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Date May 2015 Marks available 2 Reference code 15M.1.hl.TZ2.13
Level HL only Paper 1 Time zone TZ2
Command term Show that Question number 13 Adapted from N/A

Question

Show that \(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \sqrt {n + 1}  - \sqrt n \) where \(n \ge 0,{\text{ }}n \in \mathbb{Z}\).

[2]
a.

Hence show that \(\sqrt 2  - 1 < \frac{1}{{\sqrt 2 }}\).

[2]
b.

Prove, by mathematical induction, that \(\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n } \) for \(n \ge 2,{\text{ }}n \in \mathbb{Z}\).

[9]
c.

Markscheme

\(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \frac{1}{{\sqrt n  + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt {n + 1}  - \sqrt n }}\)     M1

\( = \frac{{\sqrt {n + 1}  - \sqrt n }}{{(n + 1) - n}}\)     A1

\( = \sqrt {n + 1}  - \sqrt n \)     AG

[2 marks]

a.

\(\sqrt 2  - 1 = \frac{1}{{\sqrt 2  + \sqrt 1 }}\)     A2

\( < \frac{1}{{\sqrt 2 }}\)     AG

[2 marks]

b.

consider the case \(n = 2\): required to prove that \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \)     M1

from part (b) \(\frac{1}{{\sqrt 2 }} > \sqrt 2  - 1\)

hence \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) is true for \(n = 2\)     A1

now assume true for \(n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k } \)     M1

\(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k \)

attempt to prove true for \(n = k + 1:\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

from assumption, we have that \(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k  + \frac{1}{{\sqrt {k + 1} }}\)     M1

so attempt to show that \(\sqrt k  + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

EITHER

\(\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}  - \sqrt k \)     A1

\(\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k  + \sqrt {k + 1} }}\), (from part a), which is true     A1

OR

\(\sqrt k  + \frac{1}{{\sqrt {k + 1} }} = \frac{{\sqrt {k + 1} \sqrt k  + 1}}{{\sqrt {k + 1} }}\)     A1

\( > \frac{{\sqrt k \sqrt k  + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1} \)     A1

THEN

so true for \(n = 2\) and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \ge 2\)     R1

 

Note:     Award R1 only if all previous M marks have been awarded.

[9 marks]

Total [13 marks]

c.

Examiners report

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Syllabus sections

Topic 2 - Core: Functions and equations » 2.1 » Concept of function \(f:x \mapsto f\left( x \right)\) : domain, range; image (value)

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