IB DP Further Mathematics 5.7 HL Paper 1

 

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Question

Calculate the following limit

\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} – 1}}{x}\) .

[3]
a.

Calculate the following limit

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} – 1}}{{\ln (1 + x) – x}}\) .

[5]
b.
Answer/Explanation

Markscheme

\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} – 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{1}\)     M1A1

\( = \ln 2\)     A1

[3 marks]

a.

EITHER

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} – 1}}{{\ln (1 + x) – x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}}}{{\frac{1}{{1 + x}} – 1}}\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}(1 + x)}}{{ – x}}\)     A1A1

\( = – 3\)     A1

OR

\({(1 + {x^2})^{\frac{3}{2}}} – 1 = 1 + \frac{3}{2}{x^2} +  \ldots  – 1 = \frac{3}{2}{x^2} +  \ldots \)     M1A1

\(\ln (1 + x) – x = x – \frac{1}{2}{x^2} +  \ldots  – x = – \frac{1}{2}{x^2} +  \ldots \)     M1A1

Limit \( = – 3\)     A1

[5 marks]

b.

Question

By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for \(\ln \cos x\) up to and including the term in \({x^4}\) .

[8]
a.

Consider \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \cos x}}{{{x^n}}}\) , where \(n \in \mathbb{R}\) .

Using your result from (a), determine the set of values of \(n\) for which

  (i)     the limit does not exist;

  (ii)     the limit is zero;

  (iii)     the limit is finite and non-zero, giving its value in this case.

[5]
b.
Answer/Explanation

Markscheme

attempt at repeated differentiation (at least 2)     M1

let \(f(x) = \ln \cos x\) , \(f(0) = 0\)     A1

\(f'(x) = – \tan x\) , \(f'(0) = 0\)     A1

\(f”(x) = – {\sec ^2}x\) , \(f”(0) =  – 1\)     A1

\(f”'(x) = – 2{\sec ^2}x\tan x\) , \(f”'(0) = 0\)     A1

\({f^{iv}}(x) = – 2se{c^4}x – 4se{c^2}xta{n^2}x\) , \({f^{iv}}(0) =  – 2\)       A1

the Maclaurin series is

\(\ln \cos x = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}} +  \ldots \)     M1A1

Note: Allow follow-through on final A1.

[8 marks]

a.

\(\frac{{\ln \cos x}}{{{x^n}}} = – \frac{{{x^{2 – n}}}}{2} – \frac{{{x^{4 – n}}}}{{12}} +  \ldots \)     (M1)

(i)     the limit does not exist if \(n > 2\)     A1

(ii)     the limit is zero if \(n < 2\)     A1

(iii)     if \(n = 2\) , the limit is \( – \frac{1}{2}\)     A1A1

[5 marks]

b.

Question

Find the general solution of the differential equation \((1 – {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 + xy\) , for \(\left| x \right| < 1\) .

[7]
a.

(i)     Show that the solution \(y = f(x)\) that satisfies the condition \(f(0) = \frac{\pi }{2}\) is \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\) .

(ii)     Find \(\mathop {\lim }\limits_{x \to  – 1} f(x)\) .

[6]
b.
Answer/Explanation

Markscheme

rewrite in linear form     M1

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \left( {\frac{{ – x}}{{1 – {x^2}}}} \right)y = \frac{1}{{1 – {x^2}}}\)

attempt to find integrating factor     M1

\(I = {e^{\int {\frac{{ – x}}{{1 – {x^2}}}} {\rm{d}}x}} = {e^{\frac{1}{2}\ln (1 – {x^2})}}\)     A1

\( = \sqrt {1 – {x^2}} \)     A1

multiply by \(I\) and attempt to integrate     (M1)

\(y{(1 – {x^2})^{\frac{1}{2}}} = \int {\frac{1}{{\sqrt {1 – {x^2}} }}} {\rm{d}}x\)     (A1)

\(y{(1 – {x^2})^{\frac{1}{2}}} = \arcsin x + c\)     A1

[7 marks]

a.

(i)     attempt to find c     M1

\(\frac{\pi }{2} = 0 + c\)     A1

so \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\)     AG  

(ii)     \(\mathop {\lim }\limits_{x \to  – 1} f(x) = \frac{0}{0}\) , so attempt l’Hôpital’s rule     (M1)

consider \(\mathop {\lim }\limits_{x \to  – 1} \frac{{\frac{1}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}{{\frac{{ – x}}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}\)     A1A1

\( = 1\)     A1  

[6 marks]

b.

Question

Use l’Hôpital’s rule to find \(\mathop {\lim }\limits_{x \to 0} (\csc x – \cot x)\).

Answer/Explanation

Markscheme

\(\mathop {\lim }\limits_{x \to 0} (\csc x – \cot x) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – \cos x}}{{\sin x}}} \right)\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{\cos x}}} \right)\)     M1A1

\( = 0\)     A1

Question

Let

\({I_n} = \int_1^\infty  {{x^n}{{\text{e}}^{ – x}}{\text{d}}x} \) where \(n \in \mathbb{N}\).

Using l’Hôpital’s rule, show that

\(\mathop {\lim }\limits_{x \to \infty } {x^n}{{\text{e}}^{ – x}} = 0\) where \(n \in \mathbb{N}\).

[4]
a.

Show that, for \(n \in {\mathbb{Z}^ + }\),

\[{I_n} = \alpha {{\text{e}}^{ – 1}} + \beta n{I_{n – 1}}\]

where \(\alpha \), \(\beta \) are constants to be determined.

[4]
b.i.

Determine the value of \({I_3}\), giving your answer as a multiple of \({{\text{e}}^{ – 1}}\).

[5]
b.ii.
Answer/Explanation

Markscheme

consider \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}}\)     M1

its value is \(\frac{\infty }{\infty }\) so we use l’Hôpital’s rule

\(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n{x^{n – 1}}}}{{{{\text{e}}^x}}}\)     (A1)

its value is still \(\frac{\infty }{\infty }\) so we need to differentiate numerator and denominator a further \(n – 1\) times     (R1)

this gives \(\mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{{\text{e}}^x}}}\)     A1

since the numerator is finite and the denominator \( \to \infty \), the limit is zero     AG

[4 marks]

a.

attempt at integration by parts \(\left( {{I_n} =  – \int_1^\infty  {{x^n}{\text{d}}({{\text{e}}^{ – x}})} } \right)\)     M1

\({I_n} =  – [{x^n}{{\text{e}}^{ – x}}]_1^\infty  + n\int_1^\infty  {{x^{n – 1}}{{\text{e}}^{ – x}}{\text{d}}x} \)     A1A1

\( = {{\text{e}}^{ – 1}} + n{I_{n – 1}}\)     A1

\(\alpha  = \beta  = 1\)

[9 marks]

b.i.

\({I_3} = {{\text{e}}^{ – 1}} + 3{I_2}\)     M1

\( = {{\text{e}}^{ – 1}} + 3({{\text{e}}^{ – 1}} + 2{I_1})\)     A1

\( = 4{{\text{e}}^{ – 1}} + 6({{\text{e}}^{ – 1}} + {I_0})\)     A1

\( = 4{{\text{e}}^{ – 1}} + 6{{\text{e}}^{ – 1}} + 6\int_1^\infty  {{{\text{e}}^{ – x}}{\text{d}}x} \)

\( = 10{{\text{e}}^{ – 1}} – 6[{{\text{e}}^{ – x}}]_1^\infty \)     A1

\( = 16{{\text{e}}^{ – 1}}\)     A1

[9 marks]

b.ii.

Question

QUESTION 1

[[N/A]]
.

QUESTION 1

[[N/A]]
.

Using l’Hôpital’s Rule, determine the value of\[\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – x}}{{1 – \cos x}} .\]

[6]
.
Answer/Explanation

Markscheme

MARKSCHEME 1

.

MARKSCHEME 1

.

\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – x}}{{1 – \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x – 1}}{{\sin x}}\)     M1A1A1

this still gives \(\frac{0}{0}\)

EITHER

repeat the process     M1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sec }^2}x\tan x}}{{\cos x}}\)     A1

\( = 0\)     A1

OR

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}x}}{{\sin x}}\)     M1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{{{\cos }^2}x}}\)     A1

\( = 0\)     A1

[6 marks]

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