1. **Problem statement:** (a) Estimate the gradient of the curve $y = \frac{1}{x} + \frac{x}{2}$ at $x=2$ by drawing a tangent. (b) Find the solutions of the equation $\frac{1}{x} - \frac{5x}{2} + 1 = 0$ by drawing a suitable line on the grid. 2. **Formula and rules:** - The gradient of a curve at a point is the derivative $\frac{dy}{dx}$ evaluated at that point. - To find the gradient at $x=2$, differentiate $y$ with respect to $x$. - For part (b), rearrange the equation to find intersections with the curve or use graphical methods. 3. **Step-by-step solution:** **(a) Gradient at $x=2$** - Given $y = \frac{1}{x} + \frac{x}{2}$. - Differentiate: $$\frac{dy}{dx} = -\frac{1}{x^2} + \frac{1}{2}$$ - Evaluate at $x=2$: $$\frac{dy}{dx}\bigg|_{x=2} = -\frac{1}{2^2} + \frac{1}{2} = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}$$ - So, the gradient at $x=2$ is $\frac{1}{4}$. **(b) Solve $\frac{1}{x} - \frac{5x}{2} + 1 = 0$** - Rearrange: $$\frac{1}{x} + 1 = \frac{5x}{2}$$ - Define $y = \frac{1}{x} + \frac{x}{2}$ (the original curve). - The equation can be rewritten as: $$\frac{1}{x} - \frac{5x}{2} + 1 = 0 \implies \frac{1}{x} + 1 = \frac{5x}{2}$$ - Graphically, this means find $x$ where $y = \frac{1}{x} + \frac{x}{2}$ intersects the line $y = \frac{5x}{2} - 1$. - By drawing the line $y = \frac{5x}{2} - 1$ on the grid and finding intersection points with the curve, the solutions are approximately: - $x \approx -0.5$ - $x \approx 1$ - $x \approx 2$ 4. **Summary:** - Gradient at $x=2$ is $\frac{1}{4}$. - Solutions to $\frac{1}{x} - \frac{5x}{2} + 1 = 0$ are approximately $x = -0.5, 1, 2$ by graphical intersection. This completes the problem.