# 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.


a.

Find $$f'(3)$$ and $$f”(3)$$ .


b.

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


c.

## Markscheme

METHOD 1

$$f”(x) = 3{(x – 3)^2}$$     A2     N2

METHOD 2

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

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

$$f”(x) = 3{x^2} – 18x + 27$$     A1     N2

[2 marks]

a.

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

[1 mark]

b.

METHOD 1

$${f”}$$ does not change sign at P     R1

evidence for this     R1     N0

METHOD 2

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

evidence for this     R1     N0

METHOD 3

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

indicating minimum at $$x = 3$$     R1     N0

[2 marks]

c.

## Question

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

In the following table, $$f’\left( {\frac{\pi }{2}} \right) = p$$ and $$f”\left( {\frac{\pi }{2}} \right) = q$$ . The table also gives approximate values of $$f'(x)$$ and $$f”(x)$$ near $$x = \frac{\pi }{2}$$ . Use the quotient rule to show that $$f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}$$ .


a.

Find $$f”(x)$$ .


b.

Find the value of p and of q.


c.

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


d.

## Markscheme

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

evidence of using the quotient rule     M1

correct substitution     A1

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

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

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

[5 marks]

a.

METHOD 1

appropriate approach     (M1)

e.g. $$f'(x) = – {(\sin x)^{ – 2}}$$

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

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

METHOD 2

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

evidence of choosing quotient rule     (M1)

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

$$f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$$ $$\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$$     A1     N3

[3 marks]

b.

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

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

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

[3 marks]

c.

second derivative is zero, second derivative changes sign     R1R1     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 $$x = 0$$ , $$\pm \sqrt{3}$$ $$( \pm 1.316 \ldots )$$ .

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


a.

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


b.

(i)     Find $$f”(1)$$ .

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


c.

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

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


d.

## Markscheme

substitute $$0$$ into $$f$$     (M1)

eg   $$\ln (0 + 1)$$ , $$\ln 1$$

$$f(0) = 0$$     A1 N2

[2 marks]

a.

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

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

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

eg   $$f'(x) > 0$$ , diagram of signs

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

eg   $$4{x^3} = 0$$ , $${x^3} > 0$$

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

[5 marks]

b.

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

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

$$f”(1) = 2$$     A1     N2

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

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

$$f’$$ increasing both sides of zero

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

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

correct working leading to positive value     A1

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

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

[5 marks]

c. A1A1A1     N3

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

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

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

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 $$f”(x)$$.


a.

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

Show that $$k = 3$$.


b.

Find $$f'( – 2)$$.


c.

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


d.

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


e.

## Markscheme

$$f”(x) = 6x – 2k$$     A1A1     N2

[2 marks]

a.

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

eg$$\;\;\;f”(1),{\text{ }}6(1) – 2k$$

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

correct equation     A1

eg$$\;\;\;6 – 2k = 0$$

$$k = 3$$     AG     N0

[3 marks]

b.

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

eg$$\;\;\;3{( – 2)^2} – 6( – 2) – 9$$

$$f'( – 2) = 15$$     A1     N2

[2 marks]

c.

recognizing gradient value (may be seen in equation)     M1

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

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

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

correct working     (A1)

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

$$y = 15x + 31$$     A1     N2

[4 marks]

d.

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

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

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

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

$$f”( – 1) = – 12$$     A1

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

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

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

eg$$\;\;\;$$sign chart$$\;\;\;$$ correct value of $$f’$$ for $$– 1 < x < 3$$     A1

eg$$\;\;\;f'(0) = – 9$$

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

eg$$\;\;\;f'( – 2) = 15$$

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

[3 marks]

Total [14 marks]

e.

## Question

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

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

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

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

(ii)     Find $${f^{(19)}}(x)$$.


a.

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

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


b.

(i)     Find $$h'(x)$$.

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


c.

## Markscheme

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

(ii)     valid approach     (M1)

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

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

[4 marks]

a.

(i)     $$g'(x) = k{x^{k – 1}}$$

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

(ii)     METHOD 1

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

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

$$p = 19$$ (accept $$\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}$$)     A1     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     A2

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

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

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

[5 marks]

b.

(i)     valid approach using product rule     (M1)

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

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

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

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

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

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

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

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

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

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

[7 marks]

c.

## Question

A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

Find f (x).


a.

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


b.

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


c.

## Markscheme

evidence of integration       (M1)

eg  $$\int {f’\left( x \right)}$$

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

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

attempt to substitute x = −1 into their = 0 (must have C)      M1

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

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

correct working       (A1)

eg  $$8 + C = 0,\,\,\,C = – 8$$

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

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

eg   $$– \frac{b}{{2a}}$$ (must be clear this is for f′)

correct substitution      (A1)

eg   $$\frac{{ – 18}}{{2 \times 3}}$$

p = −3       A1 N3

[4 marks]

b.

valid attempt to use f” (x) to determine concavity      (M1)

eg   f” (x) < 0, f” (−2), f” (−4),  6x + 18 ≤ 0 correct working       (A1)

eg   6x + 18 < 0, f” (−2) = 6, f” (−4) = −6 f concave down for x < −3 (do not accept ≤ −3)       A1 N2

[3 marks]

c.