# 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.

## 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 $$f(x)$$ .

## 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 $$f(x)$$ .

## 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 $$f(x)$$ .

## 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 $$t$$ .

## 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 $$h(x)$$.

## 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.

## 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 $$f(x)$$.

## 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]