Differential Calculus-2
Differential Calculus-2
Allen Star Batch
Grade 12

Question:

Let $P = x^3 - \frac{1}{x^3}, Q = x - \frac{1}{x}$ and $'a'$ is the minimum value of $\frac{P}{Q^2}$. Then the value of $[a]$ is ________. (where $[x] = $ the greatest integer $\leq x$).

Step-by-Step Solution

Key Concept: Recognize that P = x³ - 1/x³ can be factored using the identity P = Q³ + 3Q where Q = x - 1/x, allowing transformation of P/Q² into a single-variable optimization problem f(Q) = Q + 3/Q that can be minimized using calculus.
Given $Q^3 = P - 3Q$, we rewrite as $f(Q) = P/Q^3 = Q + 3/Q$. Taking the derivative: $f'(Q) = 1 - 3/Q^2$. Setting $f'(Q) = 0$ gives $Q = \sqrt{3}$. At this critical point, the minimum value of $P/Q^2$ is $2\sqrt{3} \cdot \sqrt{3} = 3$.
Correct Answer: 3

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