Question

The function $f\left(x\right)=2\log (x-3)-x^2+6x+3$ increases in the interval

• A. $(3,4)$
• B. $(-\infty ,2)$
• C. $(3,\infty )$
• D. None ot these

Answer 1

The correct option is A $(3,4)$

$f^{\prime }\left(x\right)=\frac{2}{x-3}-2x+6=\frac{2}{x-3}-2(x-3)=2.\frac{1-(x-3{)}^2}{x-3}$
$\Rightarrow f^{\prime }\left(x\right)=-2.\frac{x^2-6x+8}{x-3}=-2.\frac{(x-2)(x-4)}{x-3}$
For $f\left(x\right)$ to be increasing, $f^{\prime }\left(x\right)>0$
$\Rightarrow \frac{(x-2)(x-4)}{x-3}<0\Rightarrow x\in (-\infty ,2)\cup (3,4)$
Also for $\log (x-3)$ to be defined $x-3>0\Rightarrow x>3$
Hence interval of increasing is $(3,4)$