I. HCF of p and (p + q) is G
II. HCF of p, (p - q) is G
Select the answer using the code given below :
We are given that the Highest Common Factor (HCF) of \(p\) and \(q\) is \(G\), where \(p > q\). This means we can express \(p\) and \(q\) as:
Here, \(x\) and \(y\) are coprime integers (their HCF is 1), and since \(p > q\), we know \(x > y\).
Statement I checks the HCF of \(p\) and \((p + q)\).
Let's find \((p + q)\): \(p + q = Gx + Gy = G(x + y)\)
Now, we find the HCF of \(p\) and \((p + q)\): \(HCF(p, p + q) = HCF(Gx, G(x + y))\)
Using the property \(HCF(ka, kb) = k \times HCF(a, b)\), we get: \(HCF(p, p + q) = G \times HCF(x, x + y)\)
We know that \(HCF(a, b) = HCF(a, b - a)\). Applying this: \(HCF(x, x + y) = HCF(x, (x + y) - x) = HCF(x, y)\)
Since \(x\) and \(y\) are coprime, \(HCF(x, y) = 1\). Therefore: \(HCF(p, p + q) = G \times 1 = G\)
So, Statement I is correct.
Statement II checks the HCF of \(p\) and \((p - q)\).
Let's find \((p - q)\): \(p - q = Gx - Gy = G(x - y)\)
Now, we find the HCF of \(p\) and \((p - q)\): \(HCF(p, p - q) = HCF(Gx, G(x - y))\)
Applying the property \(HCF(ka, kb) = k \times HCF(a, b)\): \(HCF(p, p - q) = G \times HCF(x, x - y)\)
Using the property \(HCF(a, b) = HCF(a, a - b)\) or similar difference properties: \(HCF(x, x - y) = HCF(x - (x - y), x - y) = HCF(y, x - y)\)
Alternatively, \(HCF(x, x - y) = HCF(x, y)\) because any common factor of \(x\) and \(x-y\) must also divide their difference, which is \(y\). Since \(x\) and \(y\) are coprime, \(HCF(x, y) = 1\). Therefore: \(HCF(x, x - y) = 1\)
So, \(HCF(p, p - q) = G \times 1 = G\).
Statement II is also correct.
Both Statement I and Statement II are correct. Thus, the correct option includes both.
The HCF of \(x\) and \(y\) is \(H\). Consider the following statements in respect of the HCF of \(p=\frac{x^3 + y^3}{x^2-xy+y^2}\) and \(q=\frac{x^3-y^3}{x^2+xy+y^2}\) :
I. The HCF of \(p\) and \(q\) can be \(H\).
II. The HCF of \(p\) and \(q\) can be \(2H\).
Which of the statements given above is/are correct?
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