\textbf{Let}\; a \; \textbf{and} \; b \; \textbf{be the Real Parameters. One Root of the equation} \\\;x^{12}-abx+a^2=0 \; \textbf{is greater than} \;\;2,\textbf{then find minimum value of }\; |b|
712. \textbf{The minimum No. of points required to uniquely determine a parabola is}
2622) i think it is 4
1) it is easy to find that f(2)<0
thus 212 -2ab +a2 <0
or 212 +(a-b)2 <b2
for minimum value of |b| , a=b
thus b2>212
or |b| >26 =64
212)ans =4
general equation is (y-k)2=4a(x-h)
to solve this we need 3 pts.[3 variables]
But this will give 2 values of k.
We need a 4th to find 1 value of k.
262@vivek - the given equation can have a maximum of two real roots. and as the leading coeff. is +ve , thus it will almost be like an upward facing curve (like a parabola) .
now since 2 lies between the roots, f(2) <0
and if a≠b then min |b| will be slightly greater (but its limiting value will still remain the same)
212) It is three and not four
It is a second degree polynomial, so enough to know it at three points
3411) The problem should read,"if for some real value of a, one root of the equation ..."
Lets assume the converse i.e. that the given polynomial has all roots less than 2 (or greater than 2) for every real a
Since its of even degree with positive leading coefficient, this means we must have f(2)>0
i.e. 2^{12}-2ab+a^2>0
This is independent of the choice of real a. Viewing it as a quadratic in a, this means we must have determinant negative.
i.e. |b|<26.
This means that if b≥26, for some choice of a, f(2)<0 i.e. there exists an odd number of roots of f greater than 2 for some real values of a.