Step 1: Differentiate the function.

Given:

$f(x)=\sqrt{x}\ln x-x+1$

Differentiate:

$f^{\prime }(x)=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}·\frac{1}{x}-1$

$=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}-1$

$=\frac{\ln x+2}{2\sqrt{x}}-1$

Step 2: Find critical points.

Set:

$f^{\prime }(x)=0$

$\frac{\ln x+2}{2\sqrt{x}}=1$

$\ln x+2=2\sqrt{x}$

Let:

$t=\sqrt{x}$

Then:

$2\ln t+2=2t$

$\ln t+1=t$

Consider:

$g\left(t\right)=\ln t+1-t$

$g^{\prime }\left(t\right)=\frac{1}{t}-1$

Thus:

$g^{\prime }\left(t\right)=0$ at $t=1$

Also:

${\mathrm{g″}}^{}\left(t\right)=-\frac{1}{t^2}<0$

Hence:

$t=1$ gives maximum value.

Now:

$g\left(1\right)=0$

Therefore:

$t=1$ is the only solution.

Hence:

$x=1$

Step 3: Use second derivative test.

Differentiate:

$f^{\prime }\left(x\right)=\frac{\ln x+2}{2\sqrt{x}}-1$

$f^{″}\left(x\right)=-\frac{\ln x}{4x^{\frac{3}{2}}}$

At:

$x=1$

$f^{″}\left(1\right)=0$

Check sign of:

$f^{\prime }\left(x\right)$

For:

$x<1$

$f^{\prime }\left(x\right)>0$

For:

$x>1$

$f^{\prime }\left(x\right)<0$

Thus function increases before:

$x=1$

and decreases after:

$x=1$

Hence:

$x=1$ is a local maximum point.

Therefore:

(B) is correct

Step 4: Check option (A).

In:

$(0,1)$

$f^{\&DoublePrime;}\left(x\right)=-\frac{\ln x}{4x^{\frac{3}{2}}}$

Since:

$\ln x<0$

$-\ln x>0$

Thus:

$f^{″}\left(x\right)>0$

Hence:

$f^{\prime }\left(x\right)$

is increasing, not decreasing.

Therefore:

(A) is incorrect

Step 5: Identify the correct option.

Therefore:

(B)

Quick Tip: If:

$f^{\prime }\left(x\right)$

changes from positive to negative at a point, then the function has a local maximum there.