<p>The number of values of <em>k</em> for which \([x^2 - (k-2)x + k^2] \times [x^2 + kx + (2k-1)]\) is a perfect square is</p>
Step-by-Step Solution
Key Concept: For a product of two quadratics to be a perfect square, either both must be perfect squares, or they must be equal (giving a perfect square when multiplied), or one must be a perfect square and the other must equal it.
<p><strong>Step 1:</strong> Let $P_1 = x^2 - (k-2)x + k^2$ and $P_2 = x^2 + kx + (2k-1)$.</p><p><strong>Step 2:</strong> For $P_1 \times P_2$ to be a perfect square, we need $P_1 = P_2$ (making the product $P_1^2$).</p><p><strong>Step 3:</strong> Set $P_1 = P_2$:</p><p>$x^2 - (k-2)x + k^2 = x^2 + kx + (2k-1)$</p><p>$-(k-2)x + k^2 = kx + (2k-1)$</p><p>$-kx + 2x + k^2 = kx + 2k - 1$</p><p>$2x - 2kx = 2k - 1 - k^2$</p><p>$x(2 - 2k) = 2k - 1 - k^2$</p><p><strong>Step 4:</strong> For this to hold for all $x$, we need both coefficients to vanish:</p><p>Coefficient of $x$: $2 - 2k = 0 \Rightarrow k = 1$</p><p>Constant term: $2k - 1 - k^2 = 0 \Rightarrow k^2 - 2k + 1 = 0 \Rightarrow (k-1)^2 = 0 \Rightarrow k = 1$</p><p><strong>Step 5:</strong> Verify $k = 1$: $P_1 = x^2 + x + 1$ and $P_2 = x^2 + x + 1$. Both are identical, so $P_1 \times P_2 = (x^2 + x + 1)^2$ ✓</p><p><strong>Step 6:</strong> Check if both can be perfect squares individually: This gives the same condition $k = 1$.</p><p>∴ <strong>Answer: B (1 value, k = 1)</strong></p>
Correct Answer: B