Question

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

• 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 1

Answer: (C)

$f(x)=(1-x)^{2} \sin ^{2} x+x^{2} \quad \forall x \in R$
$g(x)=\int\limits_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t)$ dt $\quad \forall x \in(1, \infty)$
For statement $P$ :
$f(x)+2 x=2\left(1+x^{2}\right) \dots$(i)
$(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2+2 x^{2}$
$(1-x)^{2} \sin ^{2} x=x^{2}-2 x+2=(x-1)^{2}+1$
$(1-x)^{2}\left(\sin ^{2} x-1\right)=1$
$-(1-x)^{2} \cos ^{2} x=1$
$(1-x)^{2} \cdot \cos ^{2} x=-1$
So equation (i) will not have real solution
So, $P$ is wrong.
For statement $Q$ :
$2(1-x)^{2} \sin ^{2} x+2 x^{2}+1=2 x+2 x^{2} \ldots$(ii)
$2(1-x)^{2} \sin ^{2} x=2 x-1$
$2 \sin ^{2} x=\frac{2 x-1}{(1-x)^{2}}$ Let $h(x)=\frac{2 x-1}{(1-x)^{2}}-2 \sin ^{2} x$
Clearly $h(0)=-$ ve, $\displaystyle\lim _{x \rightarrow 1^{-}} h(x)=+\infty$
So by IVT, equation (ii) will have solution. So, $Q$ is correct.