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Date May 2021 Marks available 9 Reference code 21M.1.AHL.TZ1.12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Prove Question number 12 Adapted from N/A

Question

Let fx=1+x for x>-1.

Show that f''x=-141+x3.

[3]
a.

Use mathematical induction to prove that fnx=-14n-12n-3!n-2!1+x12-n for n, n2.

[9]
b.

Let gx=emx, m.

Consider the function h defined by hx=fx×gx for x>-1.

It is given that the x2 term in the Maclaurin series for h(x) has a coefficient of 74.

Find the possible values of m.

[8]
c.

Markscheme

attempt to use the chain rule            M1

f'x=121+x-12         A1

f''x=-141+x-32         A1

=-141+x3         AG

 

Note: Award M1A0A0 for f'x=11+x or equivalent seen

  

[3 marks]

a.

let n=2

f''x=-141+x3=-1411!0!1+x12-2         R1

 

Note: Award R0 for not starting at n=2. Award subsequent marks as appropriate.

 

assume true for n=k, (so fkx=-14k-12k-3!k-2!1+x12-k)       M1

 

Note: Do not award M1 for statements such as “let n=k” or “n=k is true”. Subsequent marks can still be awarded.

 

consider n=k+1

LHS=fk+1x=dfkxdx            M1

=-14k-12k-3!k-2!12-k1+x12-k-1 (or equivalent)         A1

 

EITHER

RHS=fk+1x=-14k2k-1!k-1!1+x12-k-1 (or equivalent)         A1

=-14k2k-12k-22k-3!k-1k-2!1+x12-k-1        A1

 

Note: Award A1 for 2k-1!k-1!=2k-12k-22k-3!k-1k-2!=22k-12k-3!k-2!

 

=-14-14k-12k-12k-22k-3!k-1k-2!1+x12-k-1        A1

=-12-14k-12k-12k-3!k-2!1+x12-k-1

 

Note: Award A1 for leading coefficient of -14.

 

=12-k-14k-12k-3!k-2!1+x12-k-1        A1

 

OR

Note: The following A marks can be awarded in any order.

 

=-14k-12k-3!k-2!1-2k21+x12-k-1

=-12-14k-12k-12k-3!k-2!1+x12-k-1        A1

 

Note: Award A1 for isolating (2k1) correctly.

 

=-12-14k-12k-1!2k-2k-2!1+x12-k-1        A1

 

Note: Award A1 for multiplying top and bottom by (k1) or 2(k1).

 

=-14-14k-12k-1!k-1k-2!1+x12-k-1        A1

 

Note: Award A1 for leading coefficient of -14.

 

=-14k2k-1!k-1!1+x12-k-1        A1

 

=-14k+1-12k+1-3!k+1-2!1+x12-k+1=RHS

 

THEN

since true for n=2, and true for n=k+1 if true for n=k, the statement is true for all, n, n2  by mathematical induction           R1

 

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

 

[9 marks]

b.

METHOD 1

hx=1+xemx

using product rule to find h'x        (M1)

h'x=1+xmemx+121+xemx         A1

h''x=m1+xmemx+121+xemx+121+xmemx-141+x3emx         A1

substituting x=0 into h''x       M1

h''0=m2+12m+12m-14=m2+m-14         A1

hx=h0+xh'0+x22!h''0+

equating x2 coefficient to 74       M1

h''02!=74 h''0=72

4m2+4m-15=0         A1

2m+52m-3=0

m=-52  or  m=32         A1

 

METHOD 2

EITHER

attempt to find f0, f'0, f''0        (M1)

fx=1+x12                    f0=1

f'x=121+x-12            f'0=12

f''x=-141+x-32      f''0=-14

fx=1+12x-18x2+         A1

 

OR

attempt to apply binomial theorem for rational exponents        (M1)

fx=1+x12=1+12x+12-122!x2

fx=1+12x-18x2+         A1

 

THEN

gx=1+mx+m22x2+        (A1)

hx=1+12x-18x2+1+mx+m22x2+        (M1)

coefficient of x2 is m22+m2-18         A1

attempt to set equal to 74 and solve             M1

m22+m2-18=74

4m2+4m-15=0          A1

2m+52m-3=0

m=-52  or  m=32         A1

 

METHOD 3

g'x=memx and g''x=m2emx        (A1)

hx=h0+xh'0+x22!h''0+

equating x2 coefficient to 74       M1

h''02!=74 h''0=72

using product rule to find h'x and h''x        (M1)

h'x=fxg'x+f'xgx

h''x=fxg''x+2f'xg'x+f''xgx         A1

substituting x=0 into h''x       M1

h''0=f0g''0+2g'0f'0+g0f''0

=1×m2+2m×12+1×-14  =m2+m-14         A1

4m2+4m-15=0          A1

2m+52m-3=0

m=-52  or  m=32         A1

 

[8 marks]

c.

Examiners report

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a.
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b.
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c.

Syllabus sections

Topic 5 —Calculus » AHL 5.12—First principles, higher derivatives
Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
Topic 1—Number and algebra
Topic 5 —Calculus

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