Question

Let \( f(x) = x\sqrt{4ax - x^2, \; a > 0 \) then \( f'(x) \) at \( x = 2a \) is:}

• A. Does not exist
• B. Zero
• C. Decreasing
• D. Increasing

Hint:

If derivative at a point is positive, the function is increasing at that pointAlways check the sign after differentiation.

Answer 1

The Correct Option is D

Step 1: Write the given function.
\[ f(x) = x\sqrt{4ax - x^2} \]

Step 2: Use product rule.
\[ f'(x) = \sqrt{4ax - x^2} + x \cdot \frac{d}{dx}\left(\sqrt{4ax - x^2}\right) \]

Step 3: Differentiate the square root term.
\[ \frac{d}{dx}\left(\sqrt{4ax - x^2}\right) = \frac{1}{2\sqrt{4ax - x^2}}(4a - 2x) \]

Step 4: Substitute back into derivative.
\[ f'(x) = \sqrt{4ax - x^2} + x \cdot \frac{4a - 2x}{2\sqrt{4ax - x^2}} \]

Step 5: Evaluate at \( x = 2a \).
First compute inside root:
\[ 4a(2a) - (2a)^2 = 8a^2 - 4a^2 = 4a^2 \]
\[ \sqrt{4a^2} = 2a \]
Now substitute:
\[ f'(2a) = 2a + (2a)\cdot \frac{4a - 4a}{2 \cdot 2a} \]
\[ = 2a + (2a)\cdot \frac{0}{4a} = 2a \]

Step 6: Analyze the sign.
Since \( a > 0 \), we have:
\[ f'(2a) = 2a > 0 \]

Step 7: Final conclusion.
\[ \boxed{\text{The function is increasing at } x = 2a} \]