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Question

Let f(x)=x48x3+ax2bx+16. If all the roots of f(x)=0 lies between [0,202], then which of the following is/are correct ?

A
(ab)2 is a constant and is equal to 64.
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B
For particular values of a and b, 2 is one of the root.
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C
f(3)=4
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D
Roots are always in A.P.
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Solution

The correct options are
A (ab)2 is a constant and is equal to 64.
B For particular values of a and b, 2 is one of the root.
D Roots are always in A.P.
x48x3+ax2bx+16=0
x1+x2+x3+x4=8 ...(1)
xixj=a ...(2)
xixjxk=b ...(3)
x1.x2.x3.x4=16 ...(4)

A.M. of x1,x2,x3,x4=x1+x2+x3+x44
=2
G.M. of x1,x2,x3,x4=4x1.x2.x3.x4
=416
=2

A.M. of x1,x2,x3,x4=G.M. of x1,x2,x3,x4
Hence, the roots are equal.
x1=x2=x3=x4=2
From eqn(2), a=24
From eqn(3), b=32

(ab)2=(2432)2=64

f(x)=x48x3+24x232x+16
=(x2)4
f(3)=1

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