Application of Derivatives

f (x) = (1−x)2sin2x+x2 for all x ∊ R
g (x) =

x

∫

1

(

2(t−1)

t+1

−lnt) f (t) dt for all x ∊ (1 , ∞)
For statement P:
f (x) + 2x = 2(1+x2) 91)
(−x)2sin2x+x2+2x = 2+2x2
(1−x)2ssin2x = x2−2x+2 = (x−1)2+1
(1−x)2(sin2x−1) = 1
−(1−x)2cos2x = 1
(1−x)2cos2x = - 1
So, Eq. (1) will not have real solution. Thus, P is wrong.
For statement Q:
2(1−x)2sin2x+2x2+1 = 2x+2x2 (2)
2(1−x)2sin2x = 2x - 1
2sin2x =

2x−1

(1−x)2

Let h (x) =

2x−1

(1−x)2

*2sin2x
Clearly, h (0) = negative,

lim

x→1−

h (x) = + ∞.
So, by IVT, Eq. (2) will have solution.
So, statement Q is correct.