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Question

If f(x)=x2; when x<0,f(x)=x; when 0<x<1,f(x)=1x; when x1, then

A
3f(12)6=32
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B
f(2)=13
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C
f(6)+f(12)=f(2)
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D
3f(4)+4=16
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Solution

The correct option is D 3f(4)+4=16
(A) 3f(12)6=3×126
=92

So, A is not correct

(B) f(2)=12

So, B is not correct

(C) f(6)+f(12)=36+12
=73212=f(2)

So, C is not correct

(D) 3f(4)+4=3×16+4
=16

Hence, D is correct.

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