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If \(P^2\) varies as R and \(Q^2\) varies as R, \((P \ne Q)\), then which of the following are correct?
1. \(P^2 + Q^2\) varies as R.
2. PQ varies as R.
3. \(P^2 - Q^2\) varies as R.
Select the correct answer using the code given below:

This question was previously asked in
CDS 2 2024 Maths Question Paper (01-Sep-2024)
The correct answer is
1 and 2 only

Variation Laws Explained: P, Q, and R

This problem delves into the concept of variation, specifically direct variation. When a quantity \(A\) varies as another quantity \(B\), it signifies a direct proportionality, mathematically represented as \(A = kB\), where \(k\) is a non-zero constant of proportionality.

The problem provides the following information:

  • \(P^2\) varies as \(R\), meaning \(P^2 = k_1 R\), where \(k_1\) is a constant. Since \(P^2\) is non-negative, \(k_1 ≥ 0\). For variation, we assume \(k_1 ≠ 0\).
  • \(Q^2\) varies as \(R\), meaning \(Q^2 = k_2 R\), where \(k_2\) is a constant. Similarly, \(k_2 ≥ 0\), and we assume \(k_2 ≠ 0\).
  • Additionally, \(P ≠ Q\).

We need to evaluate the variation of three expressions involving \(P\) and \(Q\) with respect to \(R\).

Analyzing Statements on P, Q, R Variation

Statement 1 Analysis: \(P^2 + Q^2\) and \(R\)

We substitute the given variation relationships:

  • \(P^2 = k_1 R\)
  • \(Q^2 = k_2 R\)

Adding these equations gives:

\(P^2 + Q^2 = k_1 R + k_2 R\)

Factoring out \(R\) yields:

\(P^2 + Q^2 = (k_1 + k_2) R\)

Let the combined constant be \(k_3 = k_1 + k_2\). Since \(k_1 > 0\) and \(k_2 > 0\), their sum \(k_3\) is also positive (\(k_3 > 0\)). This confirms that \(P^2 + Q^2\) varies directly as \(R\).

Result: Statement 1 is correct.

Statement 2 Analysis: \(PQ\) and \(R\) Variation

From the initial variations, we can write:

  • \(P = ±√(k_1 R)\)
  • \(Q = ±√(k_2 R)\)

Calculating the product \(PQ\):

\(PQ = (±√(k_1 R)) × (±√(k_2 R))\)

\(PQ = (±√(k_1 k_2)) R\)

Let \(k_4 = ±√(k_1 k_2)\). This \(k_4\) represents a constant value (positive or negative, depending on the signs of \(P\) and \(Q\)). Since \(k_1 > 0\) and \(k_2 > 0\), \(√(k_1 k_2)\) is a real positive number, making \(k_4\) a valid constant.

Therefore, \(PQ = k_4 R\), confirming that \(PQ\) varies directly as \(R\).

Result: Statement 2 is correct.

Statement 3 Analysis: \(P^2 - Q^2\) and \(R\) Variation

Using the given relationships:

  • \(P^2 = k_1 R\)
  • \(Q^2 = k_2 R\)

Subtracting the second from the first gives:

\(P^2 - Q^2 = k_1 R - k_2 R\)

Factoring out \(R\):

\(P^2 - Q^2 = (k_1 - k_2) R\)

This expression varies as \(R\) only if the constant of proportionality, \((k_1 - k_2)\), is non-zero. However, the problem conditions do not exclude the case where \(k_1 = k_2\). If \(k_1 = k_2\), then \(P^2 - Q^2 = (0) R = 0\). A value that is constantly zero does not exhibit variation in the context of proportionality.

Because this statement is not universally true under the given conditions, it is considered incorrect.

Result: Statement 3 is incorrect.

Overall Conclusion on Variation Statements

Our analysis shows that statements 1 and 2 hold true, while statement 3 does not hold true in all cases allowed by the problem statement.

Thus, the correct choice involves only statements 1 and 2.

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    Select the answer using the code given below :
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