∵[2x−2]=[(2x−1)−1] =[2x−1]−1 let, [2x−1]=a ⇒[2x−2]=a−1 so, 3(a−1)+1=2a−1 ⇒3a−3+1=2a−1⇒a=1 thus [2x−1]=1 ⇒1≤(2x−1)<2⇒2≤2x<3 ⇒1≤x<1.5 so, x∈[1,1.5) Now, k=2[2x−1]−1=2(1)−1=1 for x∈[1,1.5) f(x)=[k+5x]=[1+5x] at x=1,f(x)=[6]=6( minimum ) at x=1.5,f(x)=[1+5×1.5] =[8.5]=8 (maximum) so, f(x) can be 6,7,8 Hence, the range of f(x)={6,7,8}
