| Exam Board | SPS |
|---|---|
| Module | SPS SM Pure (SPS SM Pure) |
| Year | 2023 |
| Session | February |
| Marks | 9 |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Standard +0.3 This is a straightforward Newton-Raphson question requiring standard techniques: sign change verification, product/chain rule differentiation of an exponential-root composite, and two iterations of a formula. The only mild challenge is differentiating 3^x√x, but this is routine A-level calculus. Part (b)(iii) tests conceptual understanding of why x₁=0 fails (division by zero in the derivative), which is a standard textbook observation. Overall slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
7.
The function f is defined by
$$\mathrm { f } ( x ) = 3 ^ { x } \sqrt { x } - 1 \quad \text { where } x \geq 0$$
\begin{enumerate}[label=(\alph*)]
\item $\quad \mathrm { f } ( x ) = 0$ has a single solution at the point $x = \alpha$
By considering a suitable change of sign, show that $\alpha$ lies between 0 and 1\\[0pt]
[2 marks]
\item (i) Show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 ^ { x } ( 1 + x \ln 9 ) } { 2 \sqrt { x } }$$
(b) (ii) Use the Newton-Raphson method with $x _ { 1 } = 1$ to find $x _ { 3 }$, an approximation for $\alpha$.
Give your answer to five decimal places.\\
(b) (iii) Explain why the Newton-Raphson method fails to find $\alpha$ with $x _ { 1 } = 0$
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Pure 2023 Q7 [9]}}