Question

If \(\frac{d}{dx}\left\{ \frac{x-1}{x-\sqrt{x}} e^{2x+1} \right\} = \frac{x-1}{x-\sqrt{x}} e^{2x+1} f(x)\), then \(f(4) =\)

• A. 0
• B. 1
• C. \(\frac{35}{24}\)
• D. \(\frac{47}{24}\)

Hint:

Always simplify algebraic expressions before differentiating. Here, cancelling the factor \((\sqrt{x}-1)\) simplified the function significantly.

Answer 1

The Correct Option is D

Step 1: Simplify the Function: Let \(y = \frac{x-1}{x-\sqrt{x}} e^{2x+1}\). Simplify the algebraic part: \[ \frac{x-1}{x-\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2} \] So, \(y = (1 + x^{-1/2}) e^{2x+1}\).
Step 2: Differentiate \(y\): Using product rule: \[ \frac{dy}{dx} = \left( -\frac{1}{2}x^{-3/2} \right) e^{2x+1} + (1 + x^{-1/2}) \cdot e^{2x+1} \cdot 2 \] \[ \frac{dy}{dx} = e^{2x+1} \left[ -\frac{1}{2x\sqrt{x}} + 2 + \frac{2}{\sqrt{x}} \right] \]
Step 3: Equate to Given Form: Given \(\frac{dy}{dx} = \left( \frac{x-1}{x-\sqrt{x}} \right) e^{2x+1} f(x) = (1 + x^{-1/2}) e^{2x+1} f(x)\). So, \[ (1 + x^{-1/2}) f(x) = 2 + 2x^{-1/2} - \frac{1}{2}x^{-3/2} \] \[ \left( 1 + \frac{1}{\sqrt{x}} \right) f(x) = 2 + \frac{2}{\sqrt{x}} - \frac{1}{2x\sqrt{x}} \]
Step 4: Find \(f(4)\): Substitute \(x = 4\) (\(\sqrt{4} = 2\)): \[ \left( 1 + \frac{1}{2} \right) f(4) = 2 + \frac{2}{2} - \frac{1}{2(4)(2)} \] \[ \frac{3}{2} f(4) = 2 + 1 - \frac{1}{16} = 3 - \frac{1}{16} = \frac{47}{16} \] \[ f(4) = \frac{47}{16} \times \frac{2}{3} = \frac{47}{24} \]