题型:解答题 题类:期末考试 难易度:难
定义在\(D\)上的函数\(f\left( x \right)\),如果满足:对任意\(x\in D\),存在常数\(M > 0\),都有\(\left| f\left( x \right) \right|\leqslant M\)成立,则称\(f\left( x \right)\)是\(D\)上的有界函数,其中\(M\)称为\(f\left( x \right)\)的上界\(.\)已知函数\(f\left( x \right)=1+a{{\left( \dfrac{b}{2} \right)}^{x}}+{{\left( \dfrac{c}{4} \right)}^{x}}\).
\((1)\)当\(a=b=c=1\)时,求函数\(f\left( x \right)\)在\(\left( -\infty ,0 \right)\)上的值域,并判断函数\(f\left( x \right)\)在\(\left( -\infty ,0 \right)\)上是否有上界,请说明理由;
\((2)\)若\(b=c=1\),函数\(f\left( x \right)\)在\(\left[ 0,+\infty \right)\)是以\(3\)为上界的有界函数,求实数\(a\)的取值范围;
\((3)\)已知\(s\)为正整数,当\(a=1,b=-1,c=0\)时,是否存在整数\(\lambda \),使得对任意的\(n\in {{N}^{*}}\),不等式\(s\leqslant \lambda f\left( n \right)\leqslant s+2\)恒成立?若存在,求出\(\lambda \)的值;若不存在,说明理由.