Q90 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
Let $f(x) = x\sqrt{4ax - x^2}, a > 0$ then $f(x)$ at $x = 2a$ is
- ADecreasing
- Correct answerBIncreasing
- CDoes not exist
- DZero
Solution
$f'(x) = \dfrac{d}{dx}(x) \cdot \sqrt{4ax-x^2} + x \cdot \dfrac{4a-2x}{2\sqrt{4ax-x^2}}$
$= \sqrt{4ax-x^2} + \dfrac{x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{(4ax-x^2) + x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{6ax-2x^2}{\sqrt{4ax-x^2}}$
At $x = 2a$:
$f'(2a) = \dfrac{6a(2a) - 2(2a)^2}{\sqrt{4a(2a)-(2a)^2}} = \dfrac{12a^2 - 8a^2}{\sqrt{8a^2-4a^2}} = \dfrac{4a^2}{\sqrt{4a^2}} = \dfrac{4a^2}{2a} = 2a$
Since $a > 0$, $f'(2a) = 2a > 0$, so $f(x)$ is increasing at $x = 2a$.
$= \sqrt{4ax-x^2} + \dfrac{x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{(4ax-x^2) + x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{6ax-2x^2}{\sqrt{4ax-x^2}}$
At $x = 2a$:
$f'(2a) = \dfrac{6a(2a) - 2(2a)^2}{\sqrt{4a(2a)-(2a)^2}} = \dfrac{12a^2 - 8a^2}{\sqrt{8a^2-4a^2}} = \dfrac{4a^2}{\sqrt{4a^2}} = \dfrac{4a^2}{2a} = 2a$
Since $a > 0$, $f'(2a) = 2a > 0$, so $f(x)$ is increasing at $x = 2a$.
Hence, the correct answer is option: 'B'.