Question

If the points of local extremum of $f\left(x\right)=x^3-3a{x}^2+3(a^2-1)x+1$ lies between $-2\&4$, then ${}^{\prime }{a}^{\prime }$ belongs to

• A. $(-2,2)$
• B. $(-\infty ,-1)\cup (3,\infty )$
• C. $(-1,3)$
• D. $(3,\infty )$

Answer 1

Answer: (C)

$(-1,3)$

$f^{\prime }\left(x\right)=3(x^2-2ax+a^2-1)$
Clearly roots of $f\left(x\right)=0$ must be distinct and lie in the interveal $(-2,4)$.
$\therefore D>0\Rightarrow a\in R........\left(1\right)$
$f^{\prime }(-2)>0\Rightarrow a^2+4a+3>0$
$\Rightarrow a<-3ora>-1.......\left(2\right)$
$f^{\prime }\left(4\right)>0\Rightarrow a^2-8a+15>0$
$\Rightarrow a>5ora<3........\left(3\right)$
and $-2<-\frac{B}{2A}<4\Rightarrow -2<a<4........\left(4\right)$
From $\left(1\right),\left(2\right),\left(3\right)\&\left(4\right),a\in (-1,3)$