Question

For \(n\isin\N\) and \(x\isin[1,\infin)\), let
\(f_n(x)=\int\limits_{0}^{\pi}(x^2+(cos\theta)\sqrt{x^2-1})^nd\theta\)
Then which one of the following is true?

• A. f_n(x) is not a polynomial in x if n is odd and n ≥ 3.
• B. f_n(x) is not a polynomial in x if n is even and n ≥ 4.
• C. f_n(x) is a polynomial in x for all \(n\isin\N \).
• D. f_n(x) is not a polynomial in x for any n ≥ 3.

Answer 1

The Correct Option is C

To determine whether \(f_n(x)\) is a polynomial in \(x\), let's analyze the given function:

\(f_n(x)=\int\limits_{0}^{\pi}(x^2+(cos\theta)\sqrt{x^2-1})^nd\theta\)

We need to determine the nature of this function with respect to \(x\) for various values of \(n\).

First, consider the expression inside the integral: \(x^2 + (\cos \theta) \sqrt{x^2 - 1}\).

• The term \(x^2\) is clearly a polynomial term.
• The term \((\cos \theta) \sqrt{x^2 - 1}\) involves a square root and is multiplied by \(\cos \theta\).

To understand whether the complete expression \( (x^2 + (\cos \theta) \sqrt{x^2 - 1})^n \) can remain a polynomial when integrated, consider expanding it using the binomial theorem. The expansion will involve terms of the form:

• \(x^{2k} \cdot ((\cos \theta) \sqrt{x^2 - 1})^{n-k}\)

The key observation is that for each \((\cos \theta) \sqrt{x^2 - 1}\) term, the integration over a full period from \([0, \pi]\) of \(\cos \theta\) (an odd function) contributes nothing when integrating over a symmetric interval as it sums to zero.

Hence, all non-integral powers of square root terms will eventually counteract via the properties of trigonometric functions upon integration, effectively leaving the polynomial terms of \(x\).

Furthermore, the expansion will contain only even powered terms (as a result of attributing even powers of the root lead by binomial expansion symmetry).

Thus, \(f_n(x)\) reduces to a polynomial after evaluating the integral for any \(n\isin\N \).

Therefore, the statement that \(f_n(x)\) is a polynomial in \(x\) for all \(n\) is correct.

Hence, the correct answer is: \(f_n(x)\) is a polynomial in \(x\) for all \(n\isin\N \).