Show that the equation 3x3 − x2 + 3x − 1 = 0 has a root between 0 and 1. (ii) If x = 2theta and y = 2theta, find (dy)/(dx) and (d2y)/(dx2).

Given $f(x) = 3x^3 - x^2 + 3x - 1$.

$$f(0) = 3(0)^3 - (0)^2 + 3(0) - 1$$ $$f(0) = 0 - 0 + 0 - 1$$ $$f(0) = -1$$

$$f(1) = 3(1)^3 - (1)^2 + 3(1) - 1$$ $$f(1) = 3 - 1 + 3 - 1$$ $$f(1) = 4$$

Since $f(x)$ is a polynomial, it is continuous for all real numbers. We found that $f(0) = -1$ (which is negative) and $f(1) = 4$ (which is positive). Because $f(0)$ and $f(1)$ have opposite signs, by the Intermediate Value Theorem, there must be at least one root (a value of $x$ where $f(x)=0$) between $x=0$ and $x=1$.
Thus, it is shown that the equation $f(x)=0$ has a root between 0 and 1.

7. (ii)

Given $x = \sin 2\theta$ and $y = \cos 2\theta$.

$$\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin 2\theta) = 2 \cos 2\theta$$ $$\frac{dy}{d\theta} = \frac{d}{d\theta}(\cos 2\theta) = -2 \sin 2\theta$$

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2 \sin 2\theta}{2 \cos 2\theta}$$ $$\frac{dy}{dx} = -\frac{\sin 2\theta}{\cos 2\theta} = -\tan 2\theta$$

$$\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{1}{dx/d\theta}$$

First, find $\frac{d}{d\theta}\left(\frac{dy}{dx}\right)$:

$$\frac{d}{d\theta}(-\tan 2\theta) = -\sec^2(2\theta) \cdot \frac{d}{d\theta}(2\theta)$$ $$= -\sec^2(2\theta) \cdot 2$$ $$= -2 \sec^2(2\theta)$$

Now substitute this back into the formula for $\frac{d^2y}{dx^2}$:

$$\frac{d^2y}{dx^2} = (-2 \sec^2(2\theta)) \cdot \frac{1}{2 \cos 2\theta}$$ $$\frac{d^2y}{dx^2} = \frac{-2 \sec^2(2\theta)}{2 \cos 2\theta}$$

Since $\sec 2\theta = \frac{1}{\cos 2\theta}$:

$$\frac{d^2y}{dx^2} = \frac{-1}{\cos^2(2\theta) \cos 2\theta}$$ $$\frac{d^2y}{dx^2} = \frac{-1}{\cos^3(2\theta)}$$ $$\frac{d^2y}{dx^2} = \boxed{-\sec^3(2\theta)}$$