Question

Let $f\left(x\right)=(1-x{)}^{2}si{n}^{2}x+x^2$ for all $x\in R$. Consider the statements:
P: There exists some $x\in R$ such that $f\left(x\right)+2x=2(1+x^2).$
Q:There exists some $x\in R$ such that $2f\left(x\right)+1=2x(1+x).$
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

The correct option is C

P is false and Q is true

$f\left(x\right)=(1-x{)}^{2}si{n}^{2}si{n}^{2}x+x^2.\forall x\in R$
For statement P:
$f\left(x\right)+2x=2(1+x^2)$
$\Rightarrow (1-x{)}^{2}si{n}^{2}x+x^2+2x=2+2x^2$
$\Rightarrow (1-x{)}^{2}si{n}^{2}x=x^2-2x+2=(x-1{)}^2+1$
$\Rightarrow (1-x{)}^2(si{n}^{2}x-1)=1$
$\Rightarrow -(1-x{)}^{2}co{s}^{2}x=1$
$\Rightarrow (1-x{)}^{2}co{s}^{2}x=-1$
So equation (i)will not have real solution.
So, P is wrong.
For statment Q:
$2(1-x{)}^{2}si{n}^{2}x+2x^2+1=2x+2x^2$
$2(1-x{)}^{2}si{n}^{2}x=2x-1$
$2si{n}^{2}x=\frac{2x-1}{(1-x{)}^2}.Leth\left(x\right)=\frac{2x-1}{(1-x{)}^2}-2si{n}^{2}x$
Clearly, $h\left(0\right)=-1,{lim}_{x\to 1^{-}}h\left(x\right)=+\infty$
So by IVT, equation (ii)will have solution. So, Q is correct.