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Question

Let f(x)=x4+ax3+bx2+ax+1 be a polynomial, where a,bR. If b=1, then the range of a for which f(x)=0 does not have real roots is

A
(12,12)
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B
(1,1)
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C
(52,52)
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D
(3,3)
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Solution

The correct option is A (12,12)
x4+ax3+bx2+ax+1=0(1)x4+ax3x2+ax+1=0
Dividing the equation by x2,
(x2+1x2)+a(x+1x)1=0
Let x+1x=t,t(,2][2,)
Now,
(x+1x)2=t2x2+1x2=t22
Now,
t22+at1=0t2+at3=0(2)
let the roots of equation 2 be t1,t2(t1<t2)
For every t2, there exist 2 positive values of x.
For every t2, there exist 2 negative values of x.

Equation (1) will have no real roots iff equation (2) will have
(A) Non real roots or
(B) t1,t2(2,2)
Case(A) : Non real roots,
D<0a2+12<0aϕ(3)

Case (B): t1,t2(2,2)
2<t1,t2<2(i) D0a2+120aR(ii) f(2)>012a>0a<12(iii) f(2)>02a+1>0a>12(iv) 2<B2A<22<a2<24>a>4a(12,12)(4)

From equation (3) and (4),
a(12,12)

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