Question

Let $f(x) = {(1 - x)^2}{\sin ^2}x + {x^2}$ for all $x \in R$, and let $g(x) = \int\limits_1^x {\left( {\dfrac{{2(t - 1)}}{{t + 1}} - \ln t} \right)f(t)dt} $ for all $x \in (1,\infty )$.
Consider the statements:
P: There exists some $x \in R$ such that $f(x) + 2x = 2(1 + {x^2})$.
Q: There exists some $x \in R$ such that $2f(x) + 1 = 2x(1 + x)$.
Then
A. Both P and Q are true
B. P is false and Q is true
C. P is true and Q is false
D. Both P and Q are false

Answer 1

Hint: First check for statement P and then check for statement Q. Try to find whether some real solution exists or not for both the P and Q.

Complete step-by-step answer:
Given $f(x) = {(1 - x)^2}{\sin ^2}x + {x^2}$ for all $x \in R$, and $g(x) = \int\limits_1^x {\left( {\dfrac{{2(t - 1)}}{{t + 1}} - \ln t} \right)f(t)dt} $ for all $x \in (1,\infty )$.
Now considering statement P:
$f(x) + 2x = 2(1 + {x^2}) - (1)$
Put the value of f(x) in the above equation (1), we get-
$
  {(1 - x)^2}{\sin ^2}x + {x^2} + 2x = 2 + 2{x^2} \\
  {(1 - x)^2}({\sin ^2}x - 1) = 1 \\
  {(1 - x)^2}{\cos ^2}x = - 1 \\
   \Rightarrow {(1 - x)^2}{\cos ^2}x = - 1 \\
$
So, equation (1) will not have a real solution.
Hence, we can say that statement P is wrong.
For statement Q:
$2f(x) + 1 = 2x(1 + x)$
Put the value of f(x) in the above equation, we get-
$
  2{(1 - x)^2}{\sin ^2}x + 2{x^2} + 1 = 2x + 2{x^2} - (2) \\
   \Rightarrow 2{\sin ^2}x = \dfrac{{2x - 1}}{{{{(1 - x)}^2}}} \\
 $
Let $h(x) = \dfrac{{2x - 1}}{{{{(1 - x)}^2}}} - 2{\sin ^2}x$
Now, h (0) = - ve
${\lim _{x \to {1^ - }}}h(x) = + \infty $
This implies h (x) assumes both positive and negative values on R.
So, by the intermediate value property of continuous functions, h (x) vanishes at least once in R.
Hence, statement Q is true.
Therefore, the correct option is B. P is false and Q is true.

Note: Whenever such types of questions appear, always try to solve one statement first, as mentioned in the solution first we find out whether the equation P has real solution or not, and we found out that it does not have real solution so it is false and then we have checked for statement Q and then it was true.