Question

Let $\mathrm{f}\left(\mathrm{x}\right)={\left(1-x\right)}^2{\sin }^2\left(x\right)+{\mathrm{x}}^2$ for all $\mathrm{x}\in \mathrm{R}$, and let $\mathrm{g}\left(\mathrm{x}\right)=$integral from $1$ to $\mathrm{x}\left[\frac{\left[2(\mathrm{t}-1)\right]}{[\mathrm{t}+1]-\ln \mathrm{t}}\right]\mathrm{f}\left(\mathrm{t}\right)dt$ for all $\mathrm{x}\in (1,\infty ]$. Which of the following is true

• A. $\mathrm{g}$ is increasing on $(1,\infty )$
• B. $\mathrm{g}$ is decreasing on $(1,\infty )$
• C. $\mathrm{g}$ is increasing on $(1,2)$ and decreasing on$(2,\infty )$
• D. $\mathrm{g}$ is decreasing on $(1,2)$ and increasing on $(2,\infty )$

Answer 1

The correct option is B

$\mathrm{g}$ is decreasing on $(1,\infty )$

Explanation for the correct option:

Given that:

$$\begin{gathered} f\left(x\right)=(1-x{)}^2{\sin }^{2}x+x^2 \\ g(x)={\int }_1^x\left.\frac{2(t-1)}{(t+1)}-\ln t\right\}f(t)dt \end{gathered}$$

Now, $\mathrm{g}\left(\mathrm{x}\right)=\frac{2(\mathrm{t}-1)}{\mathrm{t}+1}-\mathrm{lnt}.\mathrm{f}\left(\mathrm{x}\right)$

Let $\phi =\frac{2(t-1)}{t+1}-\ln t$

${\phi }^1=2\left\{\frac{(t+1)·\left(1\right)-(t-1)\left(t\right)}{2(t+1{)}^2}\right\}-\ln t$

$=\frac{48}{(t+1{)}^2}-\frac{1}{t}$

$$\begin{gathered} {\phi }^1\left(x\right)=\frac{4}{(x+1)}-\frac{1}{x} \\ =\frac{4x-x^2-1-2x}{x(x+1{)}^2}=\frac{-x^2+2x-1}{x(x+1{)}^2} \end{gathered}$$

${\phi }^1\left(x\right)=\frac{-(x-1{)}^2}{x(x+1{)}^2}$

${\phi }^1\left(x\right)<0forallx\in (1,\infty ]$

Therefore, $\mathrm{g}$ is decreasing on $(1,\infty ]$

Hence, the correct answer is option (B).