Explanation

(a)
The given curve is $\left( 1 + x^{2} \right)y = 2 - x$ …(i)
It meets $x$-axis, where $y = 0\ \Rightarrow 0 = 2 - x \Rightarrow x = 2$
So, Eq. (i) meets $x$-axis at the point at the point (2, 0)
Also, from Eq. (i), $y = \frac{2 - x}{1 + x^{2}}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = \frac{\left( 1 + x^{2} \right)( - 1) - (2 - x)(2x)}{\left( 1 + x^{2} \right)^{2}}$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{x^{2} - 4x - 1}{\left( 1 + x^{2} \right)^{2}}$
$\therefore$ Slope of tangent at (2, 0)$= \frac{2^{2} - 4(2) - 1}{\left( 1 + 2^{2} \right)^{2}}$
$= \frac{4 - 8 - 1}{(1 + 4)^{2}} = - \frac{5}{25} = - \frac{1}{5}$
$\therefore$ Equation of tangent at (2, 0) with slope $- \frac{1}{5}$ is
$y - 0 = - \frac{1}{5}(x - 2)$
$\Rightarrow \ \ \ 5y = - x + 2$
$\Rightarrow \ \ x + 5y = 2$