1. **Problem (a):** Find the gradient of the curve $y = 3x^4 - 2x^2 + 5x - 2$ at points $(0, -2)$ and $(1, 4)$. The gradient of a curve at a point is given by the derivative $\frac{dy}{dx}$ evaluated at that point. First, differentiate: $$\frac{dy}{dx} = \frac{d}{dx}(3x^4) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(5x) - \frac{d}{dx}(2) = 12x^3 - 4x + 5$$ Evaluate at $x=0$: $$12(0)^3 - 4(0) + 5 = 5$$ Evaluate at $x=1$: $$12(1)^3 - 4(1) + 5 = 12 - 4 + 5 = 13$$ So, gradients are 5 at $(0,-2)$ and 13 at $(1,4)$. 2. **Problem (b):** Find coordinates on $y = 3x^2 - 7x + 2$ where gradient is $-1$. Differentiate: $$\frac{dy}{dx} = 6x - 7$$ Set gradient equal to $-1$: $$6x - 7 = -1$$ $$6x = 6$$ $$x = 1$$ Find $y$ at $x=1$: $$y = 3(1)^2 - 7(1) + 2 = 3 - 7 + 2 = -2$$ Coordinates are $(1, -2)$. 3. **Problem (c):** Find differential coefficient of $y = 3x^2 \sin 2x$. Use product rule: If $y = u v$, then $\frac{dy}{dx} = u'v + uv'$. Let $u = 3x^2$, $v = \sin 2x$. Calculate derivatives: $$u' = 6x$$ $$v' = 2 \cos 2x$$ Apply product rule: $$\frac{dy}{dx} = 6x \sin 2x + 3x^2 (2 \cos 2x) = 6x \sin 2x + 6x^2 \cos 2x$$ **Final answers:** (a) Gradient at $(0,-2)$ is $5$, at $(1,4)$ is $13$. (b) Coordinates where gradient is $-1$ are $(1, -2)$. (c) Differential coefficient is $$\frac{dy}{dx} = 6x \sin 2x + 6x^2 \cos 2x$$.