IBDP Maths AA: Topic SL 5.10: Indefinite integral: IB style Questions HL Paper 2

Question

The function f has a derivative given by a positive constant.  \(f'(x) = \frac{1}{x(k-x)}\) \(x\in \mathbb{R},x\neq 0,x\neq k\) where \(k\) is a positive constant.

  1. The expression for f ′(x) can be written in the form \(\frac{a}{x}+\frac{b}{k-x}\) , where \(a,b\in \mathbb{R}\)

    Find a and b in terms of k. [3]

  2. Hence, find an expression for f (x) . [3]

    Consider P , the population of a colony of ants, which has an initial value of 1200.

    The rate of change of the population can be modelled by the differential equation \(\frac{dP}{dt}=\frac{p(k-p)}{5k}\)

    where t is the time measured in days, t ≥ 0, and k is the upper bound for the population.

  3. By solving the differential equation, show that  \(P =\frac{1200k}{(k-120)e^{\frac{t}{5}}+1200}\)

    At t = 10 the population of the colony has doubled in size from its initial value.

  4. Find the value of k , giving your answer correct to four significant figures. [3]

  5. Find the value of t when the rate of change of the population is at its maximum. [3]

▶️Answer/Explanation

Ans:

(a)

\(\frac{1}{x(k-x)}= \frac{a}{x}+ \frac{b}{k-x}\) 

a (k-x)+ bx = 1 

attempt to compare coefficients OR substitute x = k  and  x = o and solve

a = \(\frac{1}{k} \) and \( b = \frac{1}{k }\)

\(f ‘(x)= \frac{1}{kx}+ \frac{1}{k(k-x)}\)

(b)

attempt to integrate their \(\frac{a}{x}+ \frac{b}{k-x}\)

\(f(x)=\frac{1}{k}\int (\frac{1}{x}+ \frac{1}{k-x})dx\)

= \(\frac{1}{k}(ln |x| – ln | k-x |)+c\)

\(\frac{1}{k}ln |\frac{x}{k-x}|+c\)

(c) attempt to separate variables and integrate both sides

5k \(\int \frac{1}{p(k-p)}dp = \int 1 dt\)

5 (ln p – ln (k-P))= t+ c

EITHER 

attempt to substitute t = o , P = 1200 into an equation involving c

c= 5( ln 1200 – ln (k-1200))

= 5ln \(\frac{1200}{k-1200}\)

5 (ln p- ln (k-p ))= t+5 \((ln 1200-ln (k-1200))\)

\(ln (\frac{p(k-1200)}{1200(k-p)})=\frac{t}{5}\)

\(\frac{p(k-1200)}{1200(k-p)}= e ^{^{\frac{t}{5}}}\)

OR

ln \(\frac{p}{k-p}= \frac{t+ c }{5 }\)

\(\frac{p }{k-p }= Ae ^{\frac{t}{5}}\)

attempt to substitute t = 0, p = 1200

\(\frac{1200}{k-1200}= A\)

\(\frac{p }{k-p }= \frac{1200 e^{\frac{t}{5}}}{k- 1200}\)

THEN

attempt to rearrange and isolate  p

\(pk – 1200p = 1200ke ^{\frac{t}{5}}- 1200 pe ^{\frac{t}{5 }}\)

OR \(pke ^{\frac{-t}{5}}-1200pe^{\frac{-t}{5}}= 1200k – 1200p\)

OR

\(\frac{K}{P}-1 = \frac{K-1200}{1200e^{\frac{t}{5}}} \)

p = \(\frac{1200k}{(k-1200)e ^{\frac{t}{5}}+1200}\)

(d)

attempt to substitute t = 10 p = 2400

2400=\(\frac{1200k}{(k-1200)e^{-2}+ 1200 }\)

k = 2845.34..

k= 2845

(e)

attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that p =\(\frac{k}{2}(=1422.67..)\)

t =1.57814…

=1.58 (days)

Question

The graph of a function f passes through the point (ln4, 20).

Given that f ′(x) = 6e2x , find f (x).

▶️Answer/Explanation

Ans:

evidence of integration

eg \(\int f'(x)dx, \int 6e^{2x}\)

correct integration (accept missing + c)

eg \(\frac{1}{2}\times 6e^{2x}, 3e^{2x}\) +c

substituting initial condition into their integrated expression (must have +c )

eg \(3e^{2\times ln4}\)+c = 20

correct application of log (\(a^{b})\)= b log a rule (seen anywhere)

eg 2ln4 = ln16, eln16, ln42

correct application of  elna= a rule (seen anywhere)

eg \(e^{ln16}= 16, (e^{ln4})^{2}=4^{2}\)

correct working

eg \(3\times 16+c= 20, 3\times (4^{2})+c= 20, c=20 c=-28\)

\(f(x)=3e^{2x}- 28\)

Question

Let \(f(x) = \int {\frac{{12}}{{2x – 5}}} {\rm{d}}x\) , \(x > \frac{5}{2}\) . The graph of \(f\) passes through (\(4\), \(0\)) .

Find \(f(x)\) .

▶️Answer/Explanation

Markscheme

attempt to integrate which involves \(\ln \)     (M1)

eg   \(\ln (2x – 5)\) , \(12\ln 2x – 5\) , \(\ln 2x\)

correct expression (accept absence of \(C\))

eg   \(12\ln (2x – 5)\frac{1}{2} + C\) , \(6\ln (2x – 5)\)     A2

attempt to substitute (4,0) into their integrated     (M1)

eg \(0 = 6\ln (2 \times 4 – 5)\) , \(0 = 6\ln (8 – 5) + C\)

\(C = – 6\ln 3\)     (A1)

\(f(x) = 6\ln (2x – 5) – 6\ln 3\) \(\left( { = 6\ln \left( {\frac{{2x – 5}}{3}} \right)} \right)\) (accept \(6\ln (2x – 5) – \ln {3^6}\) )     A1     N5

Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve \(\ln\).

[6 marks]

Question

Given that \(\int_0^5 {\frac{2}{{2x + 5}}} {\rm{d}}x = \ln k\) , find the value of k .

▶️Answer/Explanation

Markscheme

correct integration, \(2 \times \frac{1}{2}\ln (2x + 5)\)     A1A1

Note: Award A1 for \(2 \times \frac{1}{2}( = 1)\) and A1 for \(\ln (2x + 5)\) .

evidence of substituting limits into integrated function and subtracting     (M1)

e.g. \(\ln (2 \times 5 + 5) – \ln (2 \times 0 + 5)\)

correct substitution     A1

e.g. \(\ln 15 – \ln 5\)

correct working     (A1)

e.g. \(\ln \frac{{15}}{5},\ln 3\)

\(k = 3\)     A1     N3 

[6 marks]

Question

The following diagram shows the graph of \(f(x) = \frac{x}{{{x^2} + 1}}\), for \(0 \le x \le 4\), and the line \(x = 4\).

Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis and the line \(x = 4\).

Find the area of \(R\).

▶️Answer/Explanation

Markscheme

substitution of limits or function     (A1)

eg\(\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

\(\frac{1}{2}\ln ({x^2} + 1)\)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right)\)

correct working     A1

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) – \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 – 0\)

\(A = \frac{1}{2}\ln (17)\)     A1     N3

Note:     Exception to FT rule. Allow full FT on incorrect integration involving a \(\ln \) function.

[6 marks]

Question

Let \(f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}\). Given that \(f(0) = 1\), find \(f(x)\).

▶️Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

eg\(\,\,\,\,\,\)\(f(x) =  – \frac{1}{4}{({x^3} + 1)^{ – 4}} + c,{\text{ }}\frac{{ – 1}}{{4{{({x^3} + 1)}^4}}}\)

correct substitution into their integrated function (must include \(c\))     M1

eg\(\,\,\,\,\,\)\(1 = \frac{{ – 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} – \frac{1}{4} + c = 1\)

Note:     Award M0 if candidates substitute into \(f’\) or \(f’’\).

\(c = \frac{5}{4}\)     (A1)

\(f(x) =  – \frac{1}{4}{({x^3} + 1)^{ – 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ – 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} – 1}}{{4{{({x^3} + 1)}^4}}}} \right)\)     A1     N4

[6 marks]

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