Lời giải chi tiết:
(fleft( x right) = left{ matrix{ sin x,,,,,,,khi,,left| x right| le {pi over 2} hfill cr ax + b,,,,khi,,left| x right| > {pi over 2} hfill cr} right. Leftrightarrow fleft( x right) = left{ matrix{ sin x,,,,,,,khi,, – {pi over 2} le x le {pi over 2} hfill cr ax + b,,,,khi,,left[ matrix{ x > {pi over 2} hfill cr x < – {pi over 2} hfill cr} right. hfill cr} right.)
Ta có hàm số liên tục trên các khoảng (left( { – infty ; – {pi over 2}} right) cup left( { – {pi over 2};{pi over 2}} right) cup left( {{pi over 2}; + infty } right))
Để hàm số liên tục trên R thì hàm số phải liên tục tại các điểm (x = pm {pi over 2} Rightarrow left{ matrix{ mathop {lim }limits_{x to {pi over 2}} fleft( x right) = fleft( {{pi over 2}} right) hfill cr mathop {lim }limits_{x to – {pi over 2}} fleft( x right) = fleft( { – {pi over 2}} right) hfill cr} right.)
Ta có
(eqalign{ & left. matrix{ mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ + }} fleft( x right) = mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ + }} left( {ax + b} right) = a{pi over 2} + b hfill cr mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ – }} fleft( x right) = mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ – }} left( {sin x} right) = sin {pi over 2} = 1 hfill cr fleft( {{pi over 2}} right) = sin {pi over 2} = 1 hfill cr} right} Rightarrow mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ + }} fleft( x right) = mathop {lim }limits_{x to {{left( {{pi over 2}} right)}^ – }} fleft( x right) = fleft( {{pi over 2}} right) Leftrightarrow a{pi over 2} + b = 1,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,left( 1 right) cr & left. matrix{ mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ + }} fleft( x right) = mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ + }} left( {sin x} right) = sin left( { – {pi over 2}} right) = – 1 hfill cr mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ – }} fleft( x right) = mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ – }} left( {ax + b} right) = – a{pi over 2} + b hfill cr fleft( { – {pi over 2}} right) = sin {{ – pi } over 2} = – 1 hfill cr} right} Rightarrow mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ + }} fleft( x right) = mathop {lim }limits_{x to {{left( { – {pi over 2}} right)}^ – }} fleft( x right) = fleft( { – {pi over 2}} right) Leftrightarrow – a{pi over 2} + b = – 1,,,,,,left( 2 right) cr} ) Từ (1) và (2) ta có hệ phương trình (left{ matrix{ a{pi over 2} + b = 1 hfill cr – a{pi over 2} + b = – 1 hfill cr} right. Leftrightarrow left{ matrix{ a = {2 over pi } hfill cr b = 0 hfill cr} right.)
Chọn D.
