Consider the function \(f:\left(0,\infty \right)\to \left(−\infty ,\infty \right)\) given by \(f\left(x\right)=\sqrt{x}{…

Given \(f(x)=\sqrt{x}\ln x-x+1\), where \(x>0\).

\(f'(x)=\dfrac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\dfrac1x-1\)

\(f'(x)=\dfrac{\ln x}{2\sqrt{x}}+\dfrac1{\sqrt{x}}-1=\dfrac{\ln x+2}{2\sqrt{x}}-1\)

\(f'(x)=\dfrac{\ln x+2-2\sqrt{x}}{2\sqrt{x}}\)

Put \(t=\sqrt{x}\), so \(x=t^2\) and \(\ln x=2\ln t\).

\(f'(x)=\dfrac{2\ln t+2-2t}{2t}=\dfrac{\ln t+1-t}{t}\)

Now \(\ln t\le t-1\) for \(t>0\).

So \(\ln t+1-t\le0\), hence \(f'(x)\le0\).

Equality occurs only when \(t=1\), i.e. \(x=1\).

Therefore \(f\) is decreasing on \((0,\infty)\), but derivative becomes zero only at one point.

Now check option \(A\):

\(f''(x)=\dfrac{-\ln x}{4x^{3/2}}\)

For \(0<x<1\), \(\ln x<0\), so \(f''(x)>0\).

Thus \(f'\) is increasing on \((0,1)\), not decreasing. So option \(A\) is false.

Since \(f'(x)\le0\) on \((0,\infty)\), \(f\) cannot have a local maximum or local minimum. So options \(B\) and \(C\) are false.

Hence option \(D\) is true.