Question

If \(f^'(x) = \frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}}\) and \(f(0) = e\) then \(f(1) = \ldots \ldots \ldots \ldots\)

• A. \(e\)
• B. \(2e\)
• C. \(3e\)
• D. \(4e\)

Hint:

Split \(f'\) and integrate each term using \(t=\sqrt x\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(f'(x)=\dfrac{(\sqrt x+1)e^{\sqrt x}}{\sqrt x}=e^{\sqrt x}+\dfrac{e^{\sqrt x}}{\sqrt x}\).

Step 2: Key Formula or Approach
Put \(t=\sqrt x\), so \(dx=2t\,dt\).

Step 3: Detailed Explanation
\(\int\dfrac{e^{\sqrt x}}{\sqrt x}dx=\int2e^t\,dt=2e^{\sqrt x}\).
\(\int e^{\sqrt x}dx=2\int te^tdt=2(t-1)e^t=2(\sqrt x-1)e^{\sqrt x}\).
Add: \(f(x)=2\sqrt x\,e^{\sqrt x}+C\).
\(f(0)=C=e\). Then \(f(1)=2e+e=3e\).

Final Answer:
\(f(1)=3e\), option (C). \[ \boxed{3e\ \text{(C)}} \]