Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in R$ and let…

Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in R$ 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
  1. both $P$ and $Q$ are true
  2. $P$ is true and $Q$ is false
  3. $P$ is false and $Q$ is true
  4. both $P$ and $Q$ are false

Solution

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.

Asked in: JEE Advanced 2012 (Paper 2)

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