.d. [4 marks]
Add series: \( \ln (1 + \sin x) + \ln (1 – \sin x) \) (M1).
\( = \left( x – \frac{x^2}{2} + \frac{x^3}{6} – \frac{x^4}{12} + \ldots \right) + \left( -x – \frac{x^2}{2} – \frac{x^3}{6} – \frac{x^4}{12} + \ldots \right) \).
Combine: \( (x – x) + \left( -\frac{x^2}{2} – \frac{x^2}{2} \right) + \left( \frac{x^3}{6} – \frac{x^3}{6} \right) + \left( -\frac{x^4}{12} – \frac{x^4}{12} \right) + \ldots = -x^2 – \frac{x^4}{6} + \ldots \).
Since \( (1 + \sin x)(1 – \sin x) = \cos^2 x \), we have \( \ln (1 – \sin^2 x) = \ln \cos^2 x = 2 \ln \cos x \) (A1).
Thus, \( 2 \ln \cos x = -x^2 – \frac{x^4}{6} + \ldots \), so \( \ln \cos x = -\frac{x^2}{2} – \frac{x^4}{12} + \ldots \) (A1).
Hence, \( \ln \sec x = -\ln \cos x = \frac{x^2}{2} + \frac{x^4}{12} + \ldots \) (A1, AG).
A.e. [2 marks]
Find \( \lim_{x \to 0} \frac{\ln \sec x}{x \sqrt{x}} \).
From A.d: \( \ln \sec x = \frac{x^2}{2} + \frac{x^4}{12} + \ldots \).
Divide: \( \frac{\ln \sec x}{x \sqrt{x}} = \frac{\frac{x^2}{2} + \frac{x^4}{12} + \ldots}{x^{3/2}} = \frac{x^{1/2}}{2} + \frac{x^{5/2}}{12} + \ldots \) (M1).
As \( x \to 0 \): \( \frac{x^{1/2}}{2} \to 0 \), higher terms vanish, so limit is 0 (A1).
B.a. [5 marks]
Given interval [22.7, 26.1], \( n = 5 \), \( \sigma = 1.6 \).
Width: \( 26.1 – 22.7 = 3.4 \).
Mean: \( \bar{x} = \frac{22.7 + 26.1}{2} = 24.4 \).
Width formula: \( 2z \cdot \frac{\sigma}{\sqrt{n}} = 3.4 \), so \( 2z \cdot \frac{1.6}{\sqrt{5}} = 3.4 \) (M1).
Solve: \( z = \frac{3.4 \cdot \sqrt{5}}{2 \cdot 1.6} \approx 2.375 \) (A1A1).
Standard normal: \( P(Z < 2.375) \approx 0.9912 \) (A1).
Confidence level: \( 2 \cdot 0.9912 – 1 = 0.9824 \approx 98.2\% \) (A1).
B.b. [5 marks]
For 95% confidence, \( z = 1.96 \) (A1).
Width \( < 2 \): \( 2 \cdot \frac{1.96 \cdot 1.6}{\sqrt{n}} < 2 \implies \frac{3.136}{\sqrt{n}} < 1 \) (M1A1).
Solve: \( \sqrt{n} > 3.136 \implies n > 9.834 \) (A1).
Minimum \( n = 10 \) (A1).
Markscheme Answers:
A.a. \( f”(x) = \frac{-1}{1 + \sin x} \) (M1A1M1A1)
A.b. \( x – \frac{x^2}{2} + \frac{x^3}{6} – \frac{x^4}{12} + \ldots \) (A1A1A2M1A1)
A.c. \( -x – \frac{x^2}{2} – \frac{x^3}{6} – \frac{x^4}{12} + \ldots \) (M1A1)
A.d. \( \ln \sec x = \frac{x^2}{2} + \frac{x^4}{12} + \ldots \) (M1A1A1A1)
A.e. 0 (M1A1)
B.a. 98.2% (M1A1A1A1A1)
B.b. \( n = 10 \) (A1M1A1A1A1)
Total [28 marks]