Question:

<p><strong>Step 1: Find the intersection points of the two curves.</strong></p><p>Curve 1: $(1 + x^2)y = 1$ ⟹ $y = \frac{1}{1+x^2}$</p><p>Curve 2: $(x + 1)y = 1$ ⟹ $y = \frac{1}{x+1}$ (where $x \neq -1$)</p><p>At intersection: $\frac{1}{1+x^2} = \frac{1}{x+1}$</p><p>Therefore: $1 + x^2 = x + 1$ ⟹ $x^2 = x$ ⟹ $x(x-1) = 0$</p><p>So $x = 0$ or $x = 1$</p><p><strong>Step 2: Find corresponding y-coordinates.</strong></p><p>When $x = 0$: $y = \frac{1}{1+0} = 1$. Point: $(0, 1)$</p><p>When $x = 1$: $y = \frac{1}{1+1} = \frac{1}{2}$. Point: $(1, \frac{1}{2})$</p><p><strong>Step 3: Find the tangent to curve $(1+x^2)y = 1$ at these points.</strong></p><p>Differentiating $(1+x^2)y = 1$ implicitly:</p><p>$2xy + (1+x^2)\frac{dy}{dx} = 0$</p><p>$\frac{dy}{dx} = -\frac{2xy}{1+x^2}$</p><p><strong>Step 4: Calculate slope at $(0, 1)$.</strong></p><p>$\frac{dy}{dx}\bigg|_{(0,1)} = -\frac{2(0)(1)}{1+0} = 0$</p><p>Tangent line: $y - 1 = 0(x - 0)$ ⟹ $y = 1$</p><p><strong>Step 5: Calculate slope at $(1, \frac{1}{2})$.</strong></p><p>$\frac{dy}{dx}\bigg|_{(1,1/2)} = -\frac{2(1)(\frac{1}{2})}{1+1} = -\frac{1}{2}$</p><p>Tangent line: $y - \frac{1}{2} = -\frac{1}{2}(x - 1)$</p><p>$y = \frac{1}{2} - \frac{1}{2}x + \frac{1}{2} = 1 - \frac{1}{2}x$</p><p>$2y = 2 - x$ ⟹ $x + 2y = 2$</p><p><strong>Step 6: Verify the equations.</strong></p><p>The two tangent lines are: $x + 2y = 2$ and $y = 1$</p><p>∴ Answer: C</p>