(A)2x2+4x+5 2x2+4x+2+3⇒2(x+1)2+3 ∴ The minimum value of 2(x+1)2+3 is 3 . (B) Let f(x)=
x2+4x+1
x2+x+1
=
x2+x+1
x2+x+1
+
3x
x2+x+1
=1+
3x
x2+x+1
The maximum value of
3x
x2+x+1
can be 1 .
3x
x2+x+1
=
3
x+
1
x
+1
and x+
1
x
≥2 Therefore, maximum value of f(x) is 2 . (C) Given, 1≤
3x2−5x+6
x2+1
≤2 =1≤
3x2−5x+6
x2+1
and
3x2−5x+6
x2+1
≤2 ⇒x2+1≤3x2−5x+5 and 3x2−5x+6≤2x2+2 ⇒2x2−5x+4≥0 and x2−5x+4≤0 The value of x for which 2x2−5x+4≥0 are not real. ∵2x2−5x+4≥0 is possible for all real value ⇒x∈R and x2−5x+4≤0 ⇒(x−4)(x−1)≤0⇒x∈[1,4] Now common of x∈R and x∈[1,4]⇒x∈[1,4]∴b=4 (D) a=1 Correct order is = IV, III, V, II :