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g(x)=x²-x-k²+k=0
を解くと、
→ x²-x-k(k-1)=0
→ (x-k)(x+(k-1))=0
→ x=k、-k+1

①k≦-k+1のとき
すなわち、k≦1/2のとき
x²-x-k²+k<0
→ (x-k)(x+(k-1))<0
→ k<x<-k+1

②k>-k+1のとき
すなわち、k>1/2のとき
→ -k+1<x<k

成田万李

なるほど!
ほんとに分かりやすく教えてくださり、ありがとうございます🙇‍♀️

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