For the curve \(y\left(1+x^2\right)=2-x\), if \(\displaystyle\frac{dy}{dx}=\frac{1}{A}\) at the point where the curve crosses the \(x\)-axis, then the value of A is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −5
On the x-axis, y = 0, so
\(0 = 2 - x,\quad x = 2\)
The curve crosses the x-axis at (2, 0). Differentiating \(y\left(1+x^2\right) = 2 - x\) by the product rule:
\( \displaystyle 2xy + \left(1 + x^2\right)\frac{dy}{dx} = -1 \)
At (2, 0):
\( \displaystyle 0 + 5\,\frac{dy}{dx} = -1,\quad \frac{dy}{dx} = -\frac{1}{5} \)
Comparing with \(\frac{1}{A}\) gives A = −5.