Let f(x) = (1 - x)2sin2x + x2 for all x ∈ R, and let g(x) = ∫(2(t - 1)/(t + 1) - ln t)f(t)dt for t ∈ [1, x] for all x ∈ (1, ∞). Consider the statements: P: There exists some x ∈ R, such that f(x) + 2x = 2(1 + x2) Q: There exists some x ∈ R, such that 2f(x) + 1 = 2x(1 + x) (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.

Options

A. both P and Q are true
B. P is true and Q is false
C. P is false and Q is true
D. both P and Q are false

Answer

C

Explanation

`f(x)=(1-x)^(2) sin^(2) x+x^(2)" "AA x in R`
For statement P :
`f(x)+2x=2(1+x^(2))` (i)
`rArr (1-x)^(2) sin^(2) x+x^(2)+2x=2+2x^(2)`
`rArr (1-x)^(2) sin^(2) x=x^(2) -2x+2=(x-1)^(2)+1`
`rArr (1-x)^(2) (sin^(2) x-1)=1`
`rArr -(1-x)^(2) cos^(2) x=1`
`rArr (1-x)^(2) cos^(2) x=-1`
So equation (i) will not have real solution.
So, P is wrong.
For statement Q :
`2(1-x)^(2) sin^(2) x+2x^(2)+1=2x+2x^(2)` (ii)
`2(1-x)^(2) sin^(2) x=2x-1`
`2 sin^(2) x= (2x-1)/((1-x)^(2))`.
Let `h(x)=(2x-1)/((-x)^(2))-2 sin^(2) x`
Clearly, `h(0)=-1, lim_(x rarr 1^(-)) h(x) = +oo`
So by IVT, equation (ii) will have solution.
So, Q is correct.