4. Let f(x)=ax+b (where a,b are real numbers), f1(x)=f(x),fn+1(x)=f(fn(x)),n=1, 2,3,⋯, If 2a+b=−2, and fk(x)=−243x+244, then k=
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Official solution
4. 5 Detailed Explanation: We can first find f1(x),f2(x),f3(x), and by induction we get fk(x)=akx+1−a1−ak⋅b. Substituting the known values, we can solve for the result.
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