Question

Consider the statements:
$P$: There exists some $x\in R$ such that $f\left(x\right)+2x=2(1+x^2)$
$Q$: There exists some $x\in R$ such that $2f\left(x\right)+1=2x(1+x)$
Then

• A. both $P$ and $Q$ are true
• B. $P$ is false and $Q$ is true
• C. $P$ is true and $Q$ is false
• D. both $P$ and $Q$ are false

Answer 1

The correct option is B $P$ is false and $Q$ is true

$f\left(x\right)=(1-x{)}^2{\sin }^{2}x+x^2$ for all $x\in R$
$g\left(x\right)={\int }_1^x(\frac{2(t-1)}{t+1}-\ln t)f\left(t\right)dt$ for all $x\in (1,\infty )$
Now, for statement P :
$f\left(x\right)+2x=2(1+x^2)$ (i)
$(1-x{)}^2{\sin }^{2}x+x^2+2x=2+2x^2$
$(1-x{)}^2({\sin }^{2}x-1)=1$
$(1-x{)}^2{\cos }^{2}x=-1$
$\Rightarrow (1-x{)}^2{\cos }^{2}x=-1$
So equation (i) will not have real solution.
So, P is wrong.
For statement Q :
$2(1-x{)}^2{\sin }^{2}x+2x^2+1=2x+2x^2$ (ii)
$\Rightarrow 2{\sin }^{2}x=\frac{2x-1}{(1-x{)}^2}$
Let $h\left(x\right)=\frac{2x-1}{(1-x{)}^2}-2{\sin }^{2}x$
Now, $h\left(0\right)=-ve$
${lim}_{x\to 1^{-}}h\left(x\right)=+\infty$
$\Rightarrow$ $h\left(x\right)$ assumes both positive and negative values on $R$.
Also, $h\left(x\right)$ is continuous everywhere.
So, by intermediate value property of continuous functions, $h\left(x\right)$ vanishes at least once in $R$.
Hence, statement $Q$ is true.