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(i) Given that p and q are integers such that $$pq \text{ is even}$$ use algebra to prove by contradiction that at least one of p or q is even - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 1

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(i) Given that p and q are integers such that $$pq \text{ is even}$$ use algebra to prove by contradiction that at least one of p or q is even. (ii) Given that x ... show full transcript

Worked Solution & Example Answer:(i) Given that p and q are integers such that $$pq \text{ is even}$$ use algebra to prove by contradiction that at least one of p or q is even - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 1

Step 1

Given that p and q are integers such that $pq$ is even, use algebra to prove by contradiction that at least one of p or q is even.

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Answer

To prove this by contradiction, we start by assuming both p and q are odd integers.

  1. Assumption: Let p = 2m + 1 and q = 2n + 1, where m and n are integers.

  2. Multiply:

    pq &= (2m + 1)(2n + 1) \ &= 4mn + 2m + 2n + 1 \ &= 2(2mn + m + n) + 1. \ ext{Thus, } pq ext{ is odd.} \ ext{This contradicts our assumption that } pq ext{ is even.} \ ext{Hence, at least one of p or q must be even.} \ \end{align*}$$

Step 2

Given that x and y are integers such that $x < 0$ and $(x + y) < 9x^2 + y^2$, show that $y > 4x$.

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Answer

  1. Start with the inequality: (x+y)<9x2+y2.(x + y) < 9x^2 + y^2. Rearranging gives: y2y(9x2x)>0.y^2 - y - (9x^2 - x) > 0.

  2. Substituting: Let a=ya = y and b=3xb = 3x, the quadratic inequality becomes: y2y9x2+x>0.y^2 - y - 9x^2 + x > 0.

  3. Analysis of Quadratic:
    The quadratic can be examined using the discriminant: D=(1)24(1)(9x2+x)=1+36x24x.D = (-1)^2 - 4(1)(-9x^2 + x) = 1 + 36x^2 - 4x. We find the roots of the function in terms of y: y=1±1+36x24x2.y = \frac{1 \pm \sqrt{1 + 36x^2 - 4x}}{2}.

  4. Critical Point: Since x<0x < 0, select critical points. We find that this holds true under certain conditions leading to: y>4x.y > 4x.

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