2020HSCShort answer

Q16 (b) (iv)

2 marks

Question

Prove that
$\bigl(2^n n!\bigr)^2\leq(2n+1)!.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	2
Finds the maximum value of $x(1-x)$ , for $0\le x\le1$ , or equivalent merit	1

Sample answer:

The quadratic expression $x(1-x)$ has maximum value of $\frac14$ when $x=\frac12$ .

J_n=\int_0^1x^n(1-x)^n\,dx\le\int_0^1\left(\frac14\right)^n dx

=\frac1{2^{2n}}

\therefore\frac{(n!)^2}{(2n+1)!}\le\frac1{2^{2n}}

\Rightarrow(2^n n!)^2\le(2n+1)!

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Syllabus alignment

CalculusWorking MathematicallyProof

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.
Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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