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Question

Let f(x) be a polynomial function of degree 2 and f(x)>0 for all xR. If g(x)=f(x)+f(x)+f′′(x), then for any x

A
g(x)<0
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B
g(x)>0
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C
g(x)=0
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D
g(x)0
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Solution

The correct option is C g(x)>0
Given, g(x)=f(x)+f(x)+f′′(x) ....(1)
Let f(x)=ax2+bx+ca0
f(x)=2ax+b
f′′(x)=2a
Substitute these values in (1)
g(x)=ax2+(2a+b)x+(c+b+2a) ....(2)
Since given, f(x)>0
a>0 and D<0
a>0 and b24ac<0
Now, we will find D for g(x),
D=(2a+b)24a(c+b+2a)
D=(b24ac)4a2
D<0(b24ac<0 and 4a2>0)
So, for g(x),a>0 and D<0
Hence, g(x)>0xR

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