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Q.

Ltx→π2acotx−acosxcotx−cosx=

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a

log a

b

2

c

a

d

log x

answer is A.

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Detailed Solution

=Ltx→π/2acosxacotx−cosx−1cotx−cosx=acos(π/2).Ltz→0az−1z=a0⋅loga=loga (where z=cotx−cosx→0 as x→π/2)
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