Question

If $f\left(x\right)={(x^2-1)}^{n+1}(x^2+x+1),nϵN$ and $f\left(x\right)$ has a local extremum at $x=1$, then $n=$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B), (D)

The correct options are
B 3
D 5

$f\left(x\right)={(x^2-1)}^{n+1}(x^2+x+1),nϵN$
$\frac{dy}{dx}={(x^2-1)}^{n+1}(2x+1)+(n+1){(x^2-1)}^{n}2x(x^2+x+1)$
$={(x^2-1)}^n[(x^2-1)(2x+1)+2x(n+1)(x^2+x+1)]$
Since $f\left(x\right)$ has a local extremum $\frac{dy}{dx}$ must change sign i.e. $f^{\prime }\left(1^{+}\right)and{f}^{\prime }\left(1^{-}\right)$ should be of opposite signs.
Now we know that $x^2+x+1$ is always +ive $(b^2-4ac=-ive)$ and the first term tends to zero as $x\to 1.$
Hence the sign of $\frac{dy}{dx}$ will depend upon ${(x^2-1)}^n.$
But $x^2-1=(x+1)(x-1)$ is +ive for $x>1$ and -ive for $x<1.$
Hence if $n$ is even then $f^{\prime }\left(1^{+}\right)$ and $f^{\prime }\left(1^{-}\right)$ will have the same sign, but they must be of opposite sign for the existence of extremum.
Thus $n$ must be odd.
Hence (b) and (d) are the correct choices.
Ans: B,D