    Question

# Let S denotes the set of all real values of x such that (x2016+1)(1+x2+x4+⋯+x2014)=2016x2015. Then

A
Number of elements in S is 2
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B
Number of elements in S is 1
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C
Point (x,2), xS lies inside the parabola Y=X22X+2, XR
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D
Image of the point (x,2) about the line y=x lies on the line x+y=4
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Solution

## The correct options are B Number of elements in S is 1 C Point (x,2), x∈S lies inside the parabola Y=X2−2X+2, X∈R (x2016+1)(1+x2+x4+⋯+x2014)=2016x2015⇒(x+1x2015)(1+x2+x4+⋯+x2014)=2016⇒x+1x2015+x3+1x2013+⋯+x2015+1x=2016⇒(x+1x)+(x3+1x3)+⋯+(x2015+1x2015)=2016 When x>0,x+1x≥2 When x<0,x+1x≤−2 So, L.H.S. = R.H.S. only if x=1 Putting the point (1,2) in the parabola, 2>1−2+2 So, the point (1,2) lies inside the parabola. Image of the point (1,2) about the line y=x is (2,1) which does not lie on the line x+y=4  Suggest Corrections  0      Similar questions  Explore more