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Question

Let f(x)=x22x and g(x)=f(f(x)1)+f(5f(x)), then

A
g(x)<0,xR
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B
g(x)<0 for some xR
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C
g(x)0 for some xR
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D
g(x)0,xR
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Solution

The correct option is D g(x)0,xR
Given, f(x)=x22x
Now, f(f(x)1)=f(x22x1)=(x22x1)22(x22x1)
And, f(5f(x))=f(5x2+2x)=(5x2+2x)22(5x2+2x)
Hence, g(x)=f(f(x)1)+f(5f(x))g(x)=(x22x1)22(x22x1)+(5x2+2x)22(5x2+2x)g(x)=(x22x1)2+(5x2+2x)22(x22x1+5x2+2x)g(x)=(x2+2x1)2+(52x2+2x2)28
Since the first two terms are in square,
it can not be negative and if x = 9 then we also get positive value.
g(x)0xR
option D is correct

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