Q.

If limx→λ 2−λxλtan⁡πx2λ=1e  then λ is equal to

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a

b

π

c

π2

d

-2π

answer is C.

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Detailed Solution

If limx→λ 2−λxλtan⁡πx2λ=1e =limx→λ 2−λxλtan⁡πx2π=elimx→λ λtan⁡πx2π2−λx−1=elimx→λ λtan⁡πx2λ1−λx=elimx→λ λ1−λxcot⁡πx2π=elimx→λ λ2/x2−cosec2⁡πx2π⋅π2λ=elimx→λ λ2x2(−1)sin2⁡πx2λπ2λ=elimx→λ −λ2sin2⁡πx2λ2πππ2λ2x2=e limx→λ −2λπsin2⁡πx2λπx2λ2=e−2λπ=e−1⇒−λ⋅2π=−1∴ λ=π2
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If limx→λ 2−λxλtan⁡πx2λ=1e  then λ is equal to