Question

Consider the function \[ f:(0,\infty)\to(-\infty,\infty) \] given by \[ f(x)=\sqrt{x}\log_e(x)-x+1 \] Then which one of the following statements is TRUE?

• A. The derivative of the function \(f\) is decreasing in the interval \((0,1)\)
• B. The function \(f\) has a local maximum at some point \(a \in (0,\infty)\)
• C. The function \(f\) has a local minimum at some point \(b \in (0,\infty)\)
• D. The function \(f\) has neither a point of local maximum nor a point of local minimum in \((0,\infty)\)

Hint:

If: \[ f'(x) \] changes from positive to negative at a point, then the function has a local maximum there.

Answer 1

The Correct Option is B

Step 1: Differentiate the function.
Given: \[ f(x)=\sqrt{x}\ln x-x+1 \] Differentiate: \[ f'(x) = \frac1{2\sqrt{x}}\ln x + \sqrt{x}\cdot\frac1x -1 \] \[ = \frac{\ln x}{2\sqrt{x}} +\frac1{\sqrt{x}} -1 \] \[ = \frac{\ln x+2}{2\sqrt{x}}-1 \]

Step 2: Find critical points.
Set: \[ f'(x)=0 \] \[ \frac{\ln x+2}{2\sqrt{x}}=1 \] \[ \ln x+2=2\sqrt{x} \] Let: \[ t=\sqrt{x} \] Then: \[ 2\ln t+2=2t \] \[ \ln t+1=t \] Consider: \[ g(t)=\ln t+1-t \] \[ g'(t)=\frac1t-1 \] Thus: \[ g'(t)=0 \quad\text{at}\quad t=1 \] Also: \[ g''(t)=-\frac1{t^2}<0 \] Hence: \[ t=1 \] gives maximum value. Now: \[ g(1)=0 \] Therefore: \[ t=1 \] is the only solution. Hence: \[ x=1 \]

Step 3: Use second derivative test.
Differentiate: \[ f'(x)=\frac{\ln x+2}{2\sqrt{x}}-1 \] \[ f''(x) = \frac{-\ln x}{4x^{3/2}} \] At: \[ x=1 \] \[ f''(1)=0 \] Check sign of: \[ f'(x) \] For: \[ x<1 \] \[ f'(x)>0 \] For: \[ x>1 \] \[ f'(x)<0 \] Thus function increases before: \[ x=1 \] and decreases after: \[ x=1 \] Hence: \[ x=1 \] is a local maximum point. Therefore: \[ \boxed{\mathrm{(B)\ is\ correct}} \]

Step 4: Check option (A).
In: \[ (0,1) \] \[ f''(x)=\frac{-\ln x}{4x^{3/2}} \] Since: \[ \ln x<0 \] \[ -\ln x>0 \] Thus: \[ f''(x)>0 \] Hence: \[ f'(x) \] is increasing, not decreasing. Therefore: \[ \boxed{\mathrm{(A)\ is\ incorrect}} \]

Step 5: Identify the correct option.
Therefore: \[ \boxed{\mathrm{(B)}} \]