Let `f(x) = x+ log_(e)x-xlog_(e)x,x in (0,oo)`
List I contains information about zero of f(X) ,f'(x) and f''(x)
List II contains information about the limiting behaviour of f(x), f(x) and f''(x) at infinty
List III contains information about increasing /decreasing nature of f(x) and f'(x)

which o fhte following options is the only CORRECT comibination?

Options

A. (iv)(i)(S)
B. (I)(ii)(R )
C. (III)(iv)(P)
D. (II)(ii)(S)

Answer

4

Explanation

`f(X) =x+log_(e)x-log_(e)x`
`f(X)=1+(1)/(x)log_(e)x-x(1)/(x)=(1)/(x)-log_(e)x`
`f(X)=(1)/(x^(2))-(1)/(x)lt0forallx(0,oo)`
Thus f(X) is strictly decreasing function for `x in (0,oo)`
`underset(xrarroo)lim((1)/(x)-log_(e)x)=underset(xrarroo)loinf(X)=-oo`
`underset(xrarr0)lim((1)/(x)-log_(e)x)=underset(xrarroo)limf(x)=-oo`
i.e f(X) =0 has only one real root in `(0,oo)`
`f(1)=1gt0`
`f(e)=(1)/(e)-1lt0`
So f(x) =0 has one root in (1,e)
Let `f(alpha)=0` where `alpha in (1,e)`

Therefore f(x) is increasing in `(0,alpha)` and decresing in `(alpha,oo)`
`f(1)=1 and f(e^(2))+2-2e^(2)=2-e^(2)lt0`
The graph of the funciton is as shown in the given figure

So f(x)= has one root in `(1,e^(2))`
From column I:II and II ar correct
From column II:ii,iii and iv are correct
From column III:P,Q,S are correct