本帖最后由 isee 于 2017-7-31 15:39 编辑
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(Ⅰ)解:\(g(x)\)的单调递增区间是\(( - \infty , - 1)\),\((\frac{1}{4}, + \infty )\),单调递减区间是\(( - 1,\frac{1}{4})\).
(Ⅱ)证明:由\(h(x) = g(x)(m - {x_0}) - f(m)\),得\(h(m) = g(m)(m - {x_0}) - f(m)\),
\(h({x_0}) = g({x_0})(m - {x_0}) - f(m)\).
令函数\({H_1}(x) = g(x)(x - {x_0}) - f(x)\),则\({H_1}^\prime (x) = g'(x)(x - {x_0})\).由(Ⅰ)知,当\(x \in [1,2]\)时,\(g'(x) > 0\),故当\(x \in [1,{x_0})\)时,\({H_1}^\prime (x) < 0\),\({H_1}(x)\)单调递减;当\(x \in ({x_0},2]\)时,\({H_1}^\prime (x) > 0\),\({H_1}(x)\)单调递增.因此,当\(x \in [1,{x_0}) \cup ({x_0},2]\)时,\({H_1}(x) > {H_1}({x_0}) = - f({x_0}) = 0\),可得.
令函数\({H_2}(x) = g({x_0})(x - {x_0}) - f(x)\),则\({H_2}^\prime (x) = g({x_0}) - g(x)\).由(Ⅰ)知,\(g(x)\)在\([1,2]\)上单调递增,故当\(x \in [1,{x_0})\)时,\({H_2}^\prime (x) > 0\),\({H_2}(x)\)单调递增;当\(x \in ({x_0},2]\)时,\({H_2}^\prime (x) < 0\),\({H_2}(x)\)单调递减.因此,当\(x \in [1,{x_0}) \cup ({x_0},2]\)时,\({H_2}(x) < {H_2}({x_0}) = 0\),可得.
所以,\(h(m)h({x_0}) < 0\).
(III)证明:对于任意的正整数 \(p\),\(q\),且\(\frac{p}{q} \in [1,{x_0}) \cup ({x_0},2]\),
令$m = \frac{p}{q}$,函数$h(x) = g(x)(m - {x_0}) - f(m)$.
由(II)知,当\(m \in [1,{x_0})\)时,\(h(x)\)在区间\((m,{x_0})\)内有零点;
当\(m \in ({x_0},2]\)时,\(h(x)\)在区间\(({x_0},m)\)内有零点.
所以\(h(x)\)在$(1,2)$内至少有一个零点,不妨设为\({x_1}\),则$h({x_1}) = g({x_1})(\frac{p}{q} - {x_0}) - f(\frac{p}{q}) = 0$.
由(I)知$g(x)$在$[1,2]$上单调递增,故$0 < g(1) < g({x_1}) < g(2)$,
于是$|\frac{p}{q} - {x_0}|\; = \;|\frac{{f(\frac{p}{q})}}{{g({x_1})}}|\; \ge \frac{{|f(\frac{p}{q})|}}{{g(2)}} = \frac{{|2{p^4} + 3{p^3}q - 3{p^2}{q^2} - 6p{q^3} + a{q^4}|}}{{g(2){q^4}}}$.
因为当\(x \in [1,2]\)时,$g(x) > 0$,故$f(x)$在\([1,2]\)上单调递增,
所以$f(x)$在区间\([1,2]\)上除${x_0}$外没有其他的零点,而$\frac{p}{q} \ne {x_0}$,故$f(\frac{p}{q}) \ne 0$.
又因为\(p\),\(q\),$a$均为整数,所以$|2{p^4} + 3{p^3}q - 3{p^2}{q^2} - 6p{q^3} + a{q^4}|$是正整数,
从而$|2{p^4} + 3{p^3}q - 3{p^2}{q^2} - 6p{q^3} + a{q^4}|\; \ge 1$.
所以$|\frac{p}{q} - {x_0}|\; \ge \frac{1}{{g(2){q^4}}}$.所以,只要取$A = g(2)$,就有$|\frac{p}{q} - {x_0}|\; \ge \frac{1}{{A{q^4}}}$.
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看到了牛顿迭代计算零点误差的影子。。。。。 |