Understanding Variation Concepts

The question deals with the concept of direct and inverse variation. When we say 'x varies as y', it means that x is directly proportional to y. Mathematically, this relationship can be expressed as:

\[ x \propto y \] \[\text{or} \] \[ x = ky \] where \(k\) is a constant of proportionality.

We need to examine each given statement based on this initial condition \(x = ky\).

Analyzing Statement 1: \(x^2 + y^2\) varies as \(x^2 - y^2\)

Statement 1 claims that \(x^2 + y^2 \propto x^2 - y^2\). This is true if the ratio \(\frac{x^2 + y^2}{x^2 - y^2}\) is a constant.

Let's substitute \(x = ky\) into the ratio:

\[ \frac{x^2 + y^2}{x^2 - y^2} = \frac{(ky)^2 + y^2}{(ky)^2 - y^2} \] \[ = \frac{k^2y^2 + y^2}{k^2y^2 - y^2} \] \[ = \frac{y^2(k^2 + 1)}{y^2(k^2 - 1)} \] (Assuming \(y \neq 0\)) \[ = \frac{k^2 + 1}{k^2 - 1} \]

Since \(k\) is a constant, \(k^2 + 1\) and \(k^2 - 1\) are also constants (assuming \(k^2 \neq 1\)). Therefore, their ratio \(\frac{k^2 + 1}{k^2 - 1}\) is a constant.

Thus, \(x^2 + y^2\) varies as \(x^2 - y^2\).

Statement 1 is correct.

Analyzing Statement 2: \(\frac{x}{y^2}\) varies inversely as y

Statement 2 claims that \(\frac{x}{y^2}\) varies inversely as y. This means \(\frac{x}{y^2} \propto \frac{1}{y}\). This proportionality is true if the product \(\left(\frac{x}{y^2}\right) \cdot y\) is a constant.

Let's evaluate the product by substituting \(x = ky\):

\[ \left(\frac{x}{y^2}\right) \cdot y = \left(\frac{ky}{y^2}\right) \cdot y \] \[ = \frac{ky}{y} \] (Assuming \(y \neq 0\)) \[ = k \]

Since \(k\) is a constant, the product \(\left(\frac{x}{y^2}\right) \cdot y\) is a constant.

Thus, \(\frac{x}{y^2}\) varies inversely as y.

Statement 2 is correct.

Analyzing Statement 3: \(\sqrt[n]{x^2y}\) varies as \(\sqrt[2n]{x^4y^2}\)

Statement 3 claims that \(\sqrt[n]{x^2y} \propto \sqrt[2n]{x^4y^2}\). This is true if the ratio \(\frac{\sqrt[n]{x^2y}}{\sqrt[2n]{x^4y^2}}\) is a constant.

We can rewrite the roots using fractional exponents:

\[ \sqrt[n]{x^2y} = (x^2y)^{1/n} = x^{2/n}y^{1/n} \] \[ \sqrt[2n]{x^4y^2} = (x^4y^2)^{1/(2n)} = x^{4/(2n)}y^{2/(2n)} = x^{2/n}y^{1/n} \]

Notice that \(\sqrt[n]{x^2y}\) is mathematically identical to \(\sqrt[2n]{x^4y^2}\).

Let \(A = \sqrt[n]{x^2y}\) and \(B = \sqrt[2n]{x^4y^2}\). We found that \(A = B\).

Now consider the ratio \(\frac{A}{B}\):

\[ \frac{A}{B} = \frac{\sqrt[n]{x^2y}}{\sqrt[2n]{x^4y^2}} = \frac{\sqrt[n]{x^2y}}{\sqrt[n]{x^2y}} \] \[ = 1 \]

The ratio is 1, which is a constant.

Alternatively, since \(\sqrt[n]{x^2y} = \sqrt[2n]{x^4y^2}\), if we let \(Z = \sqrt[n]{x^2y}\), then \(\sqrt[2n]{x^4y^2}\) is also equal to \(Z\). The statement becomes "Z varies as Z", which is trivially true with a constant of proportionality of 1.

Thus, \(\sqrt[n]{x^2y}\) varies as \(\sqrt[2n]{x^4y^2}\).

Statement 3 is correct.

Conclusion

Based on our analysis, all three statements are correct given the condition that x varies as y.

Statement 1: \(x^2 + y^2\) varies as \(x^2 - y^2\) is correct.

Statement 2: \(\frac{x}{y^2}\) varies inversely as y is correct.

Statement 3: \(\sqrt[n]{x^2y}\) varies as \(\sqrt[2n]{x^4y^2}\) is correct.

Therefore, 1, 2, and 3 are all correct.

Revision Table: Understanding Variation

Additional Information: Properties of Proportionality

If \(a \propto b\), meaning \(a = kb\), then several properties hold:

• \(b \propto a\) (If \(a = kb\), then \(b = (1/k)a\))
• \(a^n \propto b^n\) for any real number n
• \(ca \propto cb\) for any non-zero constant c
• If \(c \propto d\), then \(ac \propto bd\) (Multiplying proportionalities)
• If \(c \propto d\), then \(\frac{a}{c} \propto \frac{b}{d}\) (Dividing proportionalities, assuming \(c, d \neq 0\))
• Componendo property: If \(a \propto b\), then \(a+b \propto b\) (Since \((a+b)/b = (kb+b)/b = k+1\), a constant)
• Dividendo property: If \(a \propto b\), then \(a-b \propto b\) (Since \((a-b)/b = (kb-b)/b = k-1\), a constant, assuming \(k \neq 1\))
• Componendo and Dividendo property: If \(a \propto b\), then \(a+b \propto a-b\) (Since \((a+b)/(a-b) = (kb+b)/(kb-b) = (k+1)/(k-1)\), a constant, assuming \(k \neq 1\))

In Statement 1, we used the Componendo and Dividendo property in reverse. If \(\frac{x^2+y^2}{x^2-y^2}\) is a constant, say C, then \(x^2+y^2 = C(x^2-y^2)\). Expanding and rearranging gives \(x^2(1-C) = y^2(-1-C)\), or \(x^2/y^2 = -(1+C)/(1-C)\). Since the right side is a constant, \(x^2/y^2\) is a constant, implying \(x/y\) is a constant (or \(x^2 \propto y^2\)). If \(x \propto y\), then \(x^2 \propto y^2\), so \(\frac{x^2}{y^2}\) is constant. This implies \(\frac{x^2/y^2 + 1}{x^2/y^2 - 1}\) is constant by Componendo and Dividendo. Multiplying numerator and denominator by \(y^2\) gives \(\frac{x^2+y^2}{x^2-y^2}\) is constant. This confirms Statement 1 from the initial condition \(x \propto y\).