Question

The values of parameter a if $\left(x\right)$ has a negative point of local minimum are

• A. $\varphi$
• B. $(-3,3)$
• C. $(-\infty ,\frac{58}{14})$
• D. None of these

Answer 1

The correct option is B $(-3,3)$

$f\left(x\right)=x^3-3(7-a)x^2-3(9-a^2)x+2$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-6(7-a)x-3(9-a^2)$
For real root $D\ge 0$.
$\Rightarrow 49+a^2+14a+9-a^2\ge 0\Rightarrow a\le \frac{58}{14}$
When point of minima is negative, point of maxima is also negative.
Hence, equation $f^{\prime }\left(x\right)=3x^2-6(7-a)x-3(9-a^2)=0$ has both roots, negative.
For which sum of roots $=2(7-a)<0$ or $a>0$, which is not possible as from $\left(1\right),a\le \frac{58}{14}$.
When point of maxima is positive, point of minima also positive.
Hence, equation $f^{\prime }\left(x\right)=3x^2-6(7-a)x-3(9-a^2)=0$ has both roots positive.
For which sum of roots $=2(7-a)>0\Rightarrow a<7$
Also product of roots is positive $\Rightarrow -(9-a^2)>0$ or $a^2>9$
or $a\in (-\infty ,-3)\cup (3,\infty )$.
From $\left(1\right),\left(2\right)$ and $\left(3\right);a\in (-\infty ,-3)\cup (3,58/14)$
For points of extrema of opposite sign, equation $\left(1\right)$ has roots of opposite sign.
$\Rightarrow a\in (-3,3)$.