It is given that \[\mathrm{f}(x) = \begin{cases} 2 - \frac{2}{\pi}x & \text{for } 0 \le x

Question

It is given that $$\mathrm{f}(x) = \begin{cases} 2 - \frac{2}{\pi}x & \text{for } 0 \le x < \pi \ \x \sin x & \text{for } \pi \le x < 2\pi \end{cases}$$ and that $\mathrm{f}(x) = \mathrm{f}(x + 2\pi)$ for all real values of $x$. Sketch the graph of $y = \mathrm{f}(x)$ for $-\pi \le x < 4\pi$. You do not need to label the coordinates of any stationary point(s).[3] Find the exact area bounded by the curve $y = \mathrm{f}(x)$, the $x$-axis and the lines $x = 0$ and $x = \frac{3\pi}{2}$.[4]

Accepted Answer

(a) Period $2\pi$; see sketch, with included points $(0,2)$, $(2\pi,2)$, $(-\pi,0)$ and open endpoints at $(0,0)$, $(2\pi,0)$, $(4\pi,0)$. (b) $2\pi+1$ square units.

2025 CJC P1 Q4

2025 CJC P1 Q4

Solution:

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