1@vivek,
how is f(1) = 0 ?
\hspace{-16}$Prove that the equation $\mathbf{2x^{n+2}+1=3x}$ has one real root in $\mathbf{(0,1)}$
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6 Answers
71
Latexifying takes time here, So I usually post this in AOPS test forum and grab an image.
See this : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=224&t=456619&p=2571236#p2571236
71Uff. .. I thought that denominator is x+3. Sorry My mistake.
1at x=0
the expression
f(x)=2xn+2-3x+1
=1>0
and
at x=1/2
=2(12)n+2-32+1
=(12)n+1-12 <0 for ν n ε N
So by IVT ,we have
one real root b/w 0 and 1/2
341for 1/2≤x<1, we have
2x^{n+2}-3x+1<2x^2-3x+1 \le 0
So, the solution in this interval lies in the sub-interval (0,1/2).
Now, as shown in the above post, IVT predicts a root in that interval.
If you look at the derivative, g(x) = f'(x)
g(x) = 2x^{n+1}(n+2)-3 < g\left(\frac{1}{2} \right) = \frac{n+2}{2^n}-3 <\frac{n+2}{2^n}-2
By Bernoulli inequality
2^{n+1} = (1+1)^{n+1} > 1+(n+1) = n+2 \Rightarrow \frac{n+2}{2^n}-2<0
and hence the function is monotonically decreasing in that interval.
