IB Math Analysis & Approaches Question bank-Topic: SL 5.5 Anti-differentiation with a boundary condition SL Paper 1

Question

Consider the function f with second derivative \(f”(x) = 3x – 1\) . The graph of f has a minimum point at A(2, 4) and a maximum point at \({\rm{B}}\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)\) .

Use the second derivative to justify that B is a maximum.

[3]
a.

Given that \(f'(x) = \frac{3}{2}{x^2} – x + p\) , show that \(p = – 4\) .

[4]
b.

Find \(f(x)\) .

[7]
c.
Answer/Explanation

Markscheme

substituting into the second derivative     M1

e.g. \(3 \times \left( { – \frac{4}{3}} \right) – 1\)

\(f”\left( { – \frac{4}{3}} \right) = – 5\)     A1

since the second derivative is negative, B is a maximum     R1     N0

[3 marks]

a.

setting \(f'(x)\) equal to zero     (M1)

evidence of substituting \(x = 2\) (or \(x = – \frac{4}{3}\) )     (M1)

e.g. \(f'(2)\)

correct substitution     A1

e.g. \(\frac{3}{2}{(2)^2} – 2 + p\) , \(\frac{3}{2}{\left( { – \frac{4}{3}} \right)^2} – \left( { – \frac{4}{3}} \right) + p\)

correct simplification

e.g. \(6 – 2 + p = 0\) , \(\frac{8}{3} + \frac{4}{3} + p = 0\) , \(4 + p = 0\)     A1

\(p = – 4\)     AG     N0

[4 marks]

b.

evidence of integration     (M1)

\(f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + c\)     A1A1A1

substituting (2, 4) or \(\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)\) into their expression     (M1)

correct equation     A1

e.g. \(\frac{1}{2} \times {2^3} – \frac{1}{2} \times {2^2} – 4 \times 2 + c = 4\) , \(\frac{1}{2} \times 8 – \frac{1}{2} \times 4 – 4 \times 2 + c = 4\) , \(4 – 2 – 8 + c = 4\)

\(f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + 10\)     A1     N4

[7 marks]

c.

Question

The graph of the function \(y = f(x)\) passes through the point \(\left( {\frac{3}{2},4} \right)\) . The gradient function of f is given as \(f'(x) = \sin (2x – 3)\) . Find \(f(x)\) .

Answer/Explanation

Markscheme

evidence of integration

e.g. \(f(x) = \int {\sin (2x – 3){\rm{d}}x} \)     (M1)

\( = – \frac{1}{2}\cos (2x – 3) + C\)     A1A1

substituting initial condition into their expression (even if C is missing)     M1

e.g. \(4 = – \frac{1}{2}\cos 0 + C\)

\(C = 4.5\)     (A1)

\(f(x) = – \frac{1}{2}\cos (2x – 3) + 4.5\)     A1     N5

[6 marks]

Question

Let \(f'(x) = 3{x^2} + 2\) . Given that \(f(2) = 5\) , find \(f(x)\) .

Answer/Explanation

Markscheme

evidence of anti-differentiation     (M1)

e.g.  \(\int {f'(x)} \) , \(\int {(3{x^2} + 2){\rm{d}}x} \)

\(f(x) = {x^3} + 2x + c\) (seen anywhere, including the answer)     A1A1

attempt to substitute (2, 5)     (M1)

e.g. \(f(2) = {(2)^3} + 2(2)\) , \(5 = 8 + 4 + c\)

finding the value of c     (A1)

e.g. \(5 = 12 + c\) , \(c = – 7\)

\(f(x) = {x^3} + 2x – 7\)     A1     N5

[6 marks]

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

A rocket moving in a straight line has velocity \(v\) km s–1 and displacement \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(v(t) = 6{{\rm{e}}^{2t}} + t\) . When \(t = 0\) , \(s = 10\) .

Find an expression for the displacement of the rocket in terms of \(t\) .

Answer/Explanation

Markscheme

evidence of anti-differentiation     (M1)

eg   \(\int {(6{{\rm{e}}^{2t}} + t)} \)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C\)     A2A1

Note: Award A2 for \(3{{\rm{e}}^{2t}}\) , A1 for \(\frac{{{t^2}}}{2}\) .

attempt to substitute (\(0\), \(10\)) into their integrated expression (even if \(C\) is missing)     (M1) 

correct working     (A1)

eg   \(10 = 3 + C\) , \(C = 7\)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\)     A1     N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of \({\rm{e}}\).

[7 marks]

Question

The graph of a function h passes through the point \(\left( {\frac{\pi }{{12}}, 5} \right)\).

Given that \(h'(x) = 4\cos 2x\), find \(h(x)\).

Answer/Explanation

Markscheme

evidence of anti-differentiation     (M1)

eg     \(\int {h'(x), \int {4\cos 2x{\text{d}}x} } \)

correct integration     (A2)

eg     \(h(x) = 2\sin 2x + c, \frac{{4\sin 2x}}{2}\)

attempt to substitute \(\left( {\frac{\pi }{{12}},5} \right)\) into their equation     (M1)

eg     \(2\sin \left( {2 \times \frac{\pi }{{12}}} \right) + c = 5,{\text{ }}2\sin \left( {\frac{\pi }{6}} \right) = 5\)

correct working     (A1)

eg     \(2\left( {\frac{1}{2}} \right) + c = 5,{\text{ }}c = 4\)

\(h(x) = 2\sin 2x + 4\)     A1     N5

[6 marks]

Question

The following diagram shows the graph of a function \(f\). There is a local minimum point at \(A\), where \(x > 0\).

The derivative of \(f\) is given by \(f'(x) = 3{x^2} – 8x – 3\).

Find the \(x\)-coordinate of \(A\).

[5]
a.

The \(y\)-intercept of the graph is at (\(0,6\)). Find an expression for \(f(x)\).

The graph of a function \(g\) is obtained by reflecting the graph of \(f\) in the \(y\)-axis, followed by a translation of \(\left({\begin{array}{*{20}{c}}m\\n\end{array}}\right)\).

[6]
b.

Find the \(x\)-coordinate of the local minimum point on the graph of \(g\).

[3]
c.
Answer/Explanation

Markscheme

recognizing that the local minimum occurs when \(f'(x) = 0\)     (M1)

valid attempt to solve \(3{x^2} – 8x – 3 = 0\)     (M1)

eg\(\;\;\;\)factorization, formula

correct working     A1

\((3x + 1)(x – 3),{\text{ }}x = \frac{{8 \pm \sqrt {64 + 36} }}{6}\)

\(x = 3\)     A2     N3

Note:     Award A1 if both values \(x = \frac{{ – 1}}{3},{\text{ }}x = 3\) are given.

[5 marks]

a.

valid approach     (M1)

\(f(x) = \int {f'(x){\text{d}}x} \)

\(f(x) = {x^3} – 4{x^2} – 3x + c\;\;\;\)(do not penalize for missing “\( + c\)”)     A1A1A1

\(c = 6\)     (A1)

\(f(x) = {x^3} – 4{x^2} – 3x + 6\)     A1     N6

[6 marks]

b.

applying reflection     (A1)

eg\(\;\;\;f( – x)\)

recognizing that the minimum is the image of \(A\)     (M1)

eg\(\;\;\;x =  – 3\)

correct expression for \(x\)     A1     N3

eg\(\;\;\; – 3 + m,{\text{ }}\left( {\begin{array}{*{20}{c}} { – 3 + m} \\ { – 12 + n} \end{array}} \right),{\text{ }}(m – 3,{\text{ }}n – 12)\)

[3 marks]

Total [14 marks]

c.

Question

Let \(f'(x) = 6{x^2} – 5\). Given that \(f(2) =  – 3\), find \(f(x)\).

Answer/Explanation

Markscheme

evidence of antidifferentiation     (M1)

eg\(\;\;\;f = \int {f’} \)

correct integration (accept absence of \(C\))     (A1)(A1)

\(f(x) = \frac{{6{x^3}}}{3} – 5x + C,{\text{ }}2{x^3} – 5x\)

attempt to substitute \((2,{\text{ }} – 3)\) into their integrated expression (must have \(C\))     M1

eg\(\;\;\;2{(2)^3} – 5(2) + C =  – 3,{\text{ }}16 – 10 + C =  – 3\)

Note:     Award M0 if substituted into original or differentiated function.

correct working to find \(C\)     (A1)

eg\(\;\;\;16 – 10 + C =  – 3,{\text{ }}6 + C =  – 3,{\text{ }}C =  – 9\)

\(f(x) = 2{x^3} – 5x – 9\)     A1     N4

[6 marks]

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