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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 {...

Let f(x)=(1−x)2sin⁡2x+x2f(x) = {(1 - x)^2}{\sin ^2}x + {x^2} for all x∈Rx \in R, and let g(x)=∫1x(2(t−1)t+1−ln⁡t)f(t)dtg(x) = \int\limits_1^x {\left( {\dfrac{{2(t - 1)}}{{t + 1}} - \ln t} \right)f(t)dt} for all x∈(1,∞)x \in (1,\infty ).
Consider the statements:
P: There exists some x∈Rx \in R such that f(x)+2x=2(1+x2)f(x) + 2x = 2(1 + {x^2}).
Q: There exists some x∈Rx \in R such that 2f(x)+1=2x(1+x)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

Explanation

Solution

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)2sin⁡2x+x2f(x) = {(1 - x)^2}{\sin ^2}x + {x^2} for all x∈Rx \in R, and g(x)=∫1x(2(t−1)t+1−ln⁡t)f(t)dtg(x) = \int\limits_1^x {\left( {\dfrac{{2(t - 1)}}{{t + 1}} - \ln t} \right)f(t)dt} for all x∈(1,∞)x \in (1,\infty ).
Now considering statement P:
f(x)+2x=2(1+x2)−(1)f(x) + 2x = 2(1 + {x^2}) - (1)
Put the value of f(x) in the above equation (1), we get-
(1−x)2sin⁡2x+x2+2x=2+2x2 (1−x)2(sin⁡2x−1)=1 (1−x)2cos⁡2x=−1 ⇒(1−x)2cos⁡2x=−1  {(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)2f(x) + 1 = 2x(1 + x)
Put the value of f(x) in the above equation, we get-
2(1−x)2sin⁡2x+2x2+1=2x+2x2−(2) ⇒2sin⁡2x=2x−1(1−x)2  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)=2x−1(1−x)2−2sin⁡2xh(x) = \dfrac{{2x - 1}}{{{{(1 - x)}^2}}} - 2{\sin ^2}x
Now, h (0) = - ve
lim⁡x→1−h(x)=+∞{\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.