21Let f(x) = x^2 - (1+x)(ln(1+x))^2
f(0)= 0
f^{,}(x) = 2x - (ln(1+x))^2 - 2(ln(1+x))
suffices to prove f^{,}(x)>0 for x>0
f^{,}(0) = 0
f"(x)= \frac{2(x- log(x+1))}{x+1}
f"(x)>0 for x>0since x>log(x+1) for x>0
1. Show that x^2>(1+x)[ln(1+x)]^2 \text{ } \forall\text{ } x>0
2. Let a+b=4, where a<2 and let g(x) be a differentiable function. If dgdx>0 for all x, prove that \int_{0}^{a}{g(x)dx}+\int_{0}^{b}{g(x)dx} increases as (b-a) increases.
3. Let h(x)=f(x)-(f(x))^2 + (f(x))^3 for every real number x. Then:
a) h is increasing whenever f is increasing
b) h is increasing whenever f is decreasing
c) h is decreasing whenever f is decreasing
d) nothing can be said in general
More may follow! :D
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7 Answers
21212) get rid of b
b = 4-a, b-a = 4- 2a>0. b-a increases when a decreases. and let
P(a)= \int_{0}^{a}(g(x))dx + \int_{0}^{4-a}(g(x))dx
g(x) is increasing. a<2\implies 2a<4\implies a<4-a
By Fundamental theorem of calculus
p^{,}(a) = g(a) - g(4-a) <0
213)h(x) = f(x) - (f(x))^2 + (f(x))^3
h^{,}(x) = f^{,}(x) - 2f(x) f^{,}(x) + 3(f(x))^{2} f^{,}(x)
h^{,}(x) = f^{,}(x)(1- 2f(x) + 3f(x)^2) but (1- 2f(x) + 3f(x)^2)>0 cz its D<0
Now is obvious why (a) and (c) should be correct...
303) The answer given is a. I couldn't think of a reason for (c) not being right. :)
2) For 2, my method was similar except a=2-c and b=2+c . Rest follows :)