(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
Question 8
(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.
Assumption: Let p = 2m + 1 and q = 2n + 1, where m and n are integers.
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
Start with the inequality:
(x+y)<9x2+y2.
Rearranging gives:
y2−y−(9x2−x)>0.
Substituting: Let a=y and b=3x, the quadratic inequality becomes:
y2−y−9x2+x>0.
Analysis of Quadratic:
The quadratic can be examined using the discriminant:
D=(−1)2−4(1)(−9x2+x)=1+36x2−4x.
We find the roots of the function in terms of y:
y=21±1+36x2−4x.
Critical Point: Since x<0, select critical points. We find that this holds true under certain conditions leading to:
y>4x.