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已知定义在\({\mathbf{R}}\)上的函数\(f\left( x \right)\)在\(\left[ {1, + \infty } \right)\)上单调递增,若\(f\left( 3 \right) = 0\),且函数\(f\left( {x + 1} \right)\)为偶函数,则不等式\(xf\left( x \right) < 0\)的解集为( )
A.\(\left( {3, + \infty } \right)\)
B.\(\left( { - \infty , - 1} \right) \cup \left( {0,3} \right)\)
C.\(\left( { - 1, + \infty } \right)\)
D.\(\left( { - 3,0} \right) \cup \left( {1, + \infty } \right)\)
参考答案:B
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