IB Mathematics: Analysis and Approaches Quiz: Recursive Sequences And Series
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Recursive Sequences And SeriesQuestion 1 of 20

A sequence is defined by the recurrence relation un+1=f(un)u_{n+1} = f(u_n). The sequence is known to converge to a limit LL, where f(L)=Lf(L) = L. Which of the following conditions on the derivative of ff at LL is sufficient to ensure convergence for a starting value sufficiently close to LL?

f′(L)=0f'(L) = 0
∣f′(L)∣<1|f'(L)| < 1
f′(L)>1f'(L) > 1
∣f′(L)∣=1|f'(L)| = 1
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Recursive Sequences And Series

Practice Recursive Sequences And Series in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Recursive Sequences And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

A sequence is defined by the recurrence relation un+1=f(un)u_{n+1} = f(u_n). The sequence is known to converge to a limit LL, where f(L)=Lf(L) = L. Which of the following conditions on the derivative of ff at LL is sufficient to ensure convergence for a starting value sufficiently close to LL?

  1. f′(L)=0f'(L) = 0
  2. ∣f′(L)∣<1|f'(L)| < 1 (correct answer)
  3. f′(L)>1f'(L) > 1
  4. ∣f′(L)∣=1|f'(L)| = 1
Explanation: For a sequence defined by un+1=f(un)u_{n+1} = f(u_n) to converge to a fixed point LL, the fixed point must be stable or 'attracting'. The general condition for a stable fixed point is that the absolute value of the derivative of the function at the fixed point is less than 1, i.e., ∣f′(L)∣<1|f'(L)| < 1. If ∣f′(L)∣>1|f'(L)| > 1, the point is unstable (repelling). If ∣f′(L)∣=1|f'(L)| = 1, the test is inconclusive. While f′(L)=0f'(L)=0 is a condition for very rapid (quadratic) convergence and implies stability, ∣f′(L)∣<1|f'(L)| < 1 is the general sufficient condition.

Question 2

A sequence is defined by un+1=kun(1−un)u_{n+1} = k u_n (1-u_n), where kk is a positive constant and the initial value u1u_1 is in the interval (0,1)(0, 1). Given that the sequence converges to a limit L>0L > 0, find the range of values for kk.

  1. 0<k≤10 < k \le 1
  2. 1<k<31 < k < 3 (correct answer)
  3. k>3k > 3
  4. 0<k<40 < k < 4
Explanation: The limit LL must be a fixed point of f(x)=kx(1−x)f(x) = kx(1-x). We solve L=kL(1−L)L = kL(1-L). One solution is L=0L=0. Since we are given L>0L>0, we can divide by LL to get 1=k(1−L)1 = k(1-L), which gives L=1−1/kL = 1 - 1/k. For L>0L>0, we must have 1−1/k>01-1/k > 0, which means k>1k>1. For the sequence to converge to this limit, the fixed point must be stable. The derivative is f′(x)=k−2kxf'(x) = k - 2kx. The condition for stability is ∣f′(L)∣<1|f'(L)| < 1. Substituting L=1−1/kL = 1-1/k, we get f′(L)=k−2k(1−1/k)=k−(2k−2)=2−kf'(L) = k - 2k(1-1/k) = k - (2k-2) = 2-k. The stability condition is ∣2−k∣<1|2-k| < 1, which means −1<2−k<1-1 < 2-k < 1. This simplifies to −3<−k<−1-3 < -k < -1, or 1<k<31 < k < 3. This range also satisfies the earlier condition k>1k>1.

Question 3

Consider the sequence defined by un+1=3−unu_{n+1} = 3 - u_n and u1=1u_1 = 1. Which statement accurately describes the sequence?

  1. It converges to 1.5.
  2. It is periodic with period 2. (correct answer)
  3. It diverges to infinity.
  4. It converges to 1.
Explanation: Let's compute the first few terms of the sequence: u1=1u_1 = 1, u2=3−u1=3−1=2u_2 = 3 - u_1 = 3 - 1 = 2, u3=3−u2=3−2=1u_3 = 3 - u_2 = 3 - 2 = 1, u4=3−u3=3−1=2u_4 = 3 - u_3 = 3 - 1 = 2. The sequence of terms is 1, 2, 1, 2, ... . The sequence oscillates between 1 and 2, so it is periodic with period 2. It does not converge. The value 1.5 is the fixed point of the function f(x)=3−xf(x) = 3-x, since L=3−L  ⟹  2L=3  ⟹  L=1.5L=3-L \implies 2L=3 \implies L=1.5, but the sequence does not approach this value.

Question 4

The behavior of a sequence un+1=f(un)u_{n+1} = f(u_n) can be analyzed using a cobweb diagram, which involves the graphs of y=f(x)y=f(x) and y=xy=x. If the sequence converges to a limit LL, what does this correspond to on the diagram?

  1. A point of intersection of y=f(x)y=f(x) and the y-axis.
  2. A local maximum or minimum of the graph of y=f(x)y=f(x).
  3. The x-coordinate of a point of intersection of y=f(x)y=f(x) and y=xy=x. (correct answer)
  4. The slope of the tangent to y=f(x)y=f(x) at x=0x=0.
Explanation: A limit LL of a sequence defined by un+1=f(un)u_{n+1}=f(u_n) must be a fixed point of the function ff. This means LL must satisfy the equation L=f(L)L = f(L). Graphically, the solutions to this equation are the x-coordinates of the points where the graph of y=f(x)y=f(x) intersects the graph of y=xy=x. A cobweb diagram visually shows the sequence of values unu_n converging towards such an intersection point if the fixed point is stable.

Question 5

A sequence is defined by un+1=1.5un(1−un)u_{n+1} = 1.5u_n(1-u_n). This sequence is known to converge to its non-zero fixed point, L=1/3L=1/3. What is the largest open interval of starting values u1u_1 for which this convergence occurs?

  1. (0,1)(0, 1) (correct answer)
  2. (−∞,∞)(-\infty, \infty)
  3. (0,1/3)(0, 1/3)
  4. (1/3,1)(1/3, 1)
Explanation: When analyzing recursive sequences with fixed points, you need to determine the basin of attraction - the set of initial values that lead to convergence at the fixed point. For the sequence un+1=1.5un(1−un)u_{n+1} = 1.5u_n(1-u_n), let's find all fixed points by solving L=1.5L(1−L)L = 1.5L(1-L). This gives us L=1.5L−1.5L2L = 1.5L - 1.5L^2, which simplifies to 1.5L2−0.5L=01.5L^2 - 0.5L = 0, so L(1.5L−0.5)=0L(1.5L - 0.5) = 0. The fixed points are L=0L = 0 and L=13L = \frac{1}{3}. To find the basin of attraction for L=13L = \frac{1}{3}, we need to determine where the sequence behavior changes. The critical points occur when the derivative f′(u)=1.5(1−2u)=0f'(u) = 1.5(1-2u) = 0, giving us u=12u = \frac{1}{2}. At this point, f(12)=1.5⋅12⋅12=38f(\frac{1}{2}) = 1.5 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{8}. For this type of logistic map, convergence to the non-zero fixed point occurs when 0<u1<10 < u_1 < 1. Values outside this interval either diverge or behave chaotically. Choice A) (0,1)(0,1) correctly identifies this interval. Choice B) (−∞,∞)(-\infty, \infty) is wrong because negative values and values greater than 1 don't converge to 13\frac{1}{3}. Choice C) (0,13)(0, \frac{1}{3}) is too restrictive - initial values between 13\frac{1}{3} and 1 also converge. Choice D) (13,1)(\frac{1}{3}, 1) misses the convergent values below 13\frac{1}{3}. Study tip: For logistic maps un+1=run(1−un)u_{n+1} = ru_n(1-u_n), the basin of attraction for the non-zero fixed point typically spans the interval (0,1)(0,1) when the parameter allows convergence.

Question 6

The population PnP_n of a species in a protected area after nn years is modelled by the recurrence relation Pn+1=Pn+0.1Pn(1−Pn1000)P_{n+1} = P_n + 0.1 P_n \left(1 - \frac{P_n}{1000}\right). If the initial population P1P_1 is 200, what is the long-term stable population?

  1. 0
  2. 200
  3. 1000 (correct answer)
  4. The population grows indefinitely.
Explanation: The long-term stable population is a stable fixed point of the recurrence relation. Let the limit be LL. Then L=L+0.1L(1−L/1000)L = L + 0.1L(1 - L/1000). This simplifies to 0=0.1L(1−L/1000)0 = 0.1L(1 - L/1000). The solutions (equilibrium populations) are L=0L=0 and 1−L/1000=01 - L/1000 = 0, which gives L=1000L=1000. To check stability, let f(P)=P+0.1P(1−P/1000)=1.1P−0.0001P2f(P) = P + 0.1P(1-P/1000) = 1.1P - 0.0001P^2. Then f′(P)=1.1−0.0002Pf'(P) = 1.1 - 0.0002P. At L=0L=0, ∣f′(0)∣=1.1>1|f'(0)| = 1.1 > 1 (unstable). At L=1000L=1000, ∣f′(1000)∣=∣1.1−0.0002(1000)∣=∣1.1−0.2∣=0.9<1|f'(1000)| = |1.1 - 0.0002(1000)| = |1.1 - 0.2| = 0.9 < 1 (stable). Therefore, the population will converge to the stable equilibrium of 1000.

Question 7

A sequence is defined by the recurrence relation un+1=f(un)u_{n+1} = f(u_n). The sequence is known to converge to a limit LL, where f(L)=Lf(L) = L. Which of the following conditions on the derivative of ff at LL is sufficient to ensure convergence for a starting value sufficiently close to LL?

  1. f′(L)=0f'(L) = 0
  2. ∣f′(L)∣<1|f'(L)| < 1 (correct answer)
  3. f′(L)>1f'(L) > 1
  4. ∣f′(L)∣=1|f'(L)| = 1
Explanation: For a sequence defined by un+1=f(un)u_{n+1} = f(u_n) to converge to a fixed point LL, the fixed point must be stable or 'attracting'. The general condition for a stable fixed point is that the absolute value of the derivative of the function at the fixed point is less than 1, i.e., ∣f′(L)∣<1|f'(L)| < 1. If ∣f′(L)∣>1|f'(L)| > 1, the point is unstable (repelling). If ∣f′(L)∣=1|f'(L)| = 1, the test is inconclusive. While f′(L)=0f'(L)=0 is a condition for very rapid (quadratic) convergence and implies stability, ∣f′(L)∣<1|f'(L)| < 1 is the general sufficient condition.

Question 8

A sequence is defined by u1=1u_1=1 and un+1=unun+1u_{n+1} = \frac{u_n}{u_n+1} for n≥1n \ge 1. Find the value of ∑n=14un\sum_{n=1}^{4} u_n.

  1. 2512\frac{25}{12} (correct answer)
  2. 1312\frac{13}{12}
  3. 3512\frac{35}{12}
  4. 44
Explanation: When you encounter a recursively defined sequence, your first step is to calculate several terms using the given recurrence relation to identify any patterns before attempting to find the sum. Given u1=1u_1 = 1 and un+1=unun+1u_{n+1} = \frac{u_n}{u_n + 1}, let's find the first four terms: u1=1u_1 = 1 u2=u1u1+1=11+1=12u_2 = \frac{u_1}{u_1 + 1} = \frac{1}{1 + 1} = \frac{1}{2} u3=u2u2+1=1212+1=1232=13u_3 = \frac{u_2}{u_2 + 1} = \frac{\frac{1}{2}}{\frac{1}{2} + 1} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} u4=u3u3+1=1313+1=1343=14u_4 = \frac{u_3}{u_3 + 1} = \frac{\frac{1}{3}}{\frac{1}{3} + 1} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} Notice the pattern: un=1nu_n = \frac{1}{n}. Now we can find the sum: ∑n=14un=1+12+13+14=1212+612+412+312=2512\sum_{n=1}^{4} u_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{25}{12} This confirms answer A is correct. Answer B (1312\frac{13}{12}) would result from incorrectly calculating just u1+u2+u3u_1 + u_2 + u_3. Answer C (3512\frac{35}{12}) might come from computational errors in the fraction arithmetic. Answer D (4) suggests someone mistakenly thought the sum equals the number of terms. Strategy tip: With recursive sequences, always compute the first few terms explicitly to spot patterns. Many IB questions reward recognizing that complex recursive definitions often simplify to familiar sequences like 1n\frac{1}{n} or geometric progressions.

Question 9

The behavior of a sequence un+1=f(un)u_{n+1} = f(u_n) can be analyzed using a cobweb diagram, which involves the graphs of y=f(x)y=f(x) and y=xy=x. If the sequence converges to a limit LL, what does this correspond to on the diagram?

  1. A point of intersection of y=f(x)y=f(x) and the y-axis.
  2. A local maximum or minimum of the graph of y=f(x)y=f(x).
  3. The x-coordinate of a point of intersection of y=f(x)y=f(x) and y=xy=x. (correct answer)
  4. The slope of the tangent to y=f(x)y=f(x) at x=0x=0.
Explanation: A limit LL of a sequence defined by un+1=f(un)u_{n+1}=f(u_n) must be a fixed point of the function ff. This means LL must satisfy the equation L=f(L)L = f(L). Graphically, the solutions to this equation are the x-coordinates of the points where the graph of y=f(x)y=f(x) intersects the graph of y=xy=x. A cobweb diagram visually shows the sequence of values unu_n converging towards such an intersection point if the fixed point is stable.

Question 10

The sequence un+1=2unu_{n+1} = 2u_n has a fixed point at L=0L=0. This fixed point is unstable. If a sequence starts with a value u1u_1 very close to 0, but not equal to 0, how will the sequence behave?

  1. The terms will get closer and closer to 0.
  2. The terms will move away from 0. (correct answer)
  3. The terms will oscillate around 0.
  4. The terms will remain constant at u1u_1.
Explanation: An unstable fixed point, also known as a repelling fixed point, is one where nearby values are pushed away by the iteration. For un+1=2unu_{n+1} = 2u_n, each term is twice the previous term. So if u1=0.01u_1 = 0.01, the sequence is 0.01,0.02,0.04,...0.01, 0.02, 0.04, ..., whose terms are moving away from 0. This is confirmed by the stability condition: f(x)=2xf(x)=2x, so f′(x)=2f'(x)=2. At L=0L=0, ∣f′(0)∣=2>1|f'(0)|=2 > 1, confirming the point is unstable.

Question 11

The recurrence relation un+1=12(un+5un)u_{n+1} = \frac{1}{2}\left(u_n + \frac{5}{u_n}\right) with u1=2u_1 = 2 is an application of Newton's method to find a root of a function g(x)=0g(x)=0. What is the exact value of the limit of this sequence?

  1. 2.252.25
  2. 55
  3. 5\sqrt{5} (correct answer)
  4. The sequence has no limit.
Explanation: The limit LL of the sequence must satisfy L=12(L+5L)L = \frac{1}{2}\left(L + \frac{5}{L}\right). This simplifies to 2L2=L2+52L^2 = L^2 + 5, which gives L2=5L^2=5, so L=±5L = \pm\sqrt{5}. Since u1=2u_1=2 is positive, all subsequent terms will be positive, so the limit must be 5\sqrt{5}. This recurrence relation is generated by applying Newton's method, xn+1=xn−g(xn)g′(xn)x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}, to find the roots of the function g(x)=x2−5g(x) = x^2 - 5. Therefore, the limit is a root of x2−5=0x^2-5=0.

Question 12

The iteration xn+1=xn−(xn2−a)x_{n+1} = x_n - (x_n^2 - a) is used to find a solution to x2=ax^2=a, where aa is a positive constant. For which values of aa will this iteration converge to a\sqrt{a} if started sufficiently close to the root?

  1. a>1a > 1
  2. a=1a = 1
  3. 0<a<10 < a < 1 (correct answer)
  4. For all a>0a > 0
Explanation: The iteration is given by the function f(x)=x−x2+af(x) = x - x^2 + a. A fixed point LL satisfies L=L−L2+aL=L-L^2+a, which simplifies to L2=aL^2=a, so L=±aL = \pm\sqrt{a}. For the iteration to converge to L=aL=\sqrt{a}, this fixed point must be stable. We check the derivative: f′(x)=1−2xf'(x) = 1 - 2x. The stability condition is ∣f′(a)∣<1|f'(\sqrt{a})| < 1. This means ∣1−2a∣<1|1 - 2\sqrt{a}| < 1, which is equivalent to −1<1−2a<1-1 < 1 - 2\sqrt{a} < 1. Subtracting 1 gives −2<−2a<0-2 < -2\sqrt{a} < 0. Dividing by -2 and reversing the inequalities gives 0<a<10 < \sqrt{a} < 1. Squaring gives 0<a<10 < a < 1.

Question 13

The sequence un+1=2unu_{n+1} = 2u_n has a fixed point at L=0L=0. This fixed point is unstable. If a sequence starts with a value u1u_1 very close to 0, but not equal to 0, how will the sequence behave?

  1. The terms will get closer and closer to 0.
  2. The terms will move away from 0. (correct answer)
  3. The terms will oscillate around 0.
  4. The terms will remain constant at u1u_1.
Explanation: An unstable fixed point, also known as a repelling fixed point, is one where nearby values are pushed away by the iteration. For un+1=2unu_{n+1} = 2u_n, each term is twice the previous term. So if u1=0.01u_1 = 0.01, the sequence is 0.01,0.02,0.04,...0.01, 0.02, 0.04, ..., whose terms are moving away from 0. This is confirmed by the stability condition: f(x)=2xf(x)=2x, so f′(x)=2f'(x)=2. At L=0L=0, ∣f′(0)∣=2>1|f'(0)|=2 > 1, confirming the point is unstable.

Question 14

A sequence is defined by un+1=1.5un(1−un)u_{n+1} = 1.5u_n(1-u_n). This sequence is known to converge to its non-zero fixed point, L=1/3L=1/3. What is the largest open interval of starting values u1u_1 for which this convergence occurs?

  1. (0,1)(0, 1) (correct answer)
  2. (−∞,∞)(-\infty, \infty)
  3. (0,1/3)(0, 1/3)
  4. (1/3,1)(1/3, 1)
Explanation: When analyzing recursive sequences with fixed points, you need to determine the basin of attraction - the set of initial values that lead to convergence at the fixed point. For the sequence un+1=1.5un(1−un)u_{n+1} = 1.5u_n(1-u_n), let's find all fixed points by solving L=1.5L(1−L)L = 1.5L(1-L). This gives us L=1.5L−1.5L2L = 1.5L - 1.5L^2, which simplifies to 1.5L2−0.5L=01.5L^2 - 0.5L = 0, so L(1.5L−0.5)=0L(1.5L - 0.5) = 0. The fixed points are L=0L = 0 and L=13L = \frac{1}{3}. To find the basin of attraction for L=13L = \frac{1}{3}, we need to determine where the sequence behavior changes. The critical points occur when the derivative f′(u)=1.5(1−2u)=0f'(u) = 1.5(1-2u) = 0, giving us u=12u = \frac{1}{2}. At this point, f(12)=1.5⋅12⋅12=38f(\frac{1}{2}) = 1.5 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{8}. For this type of logistic map, convergence to the non-zero fixed point occurs when 0<u1<10 < u_1 < 1. Values outside this interval either diverge or behave chaotically. Choice A) (0,1)(0,1) correctly identifies this interval. Choice B) (−∞,∞)(-\infty, \infty) is wrong because negative values and values greater than 1 don't converge to 13\frac{1}{3}. Choice C) (0,13)(0, \frac{1}{3}) is too restrictive - initial values between 13\frac{1}{3} and 1 also converge. Choice D) (13,1)(\frac{1}{3}, 1) misses the convergent values below 13\frac{1}{3}. Study tip: For logistic maps un+1=run(1−un)u_{n+1} = ru_n(1-u_n), the basin of attraction for the non-zero fixed point typically spans the interval (0,1)(0,1) when the parameter allows convergence.

Question 15

The population PnP_n of a species in a protected area after nn years is modelled by the recurrence relation Pn+1=Pn+0.1Pn(1−Pn1000)P_{n+1} = P_n + 0.1 P_n \left(1 - \frac{P_n}{1000}\right). If the initial population P1P_1 is 200, what is the long-term stable population?

  1. 0
  2. 200
  3. 1000 (correct answer)
  4. The population grows indefinitely.
Explanation: The long-term stable population is a stable fixed point of the recurrence relation. Let the limit be LL. Then L=L+0.1L(1−L/1000)L = L + 0.1L(1 - L/1000). This simplifies to 0=0.1L(1−L/1000)0 = 0.1L(1 - L/1000). The solutions (equilibrium populations) are L=0L=0 and 1−L/1000=01 - L/1000 = 0, which gives L=1000L=1000. To check stability, let f(P)=P+0.1P(1−P/1000)=1.1P−0.0001P2f(P) = P + 0.1P(1-P/1000) = 1.1P - 0.0001P^2. Then f′(P)=1.1−0.0002Pf'(P) = 1.1 - 0.0002P. At L=0L=0, ∣f′(0)∣=1.1>1|f'(0)| = 1.1 > 1 (unstable). At L=1000L=1000, ∣f′(1000)∣=∣1.1−0.0002(1000)∣=∣1.1−0.2∣=0.9<1|f'(1000)| = |1.1 - 0.0002(1000)| = |1.1 - 0.2| = 0.9 < 1 (stable). Therefore, the population will converge to the stable equilibrium of 1000.

Question 16

The recurrence relation un+1=12(un+5un)u_{n+1} = \frac{1}{2}\left(u_n + \frac{5}{u_n}\right) with u1=2u_1 = 2 is an application of Newton's method to find a root of a function g(x)=0g(x)=0. What is the exact value of the limit of this sequence?

  1. 2.252.25
  2. 55
  3. 5\sqrt{5} (correct answer)
  4. The sequence has no limit.
Explanation: The limit LL of the sequence must satisfy L=12(L+5L)L = \frac{1}{2}\left(L + \frac{5}{L}\right). This simplifies to 2L2=L2+52L^2 = L^2 + 5, which gives L2=5L^2=5, so L=±5L = \pm\sqrt{5}. Since u1=2u_1=2 is positive, all subsequent terms will be positive, so the limit must be 5\sqrt{5}. This recurrence relation is generated by applying Newton's method, xn+1=xn−g(xn)g′(xn)x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}, to find the roots of the function g(x)=x2−5g(x) = x^2 - 5. Therefore, the limit is a root of x2−5=0x^2-5=0.

Question 17

The iteration xn+1=xn−(xn2−a)x_{n+1} = x_n - (x_n^2 - a) is used to find a solution to x2=ax^2=a, where aa is a positive constant. For which values of aa will this iteration converge to a\sqrt{a} if started sufficiently close to the root?

  1. a>1a > 1
  2. a=1a = 1
  3. 0<a<10 < a < 1 (correct answer)
  4. For all a>0a > 0
Explanation: The iteration is given by the function f(x)=x−x2+af(x) = x - x^2 + a. A fixed point LL satisfies L=L−L2+aL=L-L^2+a, which simplifies to L2=aL^2=a, so L=±aL = \pm\sqrt{a}. For the iteration to converge to L=aL=\sqrt{a}, this fixed point must be stable. We check the derivative: f′(x)=1−2xf'(x) = 1 - 2x. The stability condition is ∣f′(a)∣<1|f'(\sqrt{a})| < 1. This means ∣1−2a∣<1|1 - 2\sqrt{a}| < 1, which is equivalent to −1<1−2a<1-1 < 1 - 2\sqrt{a} < 1. Subtracting 1 gives −2<−2a<0-2 < -2\sqrt{a} < 0. Dividing by -2 and reversing the inequalities gives 0<a<10 < \sqrt{a} < 1. Squaring gives 0<a<10 < a < 1.

Question 18

A sequence is defined by un+1=kun(1−un)u_{n+1} = k u_n (1-u_n), where kk is a positive constant and the initial value u1u_1 is in the interval (0,1)(0, 1). Given that the sequence converges to a limit L>0L > 0, find the range of values for kk.

  1. 0<k≤10 < k \le 1
  2. 1<k<31 < k < 3 (correct answer)
  3. k>3k > 3
  4. 0<k<40 < k < 4
Explanation: The limit LL must be a fixed point of f(x)=kx(1−x)f(x) = kx(1-x). We solve L=kL(1−L)L = kL(1-L). One solution is L=0L=0. Since we are given L>0L>0, we can divide by LL to get 1=k(1−L)1 = k(1-L), which gives L=1−1/kL = 1 - 1/k. For L>0L>0, we must have 1−1/k>01-1/k > 0, which means k>1k>1. For the sequence to converge to this limit, the fixed point must be stable. The derivative is f′(x)=k−2kxf'(x) = k - 2kx. The condition for stability is ∣f′(L)∣<1|f'(L)| < 1. Substituting L=1−1/kL = 1-1/k, we get f′(L)=k−2k(1−1/k)=k−(2k−2)=2−kf'(L) = k - 2k(1-1/k) = k - (2k-2) = 2-k. The stability condition is ∣2−k∣<1|2-k| < 1, which means −1<2−k<1-1 < 2-k < 1. This simplifies to −3<−k<−1-3 < -k < -1, or 1<k<31 < k < 3. This range also satisfies the earlier condition k>1k>1.

Question 19

A sequence is defined by un+1=a+unu_{n+1} = \sqrt{a + u_n}, where a>0a > 0 and u1=1u_1 = 1. The sequence converges to a limit of 5. Find the value of aa.

  1. 4
  2. 5
  3. 20 (correct answer)
  4. 24
Explanation: If the sequence converges to a limit L=5L=5, then this limit must be a fixed point of the function f(x)=a+xf(x) = \sqrt{a+x}. Therefore, L=a+LL = \sqrt{a+L}. Substituting L=5L=5, we get 5=a+55 = \sqrt{a+5}. Squaring both sides gives 25=a+525 = a+5, which yields a=20a = 20. Distractor '24' comes from the error L2=a+u1  ⟹  25=a+1L^2 = a + u_1 \implies 25 = a+1. Distractor '4' comes from the error L=a+u1  ⟹  5=a+1L = a+u_1 \implies 5 = a+1.

Question 20

A sequence is defined by u1=0u_1=0 and un+1=12−unu_{n+1} = \frac{1}{2-u_n}. The sequence converges to a limit LL. Find the value of the derivative of the function f(x)=12−xf(x) = \frac{1}{2-x} at x=Lx=L.

  1. 0
  2. 1/21/2
  3. 1 (correct answer)
  4. 2
Explanation: First, we find the limit LL by solving the fixed-point equation L=12−LL = \frac{1}{2-L}. This gives L(2−L)=1L(2-L) = 1, which simplifies to 2L−L2=12L - L^2 = 1, or L2−2L+1=0L^2 - 2L + 1 = 0. This factors as (L−1)2=0(L-1)^2 = 0, so the only fixed point and thus the limit is L=1L=1. Next, we find the derivative of f(x)=(2−x)−1f(x) = (2-x)^{-1}. Using the chain rule, f′(x)=−1(2−x)−2(−1)=1(2−x)2f'(x) = -1(2-x)^{-2}(-1) = \frac{1}{(2-x)^2}. Finally, we evaluate the derivative at the limit L=1L=1: f′(1)=1(2−1)2=112=1f'(1) = \frac{1}{(2-1)^2} = \frac{1}{1^2} = 1.