Question

Let $f\left(x\right)={\left(x^2-1\right)}^n(x^2+x+1)$, then $f\left(x\right)$ has local extremum at $x=1$ when least value of $n$:

• A. $2$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Given that: $f\left(x\right)={\left(x^2-1\right)}^n(x^2+x+1)$

Calculating the local extremum:

$$\begin{gathered} f\text{'}\left(x\right)={\left(x^2-1\right)}^n(x^2+x+1) \\ =n{(x^2-1)}^{n-1}\times 2x\times (x^2+x+1)+{\left(x^2-1\right)}^n\times (2x+1) \\ ={(x^2-1)}^{n-1}\left[2nx.(x^2+x+1)+\left(x^2-1\right).(2x+1)\right] \\ f\text{'}\left(1\right)=0^{n-1}\left[2n\left(3\right)+0\times \left(3\right)\right] \end{gathered}$$

When, $n>1,f\text{'}\left(1\right)=0$

So, the least value of $n$ according to the options given will be $2$.

Hence, the correct option is (A)