WJEC Unit 1 2023 June — Question 9 11 marks

Exam BoardWJEC
ModuleUnit 1 (Unit 1)
Year2023
SessionJune
Marks11
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Mark schemeDownload PDF ↗
TopicDifferentiation from First Principles
TypeFirst principles: polynomial with multiple terms
DifficultyModerate -0.3 Part (a) is a standard first principles differentiation of a simple quadratic—routine but requires careful algebraic manipulation worth 5 marks. Part (b)(i) is straightforward differentiation using power rule. Part (b)(ii) requires setting f'(x) > 0 and solving, which is a standard application but involves fractional powers. Overall, this is a typical AS-level calculus question slightly easier than average due to the straightforward nature of all parts, though the mark allocation suggests some working is expected.
Spec1.07g Differentiation from first principles: for small positive integer powers of x1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx

  1. Given that \(y = x^2 - 3x\), find \(\frac{dy}{dx}\) from first principles. [5]
  2. The function \(f\) is defined by \(f(x) = 4x^{\frac{3}{2}} + \frac{6}{\sqrt{x}}\) for \(x > 0\).
    1. Find \(f'(x)\). [2]
    2. When \(x > k\), \(f(x)\) is an increasing function. Determine the least possible value of \(k\). Give your answer correct to two decimal places. [4]

\begin{enumerate}[label=(\alph*)]
\item Given that $y = x^2 - 3x$, find $\frac{dy}{dx}$ from first principles. [5]

\item The function $f$ is defined by $f(x) = 4x^{\frac{3}{2}} + \frac{6}{\sqrt{x}}$ for $x > 0$.
\begin{enumerate}[label=(\roman*)]
\item Find $f'(x)$. [2]

\item When $x > k$, $f(x)$ is an increasing function. Determine the least possible value of $k$. Give your answer correct to two decimal places. [4]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 1 2023 Q9 [11]}}