Question

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

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

Answer 1

Answer: (C)

$f(x)=(1-x{)}^2{\sin }^{2}x+x^2\forall x\in R$
$g(x)=\underset{1}{\overset{x}{\int }}\left(\frac{2(t-1)}{t+1}-\ln t\right)f(t)$ dt $\forall x\in (1,\infty )$
For statement $P$ :
$f(x)+2x=2\left(1+x^2\right)\dots$(i)
$(1-x{)}^2{\sin }^{2}x+x^2+2x=2+2x^2$
$(1-x{)}^2{\sin }^{2}x=x^2-2x+2=(x-1{)}^2+1$
$(1-x{)}^2\left({\sin }^{2}x-1\right)=1$
$-(1-x{)}^2{\cos }^{2}x=1$
$(1-x{)}^2\cdot {\cos }^{2}x=-1$
So equation (i) will not have real solution
So, $P$ is wrong.
For statement $Q$ :
$2(1-x{)}^2{\sin }^{2}x+2x^2+1=2x+2x^2\dots$(ii)
$2(1-x{)}^2{\sin }^{2}x=2x-1$
$2{\sin }^{2}x=\frac{2x-1}{(1-x{)}^2}$ Let $h(x)=\frac{2x-1}{(1-x{)}^2}-2{\sin }^{2}x$
Clearly $h(0)=-$ ve, $\underset{x\to 1^{-}}{\lim }h(x)=+\infty$
So by IVT, equation (ii) will have solution. So, $Q$ is correct.