Sunday, 8 November 2015

algebra precalculus - no. of real roots of the equation 4x5+5x410x220x+40=0





(1) Find Total no. of real solution of the equation xesinx=cosx, where x(0,π2)



(2) The no. of real roots of the equation 4x5+5x410x220x+40=0




MyTry:: For (1) st part::



Let f(x)=xesinxcosx, Then f(x)=xesinxcosx+esinx+sinx>0x(0,π2)




So f(x) is Strictly Increasing function.



And when x, Then f(x) and when x+, Then f(x)+.



So f(x) intersect the X axis at exactly at one point.



So only one real roots of the equation.



But I did not understand how can i found that the roots lie between (0,π2).




Help me, Thanks



MyTry:: For (2) st part::



Let f(x)=4x5+5x410x220x+40, Then f(x)=20x4+20x320x20



Now I did not understand how can i solve after that
Help me, Thanks


Answer



For (2):




Your derivative is 20(x^4 +x^3 -x -1). Grouping the first two bracketed terms as one group and the last two terms as the second group, we see that x^4 +x^3 -x -1 = x^3*(x+1) -1(x+1) which is equal to (x^3-1)(x+1). This is equal to (x+1)(x-1)(x^2+x+1). Note that the term (x^2 +x+1) is always positive (I leave you to prove that fact), hence we conclude that the derivative of f(x) is negative if and only if -1< x < 1. Additionally, we note that the derivative of f is zero if and only if x is +1 or -1.



Now f(1)=19, which is clearly positive, and as the derivative of f(x) is positive for all x bigger than 1, we can be certain that f(x) has no roots in the interval [1,infinity).



Now f(-2) is equal to -8, and f(-1) is equal to 51. So by the Intermediate Value Theorem f must have at least one root in the interval (-2, -1). Let the smallest of these roots be denoted by a. Now assume for contradiction that f has two real roots. We know one of those two roots is a, where a is in the interval (-2,-1). By Rolle's theorem, there must be a critical point of f which lies (strictly) between a and the second root.



Now the last part of my answer is not perfectly rigorous: given that f(1) is positive and that there are only two critical points (at x=-1 and x=1), we see this is geometrically impossible, by tracing out a sketch of the curve f(x).


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