On a single diagram, sketch the graphs of $y = \frac{3x}{x - 1}$ and $y = 3x$, stating the equat

Question

On a single diagram, sketch the graphs of $y = \frac{3x}{x - 1}$ and $y = 3x$, stating the equation(s) of any asymptote(s) and the coordinates of points of intersection.[4] Hence find the area bounded by the curve $y = \frac{3x}{x - 1}$ and the lines $y = 3x$ and $y = 8$.[2] A curve has the equation $y = \sqrt{x^2 + 1}$ where $x$ is non-negative. With the aid of a sketch of the above curve, explain why the value of $S = \lim_{n \to \infty} \frac{1}{n} \left[ 1 + \sqrt{\frac{1}{n^2} + 1} + \sqrt{\frac{4}{n^2} + 1} + \dots + \sqrt{\frac{(n - 1)^2}{n^2} + 1} \right]$ is $\displaystyle \displaystyle\int_0^1 \sqrt{x^2 + 1} \, \mathrm{d}x$.[4] Hence find the value of $S$.[1]

Accepted Answer

(a)(i) Asymptotes $x=1$, $y=3$; intersections $(0,0)$, $(2,6)$ (ii) $1.13\text{ units}^2$ (b)(i) The expression is the limit of left Riemann sums (ii) $\dfrac12[\sqrt2+\ln(1+\sqrt2)]=1.15\text{ units}^2$

2025 SAJC P1 Q7

2025 SAJC P1 Q7

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