Question

Let p(x) = x⁵⁷ + 3x¹⁰ - 21x³ + x² + 21 and
\(q(x)=p(x)+\sum\limits_{j=1}^{57}p^{(j)}(x) \ \text{for all }x \in \R,\)
where p^(j)(x) denotes the j^th derivative of p(x). Then the function q admits

• A. NEITHER a global maximum NOR a global minimum on \(\R\)
• B. a global maximum but NOT a global minimum on \(\R\)
• C. a global minimum but NOT a global maximum on \(\R\)
• D. a global minimum and a global maximum on \(\R\)

Answer 1

The Correct Option is A

To solve this problem, we first need to understand the structure of the functions involved. We are given:

\(p(x) = x^{57} + 3x^{10} - 21x^3 + x^2 + 21\) 

and

\(q(x) = p(x) + \sum\limits_{j=1}^{57} p^{(j)}(x)\)

To determine the possible existence of a global maximum or minimum for \(q(x)\), we need to analyze its behavior or degree.

1. The highest degree of \(p(x)\) is 57, so the leading term is \(x^{57}\). This term will dominate the behavior of \(p(x)\) and thus \(q(x)\) as \(x \to \infty\) or \(x \to -\infty\).
2. Each derivative \(p^{(j)}(x)\) reduces the degree of the polynomial by 1. Thus, for \(j = 57\), \(p^{(57)}(x)\) results in a constant term (the derivative of \(x^0\)). Consequently, the sum of all derivatives up to \(p^{(57)}(x)\) will not change the leading term's long-term behavior, which is dominated by \(x^{57}\).
3. Since \(q(x)\) is dominated by a polynomial of odd degree (57), it will have the form of a standard \(x^{odd}\) shape: falling sharply towards \(-\infty\) when \(x \to -\infty\) and rising sharply to \(+\infty\) when \(x \to +\infty\).

This behavior implies that the function \(q(x)\) does not attain a global maximum or a global minimum, as its limits on both ends are infinite.

Hence, the correct answer is: NEITHER a global maximum NOR a global minimum on \(\R\).