Question

Let $f\left(x\right)=1+2^2{x}^2+3x^2{x}^4+4^2{x}^6....+n^2{x}^{2n-2}=+{(n+1)}^2{x}^{2n}$, then $f\left(x\right)$ has:

• A. exactly one minimum
• B. exactly one maximum
• C. at least one maximum
• D. None of these

Answer 1

The correct option is A

exactly one minimum

Determine the condition of $f\left(x\right)$

Given that:

$f\left(x\right)=1+2^2{x}^2+3x^2{x}^4+4^2{x}^6....+n^2{x}^{2n-2}=+{(n+1)}^2{x}^{2n}$

Differentiating the given equation:

$$\begin{gathered} f\text{'}\left(x\right)=2.2^{2}x+4.3x^2{x}^3+6.4^2{x}^5....+(2n-2).n^2{x}^{2n-3} \\ =x(2.2^2+4.3x^2{x}^2+6.4^2{x}^4....+(2n-2).n^2{x}^{2n-4}) \end{gathered}$$

Now, for maxima or minima,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \Rightarrow x=0 \end{gathered}$$

Therefore, $f\left(x\right)$) has only minimum which is $0.$

Hence, the correct answer is Option (A).