Consider the equation 8x4−16x3+16x2−8x+a=0(a∈R), Then the sum of
all non real roots of the equation can be
OPTIONS::(a)1(b)2(c)12(d)None
MyTry:: Let f(x)=8x4−16x3+16x2−8x+a, Then f′(x)=32x3−48x2+32x−8
And f″
So Using \bf{LMVT\;,} We get f'(x)=0 has at most 1 real roots and
f(x)=0 has at most 2 real roots
Now How can i solve it after that, Help me
Thanks
Answer
Now How can i solve it after that
Noting that f'(1/2)=0 is a key.
We have
f'(x)=8(4x^3-6x^2+4x-1)=8(2x-1)\left(2\left(x-\frac 12\right)^2+\frac 12\right)
So, f(x) is decreasing for x\lt 1/2 and is increasing for x\gt 1/2.
By the way,
f\left(\frac 12+s\right)=8s^4+4s^2-\frac 32+a
and so f(1/2-\alpha)=0 is equivalent to f(1/2+\alpha)=0.
Also, by Vieta's formula, the sum of roots is -(-16)/8=2.
If a\le 3/2, then f(1/2)\le 0, and the answer is
2-(1/2-\alpha)-(1/2+\alpha)=\color{red}{1}
If a\gt 3/2, then f(1/2)\gt 0, and the answer is \color{red}{2}.
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