Question

If $p,q,r$ are real numbers, then $f\left(x\right)=\left|\begin{array}{ccc}x+p^2& pq& pr\\ pq& x+q^2& qr\\ pr& qr& x+r^2\end{array}\right|$

• A. increases in $x<(-\frac{2}{3}(p^2+q^2+r^2\left)\right)$ or $x>0$
• B. decreases in $(-\frac{2}{3}(p^2+q^2+r^2),0)$
• C. decreases in $x<(-\frac{2}{3}(p^2+q^2+r^2\left)\right)$ or $x>0$
• D. increases in $(-\frac{2}{3}(p^2+q^2+r^2),0)$

Answer 1

Answer: (A), (B)

The correct options are
A increases in $x<(-\frac{2}{3}(p^2+q^2+r^2\left)\right)$ or $x>0$
B decreases in $(-\frac{2}{3}(p^2+q^2+r^2),0)$

f(x) $=\left|\begin{array}{ccc}x+p^2& pq& pr\\ pq& x+q^2& qr\\ pr& qr& x+r^2\end{array}\right|$
$\Rightarrow f\left(x\right)=x^3+(p^2+r^2+q^2)x^2$
$f^{\prime }\left(x\right)=3x^2+2x(p^2+q^2+r^2)=x(3x+2(p^2+q^2+r^2\left)\right)$
For increasing f'(x)>0
$\Rightarrow x>0$ and $3x+2(p^2+q^2+r^2)>0$
$x<-\frac{2}{3}(p^2+q^2+r^2)$ or $x>0$
For decreasing f'(x)<0
$\Rightarrow x<0$ and $3x+2(p^2+q^2+r^2)>0$