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f'(x) = -2/x

Integrate both sides with respect to x.

∫f'(x) = ∫-2/x

∫f'(x) = -2∫1/x

f(x) = - 2lnx + c ----------- eqn(1)

where c is the integration constant.

At f(2) = -2ln2 [that is, substitute x = 2 and equate to -2ln2]

eqn(1) becomes,

-2ln2 = -2ln(2) + c

Add 2ln2 to both sides,

-2ln2 + 2ln2 = - 2ln2 + 2ln2 + c

0 = 0 + c

c = 0

Thus, eqn(1) becomes:

f(x) = -2lnx + 0

f(x) = -2lnx, x > 0 [x > 0 since x = 2 and 2 is greate than 0]

**Ans C**

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