Moderate -0.3 This is a standard area-between-curves problem requiring finding intersection points, setting up the integral of (line - curve), and evaluating. The functions are straightforward (square root and linear), and the integration is routine. Slightly easier than average due to the simple forms involved, though the exact answer requirement and algebraic manipulation keep it close to typical difficulty.
6
\includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-08_613_865_262_632}
The diagram shows the curve with equation \(y = 5 x ^ { \frac { 1 } { 2 } }\) and the line with equation \(y = 2 x + 2\).
Find the exact area of the shaded region which is bounded by the line and the curve.
Line meets curve when: \(2x+2=5x^{\frac{1}{2}} \Rightarrow 2x-5x^{\frac{1}{2}}+2[=0]\) or \(4x^2+8x+4=25x \Rightarrow 4x^2-17x+4[=0]\) or \(x=\frac{y^2}{25} \Rightarrow 2y^2-25y+50[=0]\)
M1
Equating line and curve and rearranging so all terms on same side, condone sign errors, making valid attempt to solve by factorising, using formula or completing the square. Factors are: \((2x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}-2)\), \((4x-1)(x-4)\) and \((2y-5)(y-10)\)
\(x=\frac{1}{4}, x=4\)
A1
SC: If M1 not scored, SC B1 available for correct answers, could just be seen as limits
Area \(= \int 5x^{\frac{1}{2}}dx - \{\int(2x+2)dx\) or area of trapezium\(\}\)
*M1
Intention to integrate and subtract areas. Or integrate curve and subtract area of trapezium
\(\left[\frac{10}{3}x^{\frac{3}{2}}\right]_{\frac{1}{4}}^{4} - \left\{\left[x^2+2x\right]_{\frac{1}{4}}^{4}\) or \(\frac{1}{2}(\text{sum of their } y \text{ values})(\text{their } \frac{15}{4})\right\}\) \(=\left(\left(\frac{10}{3}\times8\right)-\left(\frac{10}{3}\times\frac{1}{8}\right)\right)-\left\{\left((16+8)-\left(\frac{1}{16}+\frac{1}{2}\right)\right)\text{or}\frac{1}{2}\left(\frac{5}{2}+10\right)\frac{15}{4}\right\}\)
DM1
Integrating (\(kx^{\frac{3}{2}}\) seen) and substituting their points of intersection (limits need to be found, not assumed to be 0 and something) or trapezium using correct formula (their \(\frac{15}{4}\), must be their \(4\) – their \(\frac{1}{4}\), but not 0)
\(\frac{45}{16}\) or \(2\frac{13}{16}\) or \(2.8125\)
A1
OE exact answer. Condone \(-\frac{45}{16}\) if corrected to \(\frac{45}{16}\). A0 for inclusion of \(\pi\). SC: If *M1 DM0 scored, SC B1 available for correct answer
## Question 6 (Alternative Method):
| Answer | Mark | Guidance |
|--------|------|----------|
| Line meets curve when: $2x+2=5x^{\frac{1}{2}} \Rightarrow 2x-5x^{\frac{1}{2}}+2[=0]$ **or** $4x^2+8x+4=25x \Rightarrow 4x^2-17x+4[=0]$ **or** $x=\frac{y^2}{25} \Rightarrow 2y^2-25y+50[=0]$ | M1 | Equating line and curve and rearranging so all terms on same side, condone sign errors, making valid attempt to solve by factorising, using formula or completing the square. Factors are: $(2x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}-2)$, $(4x-1)(x-4)$ and $(2y-5)(y-10)$ |
| $x=\frac{1}{4}, x=4$ | A1 | SC: If M1 not scored, SC B1 available for correct answers, could just be seen as limits |
| Area $= \int 5x^{\frac{1}{2}}dx - \{\int(2x+2)dx$ or area of trapezium$\}$ | *M1 | Intention to integrate and subtract areas. Or integrate curve and subtract area of trapezium |
| $\left[\frac{10}{3}x^{\frac{3}{2}}\right]_{\frac{1}{4}}^{4} - \left\{\left[x^2+2x\right]_{\frac{1}{4}}^{4}$ or $\frac{1}{2}(\text{sum of their } y \text{ values})(\text{their } \frac{15}{4})\right\}$ $=\left(\left(\frac{10}{3}\times8\right)-\left(\frac{10}{3}\times\frac{1}{8}\right)\right)-\left\{\left((16+8)-\left(\frac{1}{16}+\frac{1}{2}\right)\right)\text{or}\frac{1}{2}\left(\frac{5}{2}+10\right)\frac{15}{4}\right\}$ | DM1 | Integrating ($kx^{\frac{3}{2}}$ seen) and substituting their points of intersection (limits need to be found, not assumed to be 0 and something) or trapezium using correct formula (their $\frac{15}{4}$, must be their $4$ – their $\frac{1}{4}$, **but not 0**) |
| $\frac{45}{16}$ or $2\frac{13}{16}$ or $2.8125$ | A1 | OE exact answer. Condone $-\frac{45}{16}$ if corrected to $\frac{45}{16}$. A0 for inclusion of $\pi$. SC: If *M1 DM0 scored, SC B1 available for correct answer |
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6\\
\includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-08_613_865_262_632}
The diagram shows the curve with equation $y = 5 x ^ { \frac { 1 } { 2 } }$ and the line with equation $y = 2 x + 2$.\\
Find the exact area of the shaded region which is bounded by the line and the curve.\\
\hfill \mbox{\textit{CAIE P1 2022 Q6 [5]}}