<p>If \(f(x) = \left(x + \dfrac{1}{x}\right)^{2n} + \left(x - \dfrac{1}{x}\right)^{2n}\), then \(f(x)\) is a polynomial function which is an even function. (State whether true or false.)</p>
Step-by-Step Solution
Key Concept: When expanding (x + 1/x)^(2n) and (x - 1/x)^(2n) using binomial theorem, terms with odd powers of 1/x cancel out when added, leaving only even-powered terms in x, making f(x) a polynomial.
<p><strong>Step 1:</strong> Expand using binomial theorem:</p><p>$(x + \frac{1}{x})^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} x^{2n-r} \cdot \frac{1}{x^r} = \sum_{r=0}^{2n} \binom{2n}{r} x^{2n-2r}$</p><p>$(x - \frac{1}{x})^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} x^{2n-r} \cdot (-\frac{1}{x})^r = \sum_{r=0}^{2n} \binom{2n}{r} (-1)^r x^{2n-2r}$</p><p><strong>Step 2:</strong> Add the two expansions:</p><p>$f(x) = \sum_{r=0}^{2n} \binom{2n}{r} x^{2n-2r}(1 + (-1)^r)$</p><p>When $r$ is odd: $(1 + (-1)^r) = 0$ (terms cancel)</p><p>When $r$ is even: $(1 + (-1)^r) = 2$ (terms survive)</p><p><strong>Step 3:</strong> Therefore $f(x) = 2\sum_{k=0}^{n} \binom{2n}{2k} x^{2n-4k}$</p><p>All exponents are even integers (like $x^0, x^2, x^4, ..., x^{2n}$), so f(x) is a polynomial.</p><p><strong>Step 4:</strong> Check if even: $f(-x) = 2\sum_{k=0}^{n} \binom{2n}{2k} (-x)^{2n-4k} = 2\sum_{k=0}^{n} \binom{2n}{2k} x^{2n-4k} = f(x)$ ✓</p><p>∴ <strong>TRUE</strong> — f(x) is both a polynomial and an even function</p>
Correct Answer: A