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Date November 2016 Marks available 2 Reference code 16N.3ca.hl.TZ0.4
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term State Question number 4 Adapted from N/A

Question

Let f(x) be a function whose first and second derivatives both exist on the closed interval [0, h].

Let g(x)=f(h)f(x)(hx)f(x)(hx)2h2(f(h)f(0)hf(0)).

State the mean value theorem for a function that is continuous on the closed interval [a, b] and differentiable on the open interval ]a, b[.

[2]
a.

(i)     Find g(0).

(ii)     Find g(h).

(iii)     Apply the mean value theorem to the function g(x) on the closed interval [0, h] to show that there exists c in the open interval ]0, h[ such that g(c)=0.

(iv)     Find g(x).

(v)     Hence show that (hc)f(c)+2(hc)h2(f(h)f(0)hf(0))=0.

(vi)     Deduce that f(h)=f(0)+hf(0)+h22 f(c).

[9]
b.

Hence show that, for h>0

1cos(h)h22.

[5]
c.

Markscheme

there exists c in the open interval ]a, b[ such that     A1

f(b)f(a)ba=f(c)    A1

 

Note: Open interval is required for the A1.

 

[2 marks]

a.

(i)     g(0)=f(h)f(0)hf(0)h2h2( f(h)f(0)hf(0))

=0    A1

(ii)     g(h)=f(h)f(h)00

=0    A1

(iii)     (g(x) is a differentiable function since it is a combination of other differentiable functions f, f and polynomials.)

there exists c in the open interval ]0, h[ such that

g(h)g(0)h=g(c)    A1

g(h)g(0)h=0    A1

hence g(c)=0     AG

(iv)     g(x)=f(x)+f(x)(hx)f(x)+2(hx)h2(f(h)f(0)hf(0))     A1A1

 

Note: A1 for the second and third terms and A1 for the other terms (all terms must be seen).

 

=(hx)f(x)+2(hx)h2(f(h)f(0)hf(0))

(v)     putting x=c and equating to zero     M1

(hc)f(c)+2(hc)h2(f(h)f(0)hf(0))=g(c)=0    AG

(vi)     f(c)+2h2(f(h)f(0)hf(0))=0     A1

since hc0     R1

h22f(c)=f(h)f(0)hf(0)

f(h)=f(0)+hf(0)+h22f(c)    AG

[9 marks]

b.

letting f(x)=cos(x)     M1

f(x)=sin(x)    f(x)=cos(x)     A1

cos(h)=1+0h22cos(c)     A1

1cos(h)=h22cos(c)    (A1)

since cos(c)1     R1

1cos(h)h22    AG

 

Note: Allow f(x)=a±bcosx.

 

[5 marks]

c.

Examiners report

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b.
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c.

Syllabus sections

Topic 9 - Option: Calculus » 9.6 » Mean value theorem.

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