Question

The function $\mathrm{f}\left(\mathrm{x}\right)=2\log (\mathrm{x}-2)-{\mathrm{x}}^2+4\mathrm{x}+1$ increases on the interval

• A. $(1,2)$
• B. $\left(2,3\right)$
• C. $\left(\frac{5}{2},3\right)$
• D. $\left(2,4\right)$

Answer 1

Answer: (B), (C)

$\left(\frac{5}{2},3\right)$

Explanation for correct options:

As we know that, the necessary and sufficient condition for differentiable function defined on $\left(\mathrm{a},\mathrm{b}\right)$ to be closely increasing on $\left(\mathrm{a},\mathrm{b}\right)$ is that $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$ for all $\mathrm{x}\in \left(\mathrm{a},\mathrm{b}\right)$.

It is given that the function $\mathrm{f}\left(\mathrm{x}\right)=2\log (\mathrm{x}-2)-{\mathrm{x}}^2+4\mathrm{x}+1$.

Here, $\mathrm{f}\left(\mathrm{x}\right)$ is valid when $\mathrm{x}>2$.

So,

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{2}{\mathrm{x}-2}-2\mathrm{x}+4$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{2}{\mathrm{x}-2}-2\left(\mathrm{x}-2\right)$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=2\left(\frac{1-{\left(\mathrm{x}-2\right)}^2}{\mathrm{x}-2}\right)$

For $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$, $\mathrm{x}>2$.

Therefore,

$1-{\left(\mathrm{x}-2\right)}^2>0$,

$\left(\mathrm{x}-3\right)\left(\mathrm{x}-1\right)<0$ and $\mathrm{x}>2$

That is, $1<\mathrm{x}<3$ and $\mathrm{x}>2$

Critical points are $1,2$ and $3$.

Therefore, $\mathrm{f}\left(\mathrm{x}\right)$ is increasing in $\left(-\infty ,1\right)\cup \left(2,3\right)$ decreasing in $\left(1,2\right)\cup \left(3\infty \right)$.

$\because \left(\frac{5}{2},3\right)\subset \left(2,3\right)$ the function increases in this interval as well

Hence, options $\left(\mathrm{B}\right)$ and $\left(\mathrm{C}\right)$ are correct answer.