2019 May TZ1 Paper 2 - Q8

(a)

The inequality is equivalent to:

x^2 - 2x - 1 > 0

Completing the square:

(x - 1)^2 - 2 > 0

So:

(x - 1)^2 > 2

giving:

|x - 1| > \sqrt 2

so either x < 1 - \sqrt 2 or x > 1 + \sqrt 2.

(b)

For n = 3 we have:

2^{n + 1} = 2^4 = 16

and n^2 = 3^2 = 9. So 2^{n + 1} > n^2 in the case n = 3. Suppose that 2^{k + 1} > k^2 for some k. Then:

2^{k + 2} = 2 \times 2^{k + 1} > 2k^2

by the induction assumption. Then write:

2k^2 = k^2 + k^2

Since k \ge 3, we certainly have k > 1 + \sqrt 2, (2 > \sqrt 2) so the inequality from part (a) holds and we have:

k^2 + k^2 > k^2 + 2k + 1 = (k + 1)^2

So:

2^{k + 2} > (k + 1)^2

Since the inequality holds for n = 3 and the inequality holding for n = k means that it holds for n = k + 1, it holds for all n \in \mathbb Z, n \ge 3 by induction.

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