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
using product rule to find (M1)
A1
A1
substituting into M1
A1
equating coefficient to M1
A1
or A1
METHOD 2
EITHER
attempt to find (M1)
A1
OR
attempt to apply binomial theorem for rational exponents (M1)
A1
THEN
(A1)
(M1)
coefficient of is A1
attempt to set equal to and solve M1
A1
or A1
METHOD 3
and (A1)
equating coefficient to M1
using product rule to find and (M1)
A1
substituting into M1
A1
A1
or A1
[8 marks]