Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

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

(A) $f'(x) = 2x +\frac{250}{x^2}$
(B) $= 5$ 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:

Options:

(A) and (B) only

(B) and (C) only

(C) and (D) only

(A), (B) and (D) only

Correct Answer:

(B) and (C) only

Explanation:

The correct answer is Option (2) → (B) and (C) only

Given function:

$f(x) = x^2 + \frac{250}{x}$, defined on $\mathbb{R} - \{0\}$.

Differentiate:

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

Hence, (A) $f'(x) = 2x + \frac{250}{x^2}$ is false.

Find critical points:

$f'(x) = 0 \Rightarrow 2x - \frac{250}{x^2} = 0$

$\Rightarrow 2x^3 = 250 \Rightarrow x^3 = 125 \Rightarrow x = 5$

Thus, (B) is true.

Second derivative:

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

At $x = 5$, $f''(5) = 2 + \frac{500}{125} = 6 > 0$, hence minimum at $x = 5$.

Minimum value:

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

So, (C) is true.

As $x \to 0$, $\frac{250}{x} \to \infty$ or $-\infty$ depending on sign of $x$, and as $x \to \infty$, $f(x) \to \infty$.

Therefore, there is no maximum value. (D) is false.

Correct statements: (B) and (C)