Discussion Forum

let $f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x $\epsilon R$ and 

let $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-ln t\right) f(t) dt,$ for all x $\epsilon (1,\infty)$ 

consider the statements

P:  There exists some $x \epsilon R$ such that 

  $f(x)+2x=2(1+x^{2})$

Q:There exists some  $x \epsilon R$ such that 

   2f(x)+1=2x(1+x)

 Then 

Answer:

Explanation:

 Concept involved use of Newton leibnits formula

 i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x0).g'(x)$

  Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x

     $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$

$\forall x \epsilon (1, \infty)$

Here, $f(x)+2x=(1-x)^{2}.\sin^{2} x+x^{2}+2x$.............(i)

 where $P:f(x)+2x=2(1+x)^{2}$ ...............(ii)

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

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

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

$\Rightarrow$   $(1-x)^{2} \cos^{2}x=-1$

 which is never possible

$\therefore$  p is false

 Again h(x)=2f(x)+1-2x (1+x)

            h(1)=2f(1)+1-4=-3

 as  h(0) h(1)<0

 $\Rightarrow$  h(x) must have a solution

$\therefore$ Q is true