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.
Answer/Explanation

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 \(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) .

[5]
a.

Find \(f”(x)\) .

[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.
Answer/Explanation

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 \(x = 0\) , \( \pm \sqrt[4]{3}\) \(( \pm 1.316 \ldots )\) .

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

[2]
a.

Find the set of values of \(x\) for which \(f\) is increasing.

[5]
b.

(i)     Find \(f”(1)\) .

(ii)     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.
Answer/Explanation

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 \(f”(x)\).

[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 \(x =  – 1\).

[3]
e.
Answer/Explanation

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 \({f^{(19)}}(x)\).

[4]
a.

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

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

[5]
b.

(i)     Find \(h'(x)\).

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

[7]
c.
Answer/Explanation

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).

[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.
Answer/Explanation

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.

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