If limx→0atan3x+(1−cos2x)x+sinx+tanx=1, then the value of a is
A
4
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B
3
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C
2
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D
1
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Solution
The correct option is D1 limx→0atan3x+(1−cos2x)x+sinx+tanx=1⇒limx→0atan3x+2sin2xx+sinx+tanx=1
Dividing the numerator and denominator by x, ⇒limx→03atan3x3x+2sin2xx1+sinxx+tanxx=1⇒3a+01+1+1=1 ⇒3a3=1⇒a=1