Let $f(x)=(1-x)^2 \sin ^2 x+x^2$ for all $x \in$ IR and let $g(x)=\int_1^x\left(\frac{2(t-1)}{t+1}-\ell\right.$ nt $) f(t) d t$ for all $x \in(1, \infty)$.
Consider the statements :
P : There exists some $x \in \operatorname{IR}$ such that $f(x)+2 x=2\left(1+x^2\right)$
Q : There exists some $\mathrm{x} \in \mathrm{IR}$ such that $2 \mathrm{f}(\mathrm{x})+1=2 \mathrm{x}(1+\mathrm{x})$ Then
Select the correct option:
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
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & f(x)+2 x=(1-x)^2 \sin ^2 x+x^2+2 x \\ & \because \quad f(x)+2 x=2\left(1+x^2\right) \\ & \Rightarrow \quad(1-x)^2 \sin ^2 x+x^2+2 x=2+2 x^2\end{aligned}$ $$
\begin{aligned}
(1-x)^2 \sin ^2 x= & x^2-2 x+1+1 \\
& =(1-x)^2+1 \\
\Rightarrow \quad(1-x)^2 \cos ^2 x= & -1
\end{aligned}
$$
which can never be possible
$\mathbf{P}$ is not true
$$
\begin{aligned}
& \Rightarrow \quad \text { Let } H(x)=2 f(x)+1-2 x(1+x) \\
& H(0)=2 f(0)+1-0=1 \\
& H(1)=2 f(1)+1-4=-3 \\
& \Rightarrow \quad \text { so } H(x) \text { has a solution }
\end{aligned}
$$
so $Q$ is true
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