Moderate -0.8 This is a structured, guided question on exponential equations with clear scaffolding. Part (a) requires basic index law manipulation (2^{x+2} = 4y, 2^{3-x} = 8/y), part (b) is a straightforward 'show that' verification, and part (c) involves solving a quadratic then taking logarithms. While it requires multiple techniques, each step is routine and heavily signposted, making it easier than the average A-level question which typically demands more independent problem-solving.
\begin{enumerate}[label=(\alph*)]
\item Given that \(y = 2^x\), find expressions in terms of \(y\) for
\(2^{x+2}\), [2]
\(2^{3-x}\). [2]
\item Show that using the substitution \(y = 2^x\), the equation
$$2^{x+2} + 2^{3-x} = 33$$
can be rewritten as
$$4y^2 - 33y + 8 = 0.$$ [2]
\item Hence solve the equation
$$2^{x+2} + 2^{3-x} = 33.$$ [4]
\begin{enumerate}[label=(\alph*)]
\item Given that $y = 2^x$, find expressions in terms of $y$ for
\begin{enumerate}[label=(\roman*)]
\item $2^{x+2}$, [2]
\item $2^{3-x}$. [2]
\end{enumerate}
\item Show that using the substitution $y = 2^x$, the equation
$$2^{x+2} + 2^{3-x} = 33$$
can be rewritten as
$$4y^2 - 33y + 8 = 0.$$ [2]
\item Hence solve the equation
$$2^{x+2} + 2^{3-x} = 33.$$ [4]
</enumerate}
\hfill \mbox{\textit{OCR C1 Q6 [10]}}