If the curve x2= -4(y-a)does not intersect the curve y=[x2-x+1] (where[.] denotes the greatest integer function) in [0,1+(5)1/2/2], then- 1/3<a<1 -1<a,1 1/4<a<1 none of the above Share with your friends Share 0 Akhil Goyal answered this Let f(x)=[x2-x+1] (where[.] denotes the greatest integer function) in 0,1+52⇒f(x)=[x(x-1)+1]So, in x∈[0,1)f(x)=0 since -12<x(x-1)<0 and thus, 0<x2-x+1<1at x=1, f(x)=1Now, since 52 is a root of the equation x2-x=1, we have that:for x∈1,52, x2-x<1⇒1<x2-x+1<2⇒f(x)=1 in x∈1,52at x=52, f(x)=2So, we have:f(x)=012∀x∈[0,1)∀x∈[1,52)x=52 Following is the graph of f(x) along with various possibilities of g(x): Now, let: g(x)=a-x24. For this curve to not intersect the above curve, either of the following should be true:(1) g(0)<0(2) 1>g(1)≥0(3)2>g52≥1From (1) we have: a<0From (2) we have: a-14≥0 and a-14<1 ⇒14≤a<54From (3) we have: a-516≥1 and a-516<2 ⇒2116≤a<3716Therefore option (c) fits the criteria obtained through possibility (2). However, no option strictly follows the bounds on 'a' 0 View Full Answer