Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in \mathbb{R}$, and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$. Consider the statements : $\mathbf{P}$ : There exists some $x \in \mathbb{R}$ such that $f(x)+2 x=2\left(1+x^{2}\right)$ Q: There exists some $x \in \mathbb{R}$ such that $2 f(x)+1=2 x(1+x)$ Then

(A) both $\mathbf{P}$ and $\mathbf{Q}$ are true

(B) $\mathbf{P}$ is true and $\mathbf{Q}$ is false

(D) both $\mathbf{P}$ and $\mathbf{Q}$ are false

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Continuity and Differentiability

Let

f (x) = (1 - x)^{2} sin^{2} x + x^{2}

for all

x \in R

, and let

g (x) = \int_{1}^{x} (\frac{2 ( t - 1 )}{t + 1} - ln t) f (t) d t

for all

x \in (1, \infty)

.

Consider the statements :

P

: There exists some

x \in R

such that

f (x) + 2 x = 2 (1 + x^{2})

Q: There exists some

x \in R

such that

2 f (x) + 1 = 2 x (1 + x)

Then

Answer: (C)

For statement P:

f (x) + 2 x = 2 (1 + x^{2})

(1 - x)^{2} sin^{2} x + x^{2} + 2 x = 2 + 2 x^{2}

(1 - x)^{2} sin^{2} x = x^{2} - 2 x + 2 = (x - 1)^{2} + 1

(1 - x)^{2} (sin^{2} x - 1) = 1

- (1 - x)^{2} cos^{2} x = 1

This requires

(1 - x)^{2} cos^{2} x = - 1

, which is impossible since the left side is non-negative. So P is false.

For statement Q:

2 f (x) + 1 = 2 x (1 + x)

2 (1 - x)^{2} sin^{2} x + 2 x^{2} + 1 = 2 x + 2 x^{2}

2 (1 - x)^{2} sin^{2} x = 2 x - 1

Let

h (x) = 2 (1 - x)^{2} sin^{2} x - (2 x - 1)

. At

x = 0

h (0) = 0 - (- 1) = 1 > 0

. As

x \to 1^{-}

h (x) \to - (2 - 1) = - 1 < 0

. By the Intermediate Value Theorem, there exists some

x \in (0, 1)

where

h (x) = 0

. So Q is true.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option