Question

Let $f\left(x\right)={\left(1-x\right)}^2+{\sin }^{2}x+x^2$ for all $x\in \mathrm{ℝ}$ Consider the statements:

P: There exists some $x\in \mathrm{ℝ}$ such that $f\left(x\right)+2x=2\left(1+x^2\right)$

Q: There exists some $x\in \mathrm{ℝ}$ such that $2f\left(x\right)+1=2x\left(1+x\right)$. 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 A

Both P and Q are true

Explanation for the correct option:

Finding the value of $x$ and substituting in the respective equations for prooving the equality:

Suppose P is true then:

$\exists x\in \mathrm{ℝ}$ such that

$f\left(x\right)+2x=2\left(1+x^2\right)$

$$\begin{gathered} {\left(1-x\right)}^2+{\sin }^{2}x+x^2+2x=2\left(1+x^2\right) \\ \Rightarrow 1+x^2-2x+{\sin }^{2}x+x^2+2x=2+2x^2 \\ \Rightarrow {\sin }^{2}x-1=0 \\ \Rightarrow {\sin }^{2}x=1 \\ \Rightarrow \sin x=1 \end{gathered}$$

Since, $\sin \left(\frac{\pi }{2}\right)=1$therefore, $x=\frac{\mathrm{\pi }}{2}$

Therefore for $x=\frac{\mathrm{\pi }}{2}$, $f\left(x\right)+2x=2\left(1+x^2\right)$

Hence, P is true.

Now for statement Q:

Let $h\left(x\right)=2f\left(x\right)+1-2x\left(1+x\right)$

$$\begin{gathered} h\left(0\right)=2f\left(0\right)+1-0=3>0 \\ h\left(1\right)=2f\left(1\right)+1-4 \\ =2{\sin }^2\left(1\right)-3 \\ <0\mathbf{(}\mathbf{sin}\mathbf{1}\mathbf{<}\mathbf{1}\mathbf{)} \end{gathered}$$

Therefore by the intermediate value theorem$\exists x\in \left(0,1\right)$ such that $h\left(x\right)=0$

This means $\exists x\in \mathrm{ℝ}$ such that $2f\left(x\right)+1=2x\left(1+x\right)$

Hence, Q is also true.

Therefore, the correct answer is option (A).