341(a) This is surely correct:
Lemma: If x1 and x2 are two consecutive roots of f,
then f'(x1) f'(x2)<0
Proof: Let x1<x2
By Rolle's Theorem, \exists \ x_3 \in (x_1,x_2) such that f'(x_3)=0
Again by Lagrange's Theorem, we have
\frac{f'(x_3)-f'(x_1)}{x_3-x_1} = f"(\theta_1), \theta_1 \in (x_1,x_3)
Since f is concave f"(\theta_1)>0 \Rightarrow f'(x_1)<0
In a similar way, you arrive at f'(x_2)>0
Now look at the equation
f'(x) g(x) - f(x) g'(x) = x^4+3x^2+10>0
Hence we have f'(x_1) g(x_1)>0 and f'(x_2) g(x_2)>0
This implies that g(x_1)<0 \ \text{and} \ g(x_2)>0
From IMV, \exists \ x_0 \in (x_1,x_2) such that g(x_0)=0
341
Hari Shankar
·2010-02-26 02:39:45
Notice that we have h'(x)>0 where h(x) = f(x)/g(x). But that does not allow us to conclude (d)
341
Hari Shankar
·2010-02-26 07:46:46
Differentiating the given equation we get f"(x) g(x) - f(x) g"(x) = 4x(x^2+1)
Now we have already seen that when x1 and x2 are two consecutive roots we will have
g(x1)<0 and g(x2)>0
When f(x) = 0, the equation becomes f"(x) g(x) = 4x(x^2+1)
Remember that f"(x)>0
Hence, if g(x1)<0, then we must have x1<0 and g(x2)>0 implies x2>0
Hence Option (c) is also correct
341
Hari Shankar
·2010-02-26 09:17:51
Now, suppose that more than one root is possible for g(x) in the interval (x1, x2).
Let us consider two consecutive roots of g(x), x3 and x4
From the given equation, g'(x3) > 0 and g'(x4) > 0
Now we consider the function h(x) = \frac{f(x)}{g(x)}
In a small right interval of x3, we have g(x)>0 and hence h(x)<0. Again in a small left interval of x4, we have g(x)<0 and hence h(x)>0.
But h(x) has no discontinuities in (x3,x4) and hence h(x) = 0 for some x5 in (x3,x4).
But that can happen only if f(x5) = 0 contradicting that x1 and x2 are two consecutive roots of f(x).
Hence Option (b) is ruled out
