# IB Math Analysis & Approaches Question bank-Topic: SL 5.7 The second derivative SL Paper 1

## Question

A function f has its first derivative given by $$f'(x) = {(x – 3)^3}$$ .

Find the second derivative.

[2]
a.

Find $$f'(3)$$ and $$f”(3)$$ .

[1]
b.

The point P on the graph of f has x-coordinate $$3$$. Explain why P is not a point of inflexion.

[2]
c.

## Markscheme

METHOD 1

$$f”(x) = 3{(x – 3)^2}$$Â Â Â Â  A2Â Â Â Â  N2

METHOD 2

attempt to expand $${(x – 3)^3}$$Â Â Â  Â (M1)

e.g. $$f'(x) = {x^3} – 9{x^2} + 27x – 27$$

$$f”(x) = 3{x^2} – 18x + 27$$Â Â Â Â  A1Â Â Â Â  N2

[2 marks]

a.

$$f'(3) = 0$$ , $$f”(3) = 0$$Â Â Â  Â A1 Â  Â  N1

[1 mark]

b.

METHOD 1

$${f”}$$ does not change sign at PÂ Â Â Â  R1

evidence for thisÂ Â Â Â  R1 Â  Â  N0

METHOD 2

$${f’}$$ changes sign at P so P is a maximum/minimum (i.e. not inflexion)Â Â Â Â  R1

evidence for thisÂ Â Â Â  R1Â Â Â Â  N0

METHOD 3

finding $$f(x) = \frac{1}{4}{(x – 3)^4} + c$$Â and sketching this functionÂ Â Â Â  R1

indicating minimum at $$x = 3$$Â Â Â Â Â R1 Â  Â  N0Â

[2 marks]

c.

## Question

Let $$f(x) = \frac{{\cos x}}{{\sin x}}$$ , for $$\sin x \ne 0$$ .

In the following table, $$f’\left( {\frac{\pi }{2}} \right) = p$$Â and $$f”\left( {\frac{\pi }{2}} \right) = q$$Â . The table also gives approximate values of $$f'(x)$$ and $$f”(x)$$ near $$x = \frac{\pi }{2}$$ .

Use the quotient rule to show that $$f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}$$ .

[5]
a.

FindÂ $$f”(x)$$ .

[3]
b.

Find the value of p and of q.

[3]
c.

Use information from the table to explain why there is a point of inflexion on the graph of f where $$x = \frac{\pi }{2}$$ .

[2]
d.

## Markscheme

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x$$Â (seen anywhere)Â Â Â Â  (A1)(A1)

evidence of using the quotient ruleÂ Â Â Â  M1

correct substitutionÂ Â Â Â  A1

e.g. $$\frac{{\sin x( – \sin x) – \cos x(\cos x)}}{{{{\sin }^2}x}}$$Â , $$\frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}}$$

$$f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}$$Â Â Â Â  A1

$$f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}$$Â Â Â Â  AG Â  Â Â  N0

[5 marks]

a.

METHOD 1

appropriate approachÂ Â Â Â  (M1)

e.g. $$f'(x) = – {(\sin x)^{ – 2}}$$

$$f”(x) = 2({\sin ^{ – 3}}x)(\cos x)$$Â $$\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$$Â Â Â Â Â A1A1 Â  Â  N3

Note: Award A1 for $$2{\sin ^{ – 3}}x$$Â , A1 for $$\cos x$$Â .

METHOD 2

derivative of $${\sin ^2}x = 2\sin x\cos x$$Â (seen anywhere)Â Â Â Â  A1

evidence of choosing quotient ruleÂ Â Â Â  (M1)

e.g. $$u = – 1$$ , Â $$v = {\sin ^2}x$$ , $$f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$$

$$f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$$Â $$\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$$Â Â Â Â Â A1Â Â Â Â  N3

[3 marks]

b.

evidence of substituting $$\frac{\pi }{2}$$Â Â Â Â Â M1

e.g. $$\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}$$ , $$\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}$$

$$p = – 1$$ , Â $$q = 0$$Â Â Â  A1A1 Â  Â  N1N1

[3 marks]

c.

second derivative is zero, second derivative changes signÂ Â Â Â  R1R1Â Â Â Â  N2

[2 marks]

d.

## Question

Consider $$f(x) = \ln ({x^4} + 1)$$ .

The second derivative is given by $$f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}$$ .

The equation $$f”(x) = 0$$ has only three solutions, when $$x = 0$$ , $$\pm \sqrt[4]{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .

[2]
a.

Find the set of values of $$x$$ for which $$f$$ is increasing.

[5]
b.

(i)Â Â Â Â  Find $$f”(1)$$ .

(ii)Â Â Â Â  Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .

[5]
c.

There is a point of inflexion on the graph of $$f$$ at $$x = \sqrt[4]{3}$$ $$(x = 1.316 \ldots )$$ .

Sketch the graph of $$f$$ , for $$x \ge 0$$ .

[3]
d.

## Markscheme

substitute $$0$$ into $$f$$Â Â Â Â  (M1)

eg Â  $$\ln (0 + 1)$$Â , $$\ln 1$$

$$f(0) = 0$$Â Â Â Â  A1 N2

[2 marks]

a.

$$f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}$$ (seen anywhere)Â Â Â Â  A1A1

Note: Award A1 for $$\frac{1}{{{x^4} + 1}}$$Â and A1 for $$4{x^3}$$Â .

Â

recognizing $$f$$ increasing where $$f'(x) > 0$$Â (seen anywhere)Â Â Â Â  R1

eg Â  $$f'(x) > 0$$Â , diagram of signs

attempt to solve $$f'(x) > 0$$Â Â Â Â Â (M1)

eg Â  $$4{x^3} = 0$$Â , $${x^3} > 0$$

$$f$$ increasing for $$x > 0$$Â (accept $$x \ge 0$$Â )Â Â Â Â  A1 Â  Â  N1

[5 marks]

b.

(i)Â Â Â Â  substituting $$x = 1$$Â into $$f”$$Â Â Â Â Â (A1)

egÂ Â  $$\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}$$Â , $$\frac{{4 \times 2}}{4}$$

$$f”(1) = 2$$Â Â Â Â  A1Â Â Â Â  N2

Â

(ii)Â Â Â Â  valid interpretation of point of inflexion (seen anywhere)Â Â Â Â  R1

egÂ Â  no change of sign in $$f”(x)$$Â , no change in concavity,

$$f’$$ increasing both sides of zero

attempt to find $$f”(x)$$Â for $$x < 0$$Â Â Â Â Â (M1)

eg Â  $$f”( – 1)$$Â , $$\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}$$Â , diagram of signs

correct working leading to positive valueÂ Â Â Â  A1

egÂ Â  $$f”( – 1) = 2$$Â , discussing signs of numerator and denominator

there is no point of inflexion at $$x = 0$$Â Â Â Â Â AGÂ Â Â Â  N0

Â

[5 marks]

c.

Â Â Â Â  A1A1A1Â Â Â Â  N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Â Â Â  Only if this A1 is awarded, then award the following:

Â Â Â  A1 for curve through ($$0$$, $$0$$)Â , A1 for increasing throughout.

Â Â Â  Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

## Question

A function $$f$$ has its derivative given by $$f'(x) = 3{x^2} – 2kx – 9$$, where $$k$$ is a constant.

Find $$f”(x)$$.

[2]
a.

The graph of $$f$$ has a point of inflexion when $$x = 1$$.

Show that $$k = 3$$.

[3]
b.

Find $$f'( – 2)$$.

[2]
c.

Find the equation of the tangent to the curve of $$f$$ at $$( – 2,{\text{ }}1)$$, giving your answer in the form $$y = ax + b$$.

[4]
d.

Given that $$f'( – 1) = 0$$, explain why the graph of $$f$$ has a local maximum when $$x =Â – 1$$.

[3]
e.

## Markscheme

$$f”(x) = 6x – 2k$$Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

a.

substituting $$x = 1$$ into $$f”$$ Â  Â  (M1)

eg$$\;\;\;f”(1),{\text{ }}6(1) – 2k$$

recognizing $$f”(x) = 0\;\;\;$$(seen anywhere) Â  Â  M1

correct equation Â  Â  A1

eg$$\;\;\;6 – 2k = 0$$

$$k = 3$$ Â  Â  AG Â  Â  N0

[3 marks]

b.

correct substitution into $$f'(x)$$ Â  Â  (A1)

eg$$\;\;\;3{( – 2)^2} – 6( – 2) – 9$$

$$f'( – 2) = 15$$ Â  Â  A1 Â  Â  N2

[2 marks]

c.

recognizing gradient value (may be seen in equation) Â  Â  M1

eg$$\;\;\;a = 15,{\text{ }}y = 15x + b$$

attempt to substitute $$( – 2,{\text{ }}1)$$ into equation of a straight line Â  Â  M1

eg$$\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)$$

correct working Â  Â  (A1)

eg$$\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1$$

$$y = 15x + 31$$ Â  Â  A1 Â  Â  N2

[4 marks]

d.

METHOD 1 ($${{\text{2}}^{{\text{nd}}}}$$ derivative)

recognizing $$f” < 0\;\;\;$$(seen anywhere) Â  Â  R1

substituting $$x =Â – 1$$ into $$f”$$ Â  Â  (M1)

eg$$\;\;\;f”( – 1),{\text{ }}6( – 1) – 6$$

$$f”( – 1) =Â – 12$$ Â  Â  A1

therefore the graph of $$f$$ has a local maximum when $$x =Â – 1$$ Â  Â  AG Â  Â  N0

METHOD 2 ($${{\text{1}}^{{\text{st}}}}$$ derivative)

recognizing change of sign of $$f'(x)\;\;\;$$(seen anywhere) Â  Â  R1

eg$$\;\;\;$$sign chart$$\;\;\;$$

correct value of $$f’$$ for $$– 1 < x < 3$$ Â  Â  A1

eg$$\;\;\;f'(0) =Â – 9$$

correct value of $$f’$$ for $$x$$ value to the left of $$– 1$$ Â  Â  A1

eg$$\;\;\;f'( – 2) = 15$$

therefore the graph of $$f$$ has a local maximum when $$x =Â – 1$$ Â  Â  AG Â  Â  N0

[3 marks]

Total [14 marks]

e.

## Question

Let $$f(x) = \cos x$$.

Let $$g(x) = {x^k}$$, where $$k \in {\mathbb{Z}^ + }$$.

LetÂ $$k = 21$$ andÂ $$h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)$$.

(i) Â  Â  Find the first four derivatives of $$f(x)$$.

(ii) Â  Â  Find $${f^{(19)}}(x)$$.

[4]
a.

(i) Â  Â  Find the first three derivatives of $$g(x)$$.

(ii) Â  Â  Given that $${g^{(19)}}(x) = \frac{{k!}}{{(k – p)!}}({x^{k – 19}})$$, find $$p$$.

[5]
b.

(i) Â  Â  Find $$h'(x)$$.

(ii) Â  Â  Hence, show that $$h'(\pi ) = \frac{{ – 21!}}{2}{\pi ^2}$$.

[7]
c.

## Markscheme

(i) Â  Â  $$f'(x) = Â – \sin x,{\text{ }}f”(x) = Â – \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x$$Â Â  Â  A2 Â  Â  N2

(ii) Â  Â  valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$recognizing that 19 is one less than a multiple of 4,Â $${f^{(19)}}(x) = {f^{(3)}}(x)$$

$${f^{(19)}}(x) = \sin x$$Â Â  Â  A1 Â  Â  N2

[4 marks]

a.

(i) Â  Â Â $$g'(x) = k{x^{k – 1}}$$

$$g”(x) = k(k – 1){x^{k – 2}},{\text{ }}{g^{(3)}}(x) = k(k – 1)(k – 2){x^{k – 3}}$$Â Â  Â  A1A1 Â  Â  N2

(ii) Â  Â  METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative Â  Â  A2

eg$$\,\,\,\,\,$$$$k(k – 1)(k – 2) \ldots (k – 18) \times \frac{{(k – 19)!}}{{(k – 19)!}},{{\text{ }}_k}{P_{19}}$$

$$p = 19$$Â (accept $$\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}$$) Â  Â  A1 Â  Â  N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient Â  Â  A2

eg$$\,\,\,\,\,$$$$g” = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k – 1)(k – 2) = \frac{{k!}}{{(k – 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k – 3}})$$

$${g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k – 19)! \times 19!}},{{\text{ }}_k}{P_{19}}$$

$$p = 19$$Â (accept $$\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}$$) Â  Â  A1 Â  Â  N1

[5 marks]

b.

(i) Â  Â  valid approach using product rule Â  Â  (M1)

eg$$\,\,\,\,\,$$$$uv’ + vu’,{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}$$

correct 20th derivatives (must be seen in product rule) Â  Â  (A1)(A1)

eg$$\,\,\,\,\,$$$${g^{(20)}}(x) = \frac{{21!}}{{(21 – 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x$$

$$h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)$$ Â  Â A1 Â  Â  N3

(ii) Â  Â  substituting $$x = \pi$$ (seen anywhere) Â  Â  (A1)

eg$$\,\,\,\,\,$$$${f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi Â + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}$$

evidence of one correct value for $$\sin \pi$$ or $$\cos \pi$$Â (seen anywhere) Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\sin \pi Â = 0,{\text{ }}\cos \pi Â = Â – 1$$

evidence of correct values substituted into $$h'(\pi )$$Â Â  Â  A1

eg$$\,\,\,\,\,$$$$21!(\pi )\left( {0 – \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { – \frac{\pi }{2}} \right),{\text{ }}0 + ( – 1)\frac{{21!}}{2}{\pi ^2}$$

Note: If candidates write only the first line followed by the answer, award A1A0A0.

$$\frac{{ – 21!}}{2}{\pi ^2}$$Â Â  Â  AG Â  Â  N0

[7 marks]

c.

## Question

A function fâ€‰(x) has derivative fâ€‰â€²(x) = 3x2 + 18x. The graph of f has an x-intercept at x = âˆ’1.

FindÂ fâ€‰(x).

[6]
a.

The graph of f has a point of inflexion at x = p. Find p.

[4]
b.

Find the values of x for which the graph of f is concave-down.

[3]
c.

## Markscheme

evidence of integrationÂ  Â  Â  Â (M1)

egÂ Â $$\int {f’\left( x \right)}$$

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

egÂ  $${x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}$$

attempt to substitute x = âˆ’1 into their fÂ = 0 (must have C)Â  Â  Â  M1

egÂ Â $${\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0$$

Note: Award M0 if they substitute into original or differentiated function.

correct workingÂ  Â  Â  Â (A1)

egÂ Â $$8 + C = 0,\,\,\,C =Â – 8$$

$$f\left( x \right) = {x^3} + 9{x^2} – 8$$Â  Â  Â Â A1 N5

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)Â  Â  Â  M1

correct expression forÂ f”Â  Â  Â  (A1)

egÂ  Â 6x + 18, 6p + 18

correct workingÂ  Â  Â  (A1)

6pÂ + 18 = 0

pÂ =Â âˆ’3Â  Â  Â  Â A1 N3

METHOD 1Â (usingÂ 1st derivative)

recognizing the vertex of fâ€² is neededÂ  Â  Â  Â (M2)

egÂ  Â $$– \frac{b}{{2a}}$$Â (must be clear this is for fâ€²)

correct substitutionÂ  Â  Â Â (A1)

egÂ  Â $$\frac{{ – 18}}{{2 \times 3}}$$

pÂ =Â âˆ’3Â  Â  Â  Â A1 N3

[4 marks]

b.

valid attempt to use f”â€‰(x) to determine concavityÂ  Â  Â  (M1)

egÂ Â Â f”â€‰(x) < 0,Â f”â€‰(âˆ’2),Â f”â€‰(âˆ’4),Â  6x + 18 â‰¤ 0Â

correct workingÂ  Â  Â  Â (A1)

egÂ  Â 6xÂ + 18Â < 0,Â f”â€‰(âˆ’2) = 6,Â f”â€‰(âˆ’4) =Â âˆ’6Â

f concave down for x < âˆ’3 (do not accept xÂ â‰¤ âˆ’3)Â  Â  Â  Â A1 N2

[3 marks]

c.