Let $f(x) = x^2 + \frac{250}{x}$ be any function defined on $\mathbb{R} - \{0\}$, where $\mathbb{R}$ is the set of real numbers. Then which of the following are TRUE?

(A) $f'(x) = 2x + \frac{250}{x^2}$

(B) $x = 5$ in the only critical point of $f(x)$

(C) minimum value of $f(x)$ is 75

(D) maximum value of $f(x)$ is 50.

Choose the correct answer from the options given below:

Given $f(x) = x^2 + \frac{250}{x}$ for $x \in \mathbb{R} - \{0\}$

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Checking Statement (A):

Finding the derivative:

For $f(x) = x^2 + \frac{250}{x}$:

Derivative of $x^2$ is $2x$

Rewrite $\frac{250}{x}$ as $250x^{-1}$

Derivative of $250x^{-1}$ is $250 \times (-1) \times x^{-2} = -\frac{250}{x^2}$

Therefore:

$f'(x) = 2x - \frac{250}{x^2}$

Statement (A) claims $f'(x) = 2x + \frac{250}{x^2}$, which is incorrect (wrong sign).

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Checking Statement (B):

To find critical points, set $f'(x) = 0$:

$2x - \frac{250}{x^2} = 0$

$2x = \frac{250}{x^2}$

Multiply both sides by $x^2$:

$2x^3 = 250$

$x^3 = 125$

$x = 5$

Since $x^3 = 125$, only $x = 5$ works as $(-5)^3 = -125 \neq 125$.

Statement (B) is correct.

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Checking Statement (C):

To verify if $x = 5$ gives a minimum, find the second derivative:

$f''(x) = 2 + \frac{500}{x^3}$

At $x = 5$:

$f''(5) = 2 + \frac{500}{125} = 2 + 4 = 6 > 0$

Since $f''(5) > 0$, this is a minimum.

Calculate the minimum value:

$f(5) = 5^2 + \frac{250}{5} = 25 + 50 = 75$

Statement (C) is correct.

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Checking Statement (D):

Analyzing the behavior of $f(x)$:

As $x \to \infty$: $f(x) \to \infty$

As $x \to -\infty$: $f(x) \to \infty$

As $x \to 0^+$: $f(x) \to \infty$

As $x \to 0^-$: $f(x) \to -\infty$

The function has no maximum value.

Statement (D) is incorrect.

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Statements (B) and (C) are TRUE.

The correct answer is the option containing both B and C.