# Question 11:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $2u\,du = 2x\,dx$ | B1 | Any correct expression linking $du$ and $dx$. Could be $du = \frac{1}{2}2x(x^2+3)^{-\frac{1}{2}}dx$ or equiv in terms of $u$ |
| $\int \frac{4u(u^2-3)}{\sqrt{u^2}}\,du$ | M1* | Attempt to rewrite integrand in terms of $u$. Not just $dx = du$, unless from a clear attempt at $du$ eg using $u = x + \sqrt{3}$ |
| $\int (4u^2 - 12)\,du$ | A1 | Obtain correct integrand. Allow unsimplified expression |
| $\frac{4}{3}u^3 - 12u\,(+c)$ | M1dep* | Attempt integration. Simplify to form that can be integrated, then increase all powers by 1 |
| $\frac{4}{3}u(u^2-9)+c = \frac{4}{3}(x^2-6)\sqrt{x^2+3}+c$ **A.G.** | A1 | Obtain given answer, with at least one intermediate step seen. Need evidence of common factor (in terms of $u$ or $x$) being taken out. Condone omission of $+c$ |

**[5]**

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{4}{3}((-5\times2)-(-6\times\sqrt{3}))$ | M1 | Attempt to use limits $x=0$ and $x=1$, or $u=\sqrt{3}$ and $u=2$ in integral in terms of $u$. Correct order and subtraction. Attempt to use both limits in their integral to give two terms. DR so just stating decimal area is M0. Either using answer from (a) or their integration attempt with $+2$ or $+3$ |
| $= \frac{4}{3}(6\sqrt{3}-10)$ or $0.523$ | A1 | Obtain correct area under curve. Accept exact (inc unsimplified) or decimal. Using $+2$ gives $\frac{4}{3}(4\sqrt{2}-3\sqrt{3})$ or $0.614$ |
| $\frac{dy}{dx} = \frac{12x^2(x^2+3)^{\frac{1}{2}} - 4x^3 \cdot 2x \cdot \frac{1}{2}(x^2+3)^{-\frac{1}{2}}}{x^2+3}$ | M1 | Attempt derivative using the quotient rule. Or equiv with product rule. Need difference of two terms in numerator, at least one term correct, but allow subtraction in incorrect order. Using either $+2$ or $+3$ equation |
| Obtain correct, unsimplified, derivative | A1 | With either $+2$ or $+3$ |
| at $x=1$, $m = \frac{11}{2}$ hence $m' = -\frac{2}{11}$ | M1 | Attempt gradient of normal at $x=1$. Substitute $x=1$ and use negative reciprocal. Using $+2$ gives $m' = -\frac{3}{32}\sqrt{3}$ |
| $y - 2 = -\frac{2}{11}(x-1)$, when $y=0$, $x=12$ | M1 | Attempt to find point of intersection of normal with $x$-axis. Attempt equation of normal with their gradient and either $(1,2)$ or $(1, \frac{4}{3}\sqrt{5})$, and then use $y=0$ to find $x$ intersection |
| area $= 8\sqrt{3} - \frac{40}{3} + 11 = 8\sqrt{3} - \frac{7}{3}$ | A1 | Obtain correct area. Allow any exact (including unsimplified) or decimal equivalent. From combining a correct area under curve and a correct area of triangle (either $11$ or $\frac{64}{9}\sqrt{3}$), even if inconsistent. Can still get A1 following M0 for area under curve BC and/or $m$ found BC |

**[7]**

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