Question

For the curve \( y(1 + x^2) = 2 - x \), if \(\frac{dy}{dx} = \frac{1}{A}\) at the point where the curve crosses the x-axis, then the value of \( A \) is:

• A. 5
• B. -5
• C. -1
• D. 0

Answer 1

The Correct Option is A

The curve is given as:

\( y(1 + x^2) = 2 - x \).

At the point where the curve crosses the \( x \)-axis, \( y = 0 \). Substitute \( y = 0 \) into the equation:

\( 0(1 + x^2) = 2 - x \implies x = 2 \).

To find \( \frac{dy}{dx} \), differentiate both sides of the equation with respect to \( x \):

\[ \frac{d}{dx} \left[y(1 + x^2)\right] = \frac{d}{dx}[2 - x]. \]

Using the product rule on the left:

\[ (1 + x^2) \frac{dy}{dx} + y(2x) = -1. \]

Substitute \( y = 0 \) at \( x = 2 \) into the equation:

\[ (1 + 2^2) \frac{dy}{dx} = -1 \implies 5 \frac{dy}{dx} = -1. \]

Thus:

\[ \frac{dy}{dx} = -\frac{1}{5}. \]

Given \( \frac{dy}{dx} = \frac{1}{A} \), we find:

\[ A = 5. \]