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Question:
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Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$.
Question: Consider the statements:
$P$ : There exists some $x \in \mathrm{R}$ such that $f(x)+2 x$
$=2\left(1+x^{2}\right)$
$Q$ : There exists some $x \in \mathrm{R}$ such that $2 f(x)+1$ $=2 x(1+x)$
Then
Options:
Question: Consider the statements:
$P$ : There exists some $x \in \mathrm{R}$ such that $f(x)+2 x$
$=2\left(1+x^{2}\right)$
$Q$ : There exists some $x \in \mathrm{R}$ such that $2 f(x)+1$ $=2 x(1+x)$
Then
Solution:
2906 Upvotes
Verified Answer
The correct answer is:
$P$ is false and $Q$ is true
For the statement $P, f(x)+2 x=2\left(1+x^{2}\right)$
$\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2\left(1+x^{2}\right)$
$\Rightarrow(1-x)^{2} \sin ^{2} x=x^{2}-2 x+1+1$
$\Rightarrow(1-x)^{2} \sin ^{2} x=(1-x)^{2}+1$
$\Rightarrow(1-x)^{2} \cos ^{2} x=-1$, which is not possible for any real value of $x$.
Hence $P$ is not true.
Let $H(x)=2 f(x)+1-2 x(1+x)$
$H(0)=2 f(0)+1-0=1$
and $H(1)=2 f(1)+1-4=-3$
Hence, $H(x)$ has a solution in $(0,1)$
Therefore, $Q$ is true.
$\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2\left(1+x^{2}\right)$
$\Rightarrow(1-x)^{2} \sin ^{2} x=x^{2}-2 x+1+1$
$\Rightarrow(1-x)^{2} \sin ^{2} x=(1-x)^{2}+1$
$\Rightarrow(1-x)^{2} \cos ^{2} x=-1$, which is not possible for any real value of $x$.
Hence $P$ is not true.
Let $H(x)=2 f(x)+1-2 x(1+x)$
$H(0)=2 f(0)+1-0=1$
and $H(1)=2 f(1)+1-4=-3$
Hence, $H(x)$ has a solution in $(0,1)$
Therefore, $Q$ is true.
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