Answer

Solution

1. 1

   Both numerator and denominator vanish at $x=1$.

2. 2

   Since $f$ is odd, $\int_{-1}^{1} f(t)\,dt = 0$, so $F(1) = 0$.

3. 3

   For $G(1)$, note that $f$ odd implies $f(f(-t)) = f(-f(t)) = -f(f(t))$, so

4. 4

   $|f(f(\cdot))|$ is even. Hence $t\,|f(f(t))|$ is an odd function of $t$, and

5. 5
   $$G(1) = \int_{-1}^{1} t\,|f(f(t))|\,dt = 0.$$
6. 6

   Thus the limit is of indeterminate form $0/0$, and by L'H^opital's rule:

7. 7
   $$\frac{1}{14} = \lim_{x\to 1} \frac{F'(x)}{G'(x)} = \lim_{x\to 1} \frac{f(x)}{x\,|f(f(x))|} = \frac{f(1)}{1\cdot |f(f(1))|} = \frac{1/2}{|f(1/2)|}.$$
8. 8

   Therefore $|f(1/2)| = 14 \cdot \tfrac{1}{2} = 7$.

9. 9

   Determining the sign.

10. 10

    Since $f$ is continuous and vanishes only at one point, and $f$ odd implies

11. 11

    $f(0)=0$, that single zero must be at $0$. Thus $f$ has constant sign on

12. 12

    $(0,\infty)$. From $f(1) = \tfrac{1}{2} > 0$, we conclude $f > 0$ on $(0,\infty)$,

13. 13

    so $f(1/2) > 0$, giving $f(1/2) = 7$.

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