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Question

Let f be a twice differentiable function in (,) such that f′′(x)0 xR. If g(x)=f(x)+f(1x) and g(14)=0, then

A
g(x) is increasing in (,0)
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B
g(x) attains local minimum at x=12
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C
g′′(x)=0 has at least two roots in [0,1]
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D
g(x)<0 for all x(12,)
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Solution

The correct option is D g(x)<0 for all x(12,)
f′′(x)0 xR
f is decreasing in R
g(x)=f(x)f(1x) (1)

If x>1x i.e., x>12
then f(x)f(1x)
f(x)f(1x)0
g(x)0 for all x(12,)

If x<1x i.e., x<12
then f(x)f(1x)
g(x)0 for all x(,12)
By first derivative test,
g(x) has local maximum at x=12

g(x)=f(x)+f(1x)
Replacing x by 1x, we get
g(x)=g(1x)
Again, replacing x by x+12
g(12+x)=g(12x)
Graph of y=g(x) is symmetric about x=12
Given, g(14)=0
By symmetricity, g(34)=0
Also, from (1), g(12)=0
By Rolle's theorem, g′′(x) vanishes for at least two values in [0,1]

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