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    HLPaper 2
    1.

    Solve the inequality x2>2x+1{x^2} > 2x + 1x2>2x+1.

    [2]
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    Solution
    • This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. x<−0.414,  x>2.41x < - 0.414,\,\,x > 2.41x<−0.414,x>2.41 A1A1 (x<1−2,  x>1+2)\left( {x < 1 - \sqrt 2 ,\,\,x > 1 + \sqrt 2 } \right)(x<1−2​,x>1+2​) Note: Award A1 for −0.414, 2.41 and A1 for correct inequalities. [2 marks]
    2.

    Use mathematical induction to prove that 2n+1>n2{2^{n + 1}} > {n^2}2n+1>n2 for n∈Zn \in \mathbb{Z}n∈Z, n⩾3n \geqslant 3n⩾3.

    [7]
    Verified
    Solution

    check for n=3n= 3n=3, 16&gt; 9 so true when n=3n=3n=3 A1 assume true for n=kn= kn=k {2^{k + 1}} &gt; {k^2} M1 Note: Award M0 for statements such as “let n=kn= kn=k”. Note: Subsequent marks after this M1 are independent of this mark and can be awarded. prove true for n=k+1n= k + 1n=k+1 2k+2=2×2k+1{2^{k + 2}} = 2 \times {2^{k + 1}}2k+2=2×2k+1 {2^{k + 2}} &gt; 2{k^2} M1 2k+2=k2+k2{2^{k + 2}} = {k^2} + {k^2}2k+2=k2+k2 (M1) {2^{k + 2}} &gt; {k^2} + 2k + 1 (from part (a)) A1 which is true for k≥3k ≥ 3k≥3 R1 Note: Only award the A1 or the R1 if it is clear why. Alternate methods are possible. (K+1)2=((K+1))2{(K + 1)^2} = \left( (K + 1) \right)^2(K+1)2=((K+1))2 hence if true for n=kn= kn=k true for n=k+1n= k +1n=k+1, true for n=3n= 3n=3 so true for all n≥3n≥ 3n≥3 R1 Note: Only award the final R1 provided at least three of the previous marks are awarded. [7 marks]

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