## Question 14(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = -1 < 0$ and $f(1) = 3-1 = 2 > 0$ | M1 (1.1a) | Evaluates $f(0)=-1$ and $f(1)=2$; or evaluates two other suitable appropriate values correct to 1 sig fig |
| Change of sign implies root therefore $\alpha$ is between 0 and 1 | R1 (2.1) | Completes argument correctly stating $f(0)<0$ and $f(1)>0$ and concludes $0 < \alpha < 1$ |

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## Question 14(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = x^{\frac{1}{2}}(3^x)\ln 3 + \frac{1}{2}x^{-\frac{1}{2}}(3^x)$ | M1 (3.1a) | Uses product rule to obtain expression of the form $Ax^{\frac{1}{2}}(3^x) + Bx^{-\frac{1}{2}}(3^x)$; $A$ and/or $B$ can be positive or negative |
| $= 3^x\left(\ln 3\sqrt{x} + \frac{1}{2\sqrt{x}}\right)$ | A1 (1.1b) | Obtains fully correct $f'(x)$ |
| $= 3^x\left(\frac{2x\ln 3}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}\right)$ | | |
| $= 3^x\left(\frac{x\ln 9}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}\right)$ | | |
| $= 3^x\left(\frac{1+x\ln 9}{2\sqrt{x}}\right)$ | R1 (2.1) | Completes convincing argument with no slips to show the required result. AG |

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## Question 14(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_{n+1} = x_n - \dfrac{\dfrac{3^{x_n}\sqrt{x_n}-1}{\dfrac{3^{x_n}(1+x_n\ln 9)}{2\sqrt{x_n}}}}{}$ | M1 (1.1a) | Forms correct Newton-Raphson expression; PI by correct value of $x_2$ or $x_3$ stated to at least 3 decimal places |
| $x_{n+1} = x_n - \dfrac{2\sqrt{x_n}(3^{x_n}\sqrt{x_n}-1)}{3^{x_n}(1+x_n\ln 9)}$ | | |
| $x_2 = 0.5829716\ldots$ | | |
| $x_3 = 0.4246536\ldots$ | A1 (1.1b) | Obtains correct value of $x_3$; must be stated to five decimal places |
| $x_3 \approx 0.42465$ | | |

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## Question 14(b)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Convergence is impossible as all values of $x_n$ would equal 0 | E1 (2.4) | Explains that convergence is impossible; must use the word convergence or convergent |
| The tangent at $x=0$ is vertical, or explains all values of $x_n$ would equal 0, or demonstrates that several values of $x_n$ would be 0 | E1 (2.4) | Explains that the tangent at $x=0$ is vertical; or explains all values of $x_n$ would equal 0 |

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