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Date May 2016 Marks available 6 Reference code 16M.2.hl.TZ2.12
Level HL only Paper 2 Time zone TZ2
Command term Hence and Show that Question number 12 Adapted from N/A

Question

The functions ff and g are defined by

f(x)=ex+ex2, xR

g(x)=exex2, xR

Let h(x)=nf(x)+g(x) where nR, n>1.

Let t(x)=g(x)f(x).

(i)     Show that 14f(x)2g(x)=exe2x+3.

(ii)     Use the substitution u=ex to find ln3014f(x)2g(x)dx. Give your answer in the form πab where a, bZ+.

[9]
a.

(i)     By forming a quadratic equation in ex, solve the equation h(x)=k, where kR+.

(ii)     Hence or otherwise show that the equation h(x)=k has two real solutions provided that k>n21 and kR+.

[8]
b.

(i)     Show that t(x)=[f(x)]2[g(x)]2[f(x)]2 for xR.

(ii)     Hence show that t(x)>0 for xR.

[6]
c.

Markscheme

(i)     14(ex+ex2)2(exex2)     (M1)

=12(ex+ex)(exex)    (A1)

=1ex+3ex    A1

=exe2x+3    AG

(ii)     u=exdu=exdx     A1

exe2x+3dx=1u2+3du    M1

(when x=0, u=1 and when x=ln3, u=3)

311u2+3du[13arctan(u3)]31    M1A1

(=[13arctan(ex3)]ln30)

=π39π318    (M1)

=π318    A1

[9 marks]

a.

(i)     (n+1)e2x2kex+(n1)=0     M1A1

ex=2k±4k24(n21)2(n+1)    M1

x=ln(k±k2n2+1n+1)    M1A1

(ii)     for two real solutions, we require k>k2n2+1     R1

and we also require k2n2+1>0     R1

k2>n21    A1

k>n21 ( kR+)    AG

[8 marks]

b.

METHOD 1

t(x)=exexex+ex

t(x)=(ex+ex)2(exex)2(ex+ex)2    M1A1

t(x)=(ex+ex2)2(exex2)2(ex+ex2)2    A1

=[f(x)]2[g(x)]2[f(x)]2    AG

METHOD 2

t(x)=f(x)g(x)=g(x)f(x)[f(x)]2    M1A1

g(x)=f(x) and f(x)=g(x)     A1

=[f(x)]2[g(x)]2[f(x)]2    AG

METHOD 3

t(x)=(exex)(ex+ex)1

t(x)=1(exex)2(ex+ex)2    M1A1

=1[g(x)]2[f(x)]2    A1

=[f(x)]2[g(x)]2[f(x)]2    AG

METHOD 4

t(x)=g(x)f(x)g(x)f(x)[f(x)]2    M1A1

g(x)=f(x) and f(x)=g(x) gives t(x)=1[g(x)]2[f(x)]2     A1

=[f(x)]2[g(x)]2[f(x)]2    AG

(ii)     METHOD 1

[f(x)]2>[g(x)]2 (or equivalent)     M1A1

[f(x)]2>0    R1

hence t(x)>0, xR     AG

Note:     Award as above for use of either f(x)=ex+ex2 and g(x)=exex2 or ex+ex and exex.

METHOD 2

[f(x)]2[g(x)]2=1 (or equivalent)     M1A1

[f(x)]2>0    R1

hence t(x)>0, xR     AG

Note:     Award as above for use of either f(x)=ex+ex2 and g(x)=exex2 or ex+ex and exex.

METHOD 3

t(x)=4(ex+ex)2

(ex+ex)2>0    M1A1

4(ex+ex)2>0    R1

hence t(x)>0, xR     AG

[6 marks]

c.

Examiners report

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

Part (a)(i) was reasonably well done with most candidates able to show that 14f(x)2g(x)=exe2x+3. In part (a)(ii), a number of candidates correctly used the required substitution to obtain exe2x+3dx=1u2+3du but then thought that the antiderivative involved natural log rather than arctan.

a.

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

In part (b)(i), a reasonable number of candidates were able to form a quadratic in ex (involving parameters n and k) and then make some progress towards solving for ex in terms of n and k. Having got that far, a small number of candidates recognised to then take the natural logarithm of both sides and hence solve h(x)=k for χ. In part (b)(ii), a small number of candidates were able to show from their solutions to part (b)(i) or through the use of the discriminant that the equation h(x)=k has two real solutions provided that k>k2n2+1 and k>n21.

b.

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

It was pleasing to see the number of candidates who attempted part (c). In part (c)(i), a large number of candidates were able to correctly apply either the quotient rule or the product rule to find t(x). A smaller number of candidates were then able to show equivalence between the form of t(x) they had obtained and the form of t(x) required in the question. A pleasing number of candidates were able to exploit the property that f(x)=g(x) and g(x)=f(x). As with part (c)(i), part (c)(ii) could be successfully tackled in a number of ways. The best candidates offered concise logical reasoning to show that t(x)>0 for xR.

c.

Syllabus sections

Topic 6 - Core: Calculus » 6.2 » The product and quotient rules.
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