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Anonymous User Best Answer
T((af+bg)(x))=x(af+bg)'(x) =x{ af'(x)+bg'(x)} =a{xf'(x)}+b{xg'(x)} =aT(f)+bT(g) T is linear T is not one one onto bcz T(2)=0 T(3)=0 But 2≠3 T is not one one Pre image of constant does not exist so not onto