Question

The point \(P\) lies on the curve with equation \[ y = x\sqrt{2\log_e(x)}, \qquad x>0. \] If \[ \frac{dy}{dx}=2 \] at \(x=k\), then the value of \(k\) is equal to: 

• A. \(-\sqrt{e}\)
• B. e
• C. \(\sqrt{e}\)
• D. 2e
• E. \(\frac{e}{2}\)

Hint:

The equation \(t + \frac{1}{t} = 2\) always implies \(t = 1\) for real numbers. Recognizing this pattern saves time in solving quadratic equations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to differentiate the function using the product rule and chain rule, then set the derivative equal to 2 to solve for \(k\).

Step 2: Key Formula or Approach:
Product Rule: \((uv)' = uv' + vu'\).
Derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}} \cdot u'\).

Step 3: Detailed Explanation:
Let \(y = x \cdot (2\log x)^{1/2}\).
\[ \frac{dy}{dx} = x \cdot \left[ \frac{1}{2}(2\log x)^{-1/2} \cdot \frac{2}{x} \right] + (2\log x)^{1/2} \cdot 1 \]
Simplify the first term:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{2\log x}} + \sqrt{2\log x} \]
Given \(\frac{dy}{dx} = 2\) at \(x = k\):
\[ \frac{1}{\sqrt{2\log k}} + \sqrt{2\log k} = 2 \]
Let \(t = \sqrt{2\log k}\). Then:
\[ \frac{1}{t} + t = 2 \implies t^2 - 2t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1 \]
Substitute \(t\) back:
\[ \sqrt{2\log k} = 1 \]
Square both sides:
\[ 2\log k = 1 \implies \log k = \frac{1}{2} \]
Taking exponents:
\[ k = e^{1/2} = \sqrt{e} \]

Step 4: Final Answer:
The value of \(k\) is \(\sqrt{e}\).