Moderate -0.8 This is a straightforward integration problem requiring only basic knowledge of power rule integration (rewriting 1/√x as x^(-1/2) and integrating to get 2√x) and using two given points to find two constants. It's simpler than average A-level questions as it involves a single standard technique with no problem-solving insight required.
3 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { \sqrt { } x }\), where \(k\) is a constant. The points \(P ( 1 , - 1 )\) and \(Q ( 4,4 )\) lie on the curve. Find the equation of the curve.
Substitutes both points into an integrated expression with a \('+c'\) and solve as far as a value for one variable
M1
Expect to see \(-1 = 2k + c\) and \(4 = 4k + c\)
\(k = 2\frac{1}{2}\) and \(c = -6\)
A1
WWW
\(y = 5\sqrt{x} - 6\)
A1
OE. From correct values of both \(k\) & \(c\) and correct integral.
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(y=)\dfrac{kx^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\left(=\dfrac{k\sqrt{x}}{\frac{1}{2}}\right)(+c)$ | B1 | OE |
| Substitutes both points into an integrated expression with a $'+c'$ and solve as far as a value for one variable | M1 | Expect to see $-1 = 2k + c$ and $4 = 4k + c$ |
| $k = 2\frac{1}{2}$ and $c = -6$ | A1 | WWW |
| $y = 5\sqrt{x} - 6$ | A1 | OE. From correct values of both $k$ & $c$ and correct integral. |
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3 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { \sqrt { } x }$, where $k$ is a constant. The points $P ( 1 , - 1 )$ and $Q ( 4,4 )$ lie on the curve. Find the equation of the curve.\\
\hfill \mbox{\textit{CAIE P1 2019 Q3 [4]}}