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.
Use mathematical induction to prove that f(n)(x)=(-14)n-1(2n-3)!(n-2)!(1+x)12-n for n∈ℤ, n≥2.
Let g(x)=emx, m∈ℚ.
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.
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)=1√1+x or equivalent seen
[3 marks]
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 (2k−1) 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 (k−1) or 2(k−1).
=(-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∈ℤ, n≥2 by mathematical induction R1
Note: To obtain the final R1, at least four of the previous marks must have been awarded.
[9 marks]
METHOD 1
h(x)=√1+x emx
using product rule to find h'(x) (M1)
h'(x)=√1+x memx+12√1+xemx A1
h''(x)=m(√1+x memx+12√1+xemx)+12√1+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]