Question
The values of constants a and b so thatlimx→∞(x2+1x+1−ax−b)=12,are
Answer: Option A
:
A
limx→∞(x2+1x+1−ax−b)=12⇒limx→∞(x2+1)−(ax+b)(x+1)x+1=12⇒limx→∞x2(1−a)−(a+b)x−b+1x+1=12⇒1−a=0anda+b=−12∴a=1b=−32
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:
A
limx→∞(x2+1x+1−ax−b)=12⇒limx→∞(x2+1)−(ax+b)(x+1)x+1=12⇒limx→∞x2(1−a)−(a+b)x−b+1x+1=12⇒1−a=0anda+b=−12∴a=1b=−32
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