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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 f(x)=1+x for x>-1.

Show that f''(x)=-14(1+x)3.

[3]
a.

Use mathematical induction to prove that f(n)(x)=(-14)n-1(2n-3)!(n-2)!(1+x)12-n for n, n2.

[9]
b.

Let g(x)=emx, m.

Consider the function h defined by h(x)=f(x)×g(x) 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)=12(1+x)-12         A1

f''(x)=-14(1+x)-32         A1

=-14(1+x)3         AG

 

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

  

[3 marks]

a.

let n=2

f''(x)=(-14(1+x)3=)(-14)11!0!(1+x)12-2         R1

 

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

 

assume true for n=k, (so f(k)(x)=(-14)k-1(2k-3)!(k-2)!(1+x)12-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=f(k+1)(x)=d(f(k)(x))dx            M1

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

 

EITHER

RHS=f(k+1)(x)=(-14)k(2k-1)!(k-1)!(1+x)12-k-1 (or equivalent)         A1

=(-14)k(2k-1)(2k-2)(2k-3)!(k-1)(k-2)!(1+x)12-k-1        A1

 

Note: Award A1 for (2k-1)!(k-1)!=(2k-1)(2k-2)(2k-3)!(k-1)(k-2)!(=2(2k-1)(2k-3)!(k-2)!)

 

=(-14)(-14)k-1(2k-1)(2k-2)(2k-3)!(k-1)(k-2)!(1+x)12-k-1        A1

(=(-12)(-14)k-1(2k-1)(2k-3)!(k-2)!(1+x)12-k-1)

 

Note: Award A1 for leading coefficient of -14.

 

=(12-k)(-14)k-1(2k-3)!(k-2)!(1+x)12-k-1        A1

 

OR

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

 

=(-14)k-1(2k-3)!(k-2)!(1-2k2)(1+x)12-k-1

=(-12)(-14)k-1(2k-1)(2k-3)!(k-2)!(1+x)12-k-1        A1

 

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

 

=(-12)(-14)k-1(2k-1)!(2k-2)(k-2)!(1+x)12-k-1        A1

 

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

 

=(-14)(-14)k-1(2k-1)!(k-1)(k-2)!(1+x)12-k-1        A1

 

Note: Award A1 for leading coefficient of -14.

 

=(-14)k(2k-1)!(k-1)!(1+x)12-k-1        A1

 

=(-14)(k+1)-1(2(k+1)-3)!((k+1)-2)!(1+x)12-(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

h(x)=1+xemx

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

h'(x)=1+xmemx+121+xemx         A1

h''(x)=m(1+xmemx+121+xemx)+121+xmemx-14(1+x)3emx         A1

substituting x=0 into h''(x)       M1

h''(0)=m2+12m+12m-14(=m2+m-14)         A1

h(x)=h(0)+xh'(0)+x22!h''(0)+

equating x2 coefficient to 74       M1

h''(0)2!=74 (h''(0)=72)

4m2+4m-15=0         A1

(2m+5)(2m-3)=0

m=-52  or  m=32         A1

 

METHOD 2

EITHER

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

f(x)=(1+x)12                    f(0)=1

f'(x)=12(1+x)-12            f'(0)=12

f''(x)=-14(1+x)-32      f''(0)=-14

f(x)=1+12x-18x2+         A1

 

OR

attempt to apply binomial theorem for rational exponents        (M1)

f(x)=(1+x)12=1+12x+(12)(-12)2!x2

f(x)=1+12x-18x2+         A1

 

THEN

g(x)=1+mx+m22x2+        (A1)

h(x)=(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+5)(2m-3)=0

m=-52  or  m=32         A1

 

METHOD 3

g'(x)=memx and g''(x)=m2emx        (A1)

h(x)=h(0)+xh'(0)+x22!h''(0)+

equating x2 coefficient to 74       M1

h''(0)2!=74 (h''(0)=72)

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

h'(x)=f(x)g'(x)+f'(x)g(x)

h''(x)=f(x)g''(x)+2f'(x)g'(x)+f''(x)g(x)         A1

substituting x=0 into h''(x)       M1

h''(0)=f(0)g''(0)+2g'(0)f'(0)+g(0)f''(0)

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

4m2+4m-15=0          A1

(2m+5)(2m-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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