Question

For a given function \( y = f(x) \), \( \delta y \) denotes the actual error in \( y \) corresponding to actual error \( \delta x \) in \( x \) and \( dy \) denotes the approximate value of \( \delta y \). If \( y = f(x) = 2x^2 - 3x + 4 \) and \( \delta x = 0.02 \), then the value of \( \delta y - dy \) when \( x = 5 \) is: \vspace{0.5cm}

• A. \( 0.0008 \)
• B. \( 0.008 \)
• C. \( 0.0004 \)
• D. \( 0.004 \)]

Hint:

For small changes in \( x \), the differential \( dy \) provides a good approximation for \( \delta y \). The difference \( \delta y - dy \) arises due to higher-order terms ignored in the differential approximation.

Answer 1

The Correct Option is A

Step 1: Calculate \( dy \) (Approximate Change in \( y \)) The differential \( dy \) is given by: \[ dy = \frac{dy}{dx} \delta x \] First, find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{d}{dx} (2x^2 - 3x + 4) = 4x - 3 \] Substituting \( x = 5 \): \[ \frac{dy}{dx} = 4(5) - 3 = 20 - 3 = 17 \] Now, calculate \( dy \): \[ dy = 17 \times 0.02 = 0.34 \]]

Step 2: Calculate \( \delta y \) (Actual Change in \( y \)) \[ \delta y = f(5 + 0.02) - f(5) \] First, compute \( f(5) \): \[ f(5) = 2(5)^2 - 3(5) + 4 = 50 - 15 + 4 = 39 \] Now compute \( f(5.02) \): \[ f(5.02) = 2(5.02)^2 - 3(5.02) + 4 \] Approximating \( (5.02)^2 \approx 25.2004 \), we get: \[ f(5.02) = 2(25.2004) - 3(5.02) + 4 = 50.4008 - 15.06 + 4 = 39.3408 \] Thus, \[ \delta y = 39.3408 - 39 = 0.3408 \]

Step 3: Compute \( \delta y - dy \) \[ \delta y - dy = 0.3408 - 0.34 = 0.0008 \]