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″(x)=96x2−96x+32=96[x2−x+13]=96[(x−12)2+112]>0∀x∈R
So Using 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(4x3−6x2+4x−1)=8(2x−1)(2(x−12)2+12)
So, f(x) is decreasing for x<1/2 and is increasing for x>1/2.
By the way,
f(12+s)=8s4+4s2−32+a
and so f(1/2−α)=0 is equivalent to f(1/2+α)=0.
Also, by Vieta's formula, the sum of roots is −(−16)/8=2.
If a≤3/2, then f(1/2)≤0, and the answer is
2−(1/2−α)−(1/2+α)=1
If a>3/2, then f(1/2)>0, and the answer is 2.
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