Answer:
Option C
Explanation:
Concept involved use of Newton leibnits formula
i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x0).g'(x)$
Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x
$g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$
$\forall x \epsilon (1, \infty)$
Here, $f(x)+2x=(1-x)^{2}.\sin^{2} x+x^{2}+2x$.............(i)
where $P:f(x)+2x=2(1+x)^{2}$ ...............(ii)
$\therefore$ $2(1+x^{2})=(1-x)^{2}\sin^{2}x+x^{2}+2x$
$\Rightarrow$ $(1-x)^{2} \sin^{2}x=x^{2}-2x+2$
$\Rightarrow$ $(1-x)^{2}\sin^{2}x=(1-x)^{2}+1$
$\Rightarrow$ $(1-x)^{2} \cos^{2}x=-1$
which is never possible
$\therefore$ p is false
Again h(x)=2f(x)+1-2x (1+x)
h(1)=2f(1)+1-4=-3
as h(0) h(1)<0
$\Rightarrow$ h(x) must have a solution
$\therefore$ Q is true