Let ∣∣∣2secx3tanxextan−1x∣∣∣=A+Bx+Cx2+⋯(A,B,Care real constants), then which of the following(s) is/are true?
A
A=0
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B
B=1
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C
A+B=1
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D
A−B=1
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Solution
The correct option is DA−B=1 Given : ∣∣∣2secx3tanxextan−1x∣∣∣=A+Bx+Cx2+⋯
Put x=0 on both sides, ⇒A=∣∣∣2010∣∣∣=0
Differentiating both sides w.r.t. x, ⇒∣∣∣2secxtanx3sec2xextan−1x∣∣∣+∣∣
∣∣2secx3tanxex11+x2∣∣
∣∣=0+B+2Cx+⋯
Now putting x=0 on both sides, B=∣∣∣0310∣∣∣+∣∣∣2011∣∣∣=−1 ⇒A+B=−1 ⇒A−B=1