xℓnxff′(x)+lnxf(x)+f(x)≥1,x∈[2,4] And f(2)=
1
2
,f(4)=
1
4
Now xlnx,
dy
dx
+(ln+1)y≥1
d
dx
(y⋅xlnx)≥1
d
dx
(f(x)⋅xlnx)≥1 ⇒
d
dx
(xlnxf(x)−x)≥0,x∈[2,4] ⇒ The function g(x)=xlnxf(x)−x is increasing in [2,4] And g(2)=2ln2f(2)−2=ln2−2 g(2)=4ln4f(4)−4=ln4−4 =2(ln2−2) Now g(2)≤g(x)≤g(4) Ln2−2≤xlnxf(x)−x≤2(ln2−2)