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Date May 2016 Marks available 14 Reference code 16M.2.hl.TZ0.7
Level HL only Paper 2 Time zone TZ0
Command term Explain, Hence, and Show that Question number 7 Adapted from N/A

Question

Consider the functions fn(x)=secn(x), |x|<π2 and gn(x)=fn(x)tanx.

Show that

(i)     dfn(x)dx=ngn(x);

(ii)     dgn(x)dx=(n+1)fn+2(x)nfn(x).

[5]
a.

(i)     Use these results to show that the Maclaurin series for the function f5(x) up to and including the term in x4 is 1+52x2+8524x4.

(ii)     By considering the general form of its higher derivatives explain briefly why all coefficients in the Maclaurin series for the function f5(x) are either positive or zero.

(iii)     Hence show that sec5(0.1)>1.02535.

[14]
b.

Markscheme

(i)     dfn(x)dx=nsecn1(x)sec(x)tan(x)     M1A1

=ngn(x)    AG

 

(ii)     dgn(x)dx=dfn(x)dxtan(x)+fn(x)sec2(x)     M1

ngn(x)tan(x)+fn+2(x) or equivalent     A1

nfn(x)tan2(x)+fn+2(x) or equivalent     A1

=(n+1)fn+2(x)nfn(x)    AG

 

Note:     Award M1A1 for the correct differentiation of a product and A1 for an intermediate result clearly leading to the AG.

 

[5 marks]

a.

(i)     f5(0)=1     A1

df5dx(0)=5g5(0)=0    A1

d2f5dx2(0)=5(6f7(0)5f5(0))=5    A1

d3f5dx3=30df7dx25df5dx    M1

hence d3f5dx3(0)=30×025×0=0     A1

d4f5dx4=30d2f7dx225d2f5dx2=210(8f97f7)25d2f5dx2    M1A1

hence d4f5dx4(0)=210125=85     A1

hence f5(x)1+52x2+8524x4     AG

 

(ii)     each derivative of fm(x) is a sum of terms of the form Asecp(x)tanq(x)     A1

where A0     A1

when x = 0 is substituted the result is the sum of positive and/or zero terms     R1

 

(iii)     since the full series represents {f_5}(x), the truncated series is a lower bound (or some equivalent statement)     R1

hence {\sec ^5}(0.1) > 1 + \frac{5}{2}{0.1^2} + \frac{{85}}{{24}}{0.1^4}     M1

= 1.025354    A1

> 1.02535    AG

[14 marks]

b.

Examiners report

Part (a) was generally answered, albeit either with an excess of algebraic manipulation or with too little – candidates need to realise that when an answer is given in the question, they need to convincingly reach that answer.

a.

In part (b)(i), the results of part (a) were well used for up to the quadratic term. The obtaining of the cubic term, and more so the quartic term, was often not convincing. In part (ii), poor communication let down many candidates. In answering part (iii), many candidates failed to realise that in order to prove the stated inequality, they needed to actually write down the number 1.025354…, which is clearly greater than 1.02535.

b.

Syllabus sections

Topic 5 - Calculus » 5.6 » Taylor polynomials; the Lagrange form of the error term.

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