1. **State the problem:** We have the function $f(x) = 2x^2$ and a tangent line $T$ at $x=1$. We need to: (a) Show the equation of $T$ is $y = 4x - 2$. (b) Find the $x$-intercept of $T$. (c) Find the area of the shaded region $R$ enclosed by $f$, $T$, and the $x$-axis. 2. **(a) Find the equation of the tangent line $T$ at $x=1$:** - The derivative of $f(x)$ gives the slope of the tangent: $$f'(x) = \frac{d}{dx}(2x^2) = 4x$$ - At $x=1$, slope $m = f'(1) = 4(1) = 4$. - The point on the curve at $x=1$ is: $$f(1) = 2(1)^2 = 2$$ - Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2 = 4(x - 1)$$ $$y - 2 = 4x - 4$$ - Simplify: $$y = 4x - 2$$ 3. **(b) Find the $x$-intercept of $T$:** - The $x$-intercept is where $y=0$: $$0 = 4x - 2$$ $$4x = 2$$ $$x = \frac{2}{4} = \frac{1}{2}$$ 4. **(c)(i) Expression for the area of $R$:** - The region $R$ is bounded by $f(x)$, $T$, and the $x$-axis between $x=0$ and $x=1$. - The area can be found by integrating the difference between the line and the curve from $x=\frac{1}{2}$ to $x=1$, plus the area under the curve from $x=0$ to $x=\frac{1}{2}$. - But since the line $T$ crosses the $x$-axis at $x=\frac{1}{2}$, the shaded region is between $x=0$ and $x=1$ bounded by $f(x)$ and $T$ above the $x$-axis. - The area is: $$\text{Area} = \int_0^{\frac{1}{2}} f(x) \, dx + \int_{\frac{1}{2}}^1 (T(x) - f(x)) \, dx$$ 5. **(c)(ii) Calculate the area:** - Calculate each integral: $$\int_0^{\frac{1}{2}} 2x^2 \, dx = 2 \int_0^{\frac{1}{2}} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^{\frac{1}{2}} = 2 \times \frac{(\frac{1}{2})^3}{3} = 2 \times \frac{\frac{1}{8}}{3} = \frac{2}{24} = \frac{1}{12}$$ $$\int_{\frac{1}{2}}^1 (4x - 2 - 2x^2) \, dx = \int_{\frac{1}{2}}^1 (4x - 2 - 2x^2) \, dx$$ - Integrate term by term: $$\int 4x \, dx = 2x^2$$ $$\int -2 \, dx = -2x$$ $$\int -2x^2 \, dx = -\frac{2x^3}{3}$$ - So: $$\left[ 2x^2 - 2x - \frac{2x^3}{3} \right]_{\frac{1}{2}}^1$$ - Evaluate at $x=1$: $$2(1)^2 - 2(1) - \frac{2(1)^3}{3} = 2 - 2 - \frac{2}{3} = -\frac{2}{3}$$ - Evaluate at $x=\frac{1}{2}$: $$2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - \frac{2\left(\frac{1}{2}\right)^3}{3} = 2 \times \frac{1}{4} - 1 - \frac{2 \times \frac{1}{8}}{3} = \frac{1}{2} - 1 - \frac{\frac{1}{4}}{3} = \frac{1}{2} - 1 - \frac{1}{12} = -\frac{1}{2} - \frac{1}{12} = -\frac{6}{12} - \frac{1}{12} = -\frac{7}{12}$$ - Subtract: $$-\frac{2}{3} - \left(-\frac{7}{12}\right) = -\frac{2}{3} + \frac{7}{12} = -\frac{8}{12} + \frac{7}{12} = -\frac{1}{12}$$ - The negative sign means the line is below the curve in this interval, so the area is positive: $$\frac{1}{12}$$ 6. **Total area:** $$\text{Area} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$$ **Final answers:** - (a) Equation of $T$ is $y = 4x - 2$. - (b) $x$-intercept of $T$ is $x = \frac{1}{2}$. - (c) Area of $R$ is $\frac{1}{6}$.