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Question

Let a,b,c be the three sides of a triangle and f(x)=b2x2+(b2+c2a2x+c2) then which of the following is /are CORRECT?

A
f(x)>0xϵR
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B
f(x)>0<xϵR
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C
f(1)>0
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D
f(a+b+c)=0
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Solution

The correct option is D f(x)>0xϵR

We have,

f(x)=b2x2+(b2+c2a2x+c2)

f(x)=b2x2a2x+b2+2c2

Comparing that,

f(x)=Ax2+Bx+C

A=b2,B=a2,C=b2+2c2

Then, We know that,

D=B24AC

=(a2)24b2(b2+2c2)

=a44b48b2c2

Hence, a,bandc are constant term

Which is also greater than zero.

So, f(x)>0xR

Hence, this is the answer.

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