Let F(x) = f(x) + 2x – 2 – 2x²

F(x) = (1 – x)² sin²x + x² + 2x – 2 – 2x²

F(x) = (1 – x)² sin²x – (1 – x)² – 1

F(x) = – (1 – x)² cos²x – 1 = – 1 [(1 – x)²cos²x + 1]

F(x) < 0 $\forall \mathrm{x}\in \mathrm{R}\Rightarrow$ F (x) = 0 is not possible for any x $\in$ R.  So P is false

Let G(x) = 2f(x) + 1 – 2x – 2x²

G(x) = 2(1 – x)² sin²x + 2x² + 1 – 2x – 2x²

G(x) = 2 [1 – 2x + x²] sin²x + (1 – 2x)

G(x) = (2sin²x + 1) (1 – 2x) + 2x² sin²x

Now   G (x) = 0           $\Rightarrow$ sin2x = $\frac{2x-1}{2{(x-1)}^2}$

As  $\frac{2\mathrm{x}-1}{2{(\mathrm{x}-1)}^2}\in$ [– 1/2, $\infty$), so Q is true.

Hence P is false and Q is true.

 $\mathrm{g}\left(\mathrm{x}\right)=\underset{1}{\overset{\mathrm{x}}{\int }}(\frac{2\text{\hspace{0.17em}}(\mathrm{t}-1)}{\mathrm{t}+1}-\ln \mathrm{t})\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$

 $\mathrm{g}\text{'}\left(\mathrm{x}\right)=(\frac{2(\mathrm{x}-1)}{\mathrm{x}+1}-\ln \mathrm{x})\mathrm{f}\left(\mathrm{x}\right)$

 $\mathrm{g}\text{'}\left(\mathrm{x}\right)=[2-\frac{4}{(\mathrm{x}+1)}-\ln \mathrm{x}]\mathrm{f}\left(\mathrm{x}\right)$

Consider h(x) = 2 –  $\frac{4}{\mathrm{x}+1}$  – ln x

h'(x) =  $\frac{4}{{(\mathrm{x}+1)}^2}-\frac{1}{\mathrm{x}}$

h'(x) =  $\frac{4\mathrm{x}-{\mathrm{x}}^2-2\mathrm{x}-1}{\mathrm{x}{(\mathrm{x}+1)}^2}$

h'(x) =  $\frac{-{(\mathrm{x}-1)}^2}{\mathrm{x}{(\mathrm{x}+1)}^2}$  < 0

 $\therefore$ h(x) is a decreasing function  $\forall \mathrm{x}\in$ (1, $\infty$)

h(x) < h(1) = 0

 $\therefore$ h(x) < 0 $\forall \mathrm{x}\in$ (1, $\infty$)

 $\therefore$ g(x) < 0 as f(x) > 0  $\forall \mathrm{x}\in \mathrm{R}\Rightarrow$ g(x) is decreasing in x  $\in$ (1,  $\infty$). Ans.