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Question

Let f(x)=(1x)2sin2x+x2 for all xR. Consider the statements:
P: There exists some xR such that f(x)+2x=2(1+x2).
Q:There exists some xR such that 2f(x)+1=2x(1+x).
Then


A

both P and Q are true

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B

P is true and Q is false

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C

P is false and Q is true

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D

both P and Q are false

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Solution

The correct option is C

P is false and Q is true


f(x)=(1x)2sin2sin2x+x2.xR
For statement P:
f(x)+2x=2(1+x2)
(1x)2sin2x+x2+2x=2+2x2
(1x)2sin2x=x22x+2=(x1)2+1
(1x)2(sin2x1)=1
(1x)2cos2x=1
(1x)2cos2x=1
So equation (i)will not have real solution.
So, P is wrong.
For statment Q:
2(1x)2sin2x+2x2+1=2x+2x2
2(1x)2sin2x=2x1
2sin2x=2x1(1x)2.Leth(x)=2x1(1x)22sin2x
Clearly, h(0)=1,limx1h(x)=+
So by IVT, equation (ii)will have solution. So, Q is correct.


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