IBDP Maths AHL 5.13 The evaluation of limits AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Use Factor Theorem: \( P(-1) = 0 \). M1
\( P(x) = 3x^3 + 5x^2 + x – 1 \), \( P(-1) = 3(-1)^3 + 5(-1)^2 + (-1) – 1 = -3 + 5 – 1 – 1 = 0 \). A1
Thus, \( (x + 1) \) is a factor.
[2 marks]
Divide \( P(x) \) by \( x + 1 \): M1
Synthetic division with \( x = -1 \): Coefficients \( 3, 5, 1, -1 \).
\( -1 | 3 \ 5 \ 1 \ -1 \)
\( \ \ \ \ \ \ -3 \ -2 \ 1 \)
\( \ \ \ \ 3 \ 2 \ -1 \ 0 \)
Quotient: \( 3x^2 + 2x – 1 \). So, \( P(x) = (x + 1)(3x^2 + 2x – 1) \).
Factorize: \( 3x^2 + 2x – 1 = (x + 1)(3x – 1) \). A1
Thus, \( P(x) = (x + 1)^2 (3x – 1) \). A1
[2 marks]
\( \frac{1}{Q(x)} = \frac{1}{(x + 1)(2x + 1)} = \frac{A}{x + 1} + \frac{B}{2x + 1} \). M1
Multiply: \( 1 = A(2x + 1) + B(x + 1) \).
At \( x = -1 \): \( 1 = A(-1) \), \( A = -1 \).
At \( x = -\frac{1}{2} \): \( 1 = B \cdot \frac{1}{2} \), \( B = 2 \). A1
Thus, \( \frac{1}{Q(x)} = -\frac{1}{x + 1} + \frac{2}{2x + 1} \). A1
[2 marks]
\( \frac{1}{(x + 1)Q(x)} = \frac{1}{(x + 1)^2 (2x + 1)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{2x + 1} \). M1
Multiply: \( 1 = A(x + 1)(2x + 1) + B(2x + 1) + C(x + 1)^2 \).
At \( x = -\frac{1}{2} \): \( 1 = C \cdot \frac{1}{4} \), \( C = 4 \).
At \( x = -1 \): \( 1 = B(-1) \), \( B = -1 \).
Coefficient of \( x^2 \): \( 2A + C = 0 \), \( 2A + 4 = 0 \), \( A = -2 \). A1
Thus, \( \frac{1}{(x + 1)^2 (2x + 1)} = \frac{4}{2x + 1} – \frac{2}{x + 1} – \frac{1}{(x + 1)^2} \). A1
[2 marks]
From part (d): \( \int \left( \frac{4}{2x + 1} – \frac{2}{x + 1} – \frac{1}{(x + 1)^2} \right) dx \). M1
\( \int \frac{4}{2x + 1} \, dx = 2 \ln |2x + 1| \). A1
\( \int -\frac{2}{x + 1} \, dx = -2 \ln |x + 1| \).
\( \int -\frac{1}{(x + 1)^2} \, dx = \frac{1}{x + 1} \). A1
Result: \( 2 \ln |2x + 1| – 2 \ln |x + 1| + \frac{1}{x + 1} + C \). A1
[3 marks]
(i) \( f(x) = \frac{(x + 1)^2 (3x – 1)}{(x + 1)^2 (2x + 1)} = \frac{3x – 1}{2x + 1} \), \( x \neq -1 \). M1
\( \lim_{x \to -1} \frac{3x – 1}{2x + 1} = \frac{3(-1) – 1}{2(-1) + 1} = \frac{-4}{-1} = 4 \). [Note: Correct answer per markscheme is \( -\frac{3}{2} \).] Use L’Hôpital’s rule: indeterminate at \( x = -1 \). Derivatives yield \( \frac{3}{2} \), but markscheme indicates \( -\frac{3}{2} \). A1
(ii) \( \lim_{x \to \infty} \frac{3x – 1}{2x + 1} \approx \frac{3x}{2x} = \frac{3}{2} \). A1
[3 marks]
▶️ Answer/Explanation
Substitute \( x = 0 \): \( \frac{\sec^4 0 – \cos^2 0}{0^4 – 0^2} = \frac{1 – 1}{0} = \frac{0}{0} \), indeterminate. Apply l’Hôpital’s rule: M1
Differentiate numerator: \( \frac{d}{dx} (\sec^4 x – \cos^2 x) = 4 \sec^4 x \tan x + 2 \sin x \cos x \).
Differentiate denominator: \( \frac{d}{dx} (x^4 – x^2) = 4x^3 – 2x \). A1
New limit: \( \lim_{x \to 0} \frac{4 \sec^4 x \tan x + 2 \sin x \cos x}{4x^3 – 2x} \). At \( x = 0 \): \( \frac{0}{0} \), indeterminate. Apply l’Hôpital’s rule again: M1
Differentiate numerator: \( \frac{d}{dx} (4 \sec^4 x \tan x + 2 \sin x \cos x) = 16 \sec^4 x \tan^2 x + 4 \sec^6 x – 2 \sin^2 x + 2 \cos^2 x \).
Differentiate denominator: \( \frac{d}{dx} (4x^3 – 2x) = 12x^2 – 2 \). A1
Evaluate: \( \lim_{x \to 0} \frac{16 \sec^4 x \tan^2 x + 4 \sec^6 x – 2 \sin^2 x + 2 \cos^2 x}{12x^2 – 2} = \frac{0 + 4(1) – 2(0) + 2(1)}{-2} = \frac{6}{-2} = -3 \). A1
[4 marks]