(i) Given, x2−8kx+16(k2−k+1)=0Now, D=64{k2−(k2−k+1)}=64(k−1)>0k>1(ii) −2ab​>4⇒28k​>4⇒k>1(iii) f(4)≥0⇒16−32k+16(k2−k+1)≥0⇒k2−3k+2≥0⇒(k−2)(k−1)≥0⇒k≤1 or k≥2​Hence,k=2