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Q.

Let fx=4x5     for x2xk     for x>2 If limx2f(x) exists  then k is

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a

1

b

-1

c

2

d

-2

answer is B.

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Detailed Solution

limx2f(x)=limx2+fx
limx24x5=limx2+x-k
4(2)5=2kk=1

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