The equation of the tangent to the curve $\left(1+x^2\right) y=2-x$, where it crosses the X-axis, is

The curve $(1 + x^2)y = 2 - x$ intersects the X-axis when $y = 0$, yielding $x = 2$ at the point of tangency $(2, 0)$.

Differentiating implicitly using the product rule: $2x \cdot y + (1 + x^2) \frac{dy}{dx} = -1$.

At $(2, 0)$, this simplifies to $5 \frac{dy}{dx} = -1$, so the slope is $m = -\frac{1}{5}$.

Using point-slope form: $y = -\frac{1}{5}(x - 2)$, which simplifies to $x + 5y = 2$.

The correct answer is: A