Question

The quadratic polynomial \(p(x)\) has roots \(1\) and \(α\), while quadratic polynomial \(q(x)\) has roots \(1\) and \(β\). Let \(α\) and \(β\) be the roots of \(r(x) = p(x)+q(x)\). Then \(\underset{x\rightarrow \infty }{lim}[\sqrt{p(x)}-\sqrt{q(x)}] =\)

• A. \(0\)
• B. \(-1\)
• C. \(1\)
• D. \(\frac{1}{2}\)

Hint:

Show that the root condition forces \(\alpha=\beta=1\), so \(p(x)=q(x)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We take \(p\) and \(q\) as monic quadratics, as the question intends. Then \(p(x) = (x-1)(x-\alpha)\) and \(q(x) = (x-1)(x-\beta)\). The roots of \(r = p+q\) are required to be \(\alpha\) and \(\beta\).

Step 2: Form r(x):
\[ r(x) = (x-1)\big[(x-\alpha)+(x-\beta)\big] = (x-1)(2x-\alpha-\beta) \]
So the roots of \(r\) are \(1\) and \(\frac{\alpha+\beta}{2}\).

Step 3: Match the roots:
The set \(\{\alpha,\beta\}\) must equal \(\{1, \frac{\alpha+\beta}{2}\}\).
Case 1: \(\alpha = 1\). Then \(\beta = \frac{1+\beta}{2}\), which gives \(\beta = 1\).
Case 2: \(\alpha = \frac{\alpha+\beta}{2}\), so \(\alpha = \beta\), and then \(\beta = 1\) as the other root, so again \(\alpha=\beta=1\).
In both cases \(\alpha=\beta=1\).

Step 4: Evaluate the limit:
Now \(p(x) = q(x) = (x-1)^2\), so \(\sqrt{p(x)} - \sqrt{q(x)} = |x-1| - |x-1| = 0\) for every \(x\).
\[ \lim_{x\to\infty}\big[\sqrt{p(x)} - \sqrt{q(x)}\big] = 0 \]

Step 5: Why the other options are wrong:
A nonzero limit such as \(\pm 1\) or \(\frac{1}{2}\) would need the two quadratics to differ in their linear term, which would happen only if \(\alpha \ne \beta\). The root condition rules this out.

Final Answer:
The two polynomials are identical, so the limit is \(0\), option (A). \[ \boxed{0} \]