Question

Let $f\left(x\right)=(1-x{)}^2{\sin }^{2}x+x^2$ for all $x\in R,$ and let $g\left(x\right)=\underset{1}{\overset{x}{\int }}(\frac{2(t-1)}{t+1}-\ln t)f\left(t\right)dt$ for all $x\in (1,\infty )$
Consider the statements:
$P:\text{There exists some}x\in IR\text{such that}f\left(x\right)+2x=2(1+x^2)$
$Q:\text{There exists some}x\in IR\text{such that}2f\left(x\right)+1=2x(1+x)$
Then

• A. $\text{both P and Q are true}$
• B. $\text{P is true and Q is false}$
• C. $\text{P is false and Q is true}$
• D. $\text{both P and Q are false}$

Answer 1

The correct option is C $\text{P is false and Q is true}$

$P:\left({\sin }^{2}x\right)(1-x{)}^2+x^2+2x=2+2x^2$
$\Rightarrow \left({\sin }^{2}x\right)(1-x{)}^2=x^2-2x+2$
$\Rightarrow {\sin }^{2}x=\frac{(1-x{)}^2+1}{(1-x{)}^2}=1+\frac{1}{(1-x{)}^2},$ which is greater than $1\Rightarrow$ No solution
$\therefore P$ is false.

$f\left(x\right)=(1-x{)}^2{\sin }^{2}x+x^2$

$Q:2f\left(x\right)+1=2x(1+x)$
$Q:2{\sin }^{2}x=\frac{2x-1}{(1-x{)}^2}$
$0\le \frac{2x-1}{2(1-x{)}^2}\le 1.$
This inequality is satisfied by some value of $x$