Standard +0.3 This is a straightforward tangent problem requiring differentiation of a simple power function, equating the derivative to the gradient k, and solving simultaneous equations. The algebraic manipulation is routine and the question structure is standard for AS-level, making it slightly easier than average.
5 The line with equation \(y = k x - k\), where \(k\) is a positive constant, is a tangent to the curve with equation \(y = - \frac { 1 } { 2 x }\).
Find, in either order, the value of \(k\) and the coordinates of the point where the tangent meets the curve. [5]
\(kx - k = -\frac{1}{2x} \Rightarrow 2kx^2 - 2kx + 1 [= 0]\) OR quadratic in \(y\): \(x = \frac{y+k}{k} \Rightarrow y = -\frac{1}{2\!\left(\frac{y+k}{k}\right)} \Rightarrow 2y^2 + 2ky + k = 0\)
\*M1
OE e.g. \(kx^2 - kx + \frac{1}{2}[=0]\), \(x^2 - x + \frac{1}{2k}[=0]\). Equate line and curve to form 3-term quadratic (all terms on one side).
\(b^2 - 4ac[=0] \Rightarrow ([-]2k)^2 - 4(2k)(1)[=0]\) or \(4k^2 - 8k\ [=0] \Rightarrow 4k(k-2)=0\) OR using equation in \(y\): \((2k)^2 - 4(2)(k) = 0\)
DM1
Use discriminant correctly with their \(a,b,c\) not in quadratic formula. DM0 if \(x\) still present. May see \(k^2 - 4(k)\!\left(\frac{1}{2}\right)=0\) or \(1 - 4\!\left(\frac{1}{2k}\right)=0\).
\(k = 2\) only
A1
If DM0 then \(k=2\), award A0 XP then B0 B0. Allow A1 even if divides by \(k\) to solve. If \(k=0\) also present but uses \(k=2\), award A1.
\(\frac{dy}{dx} = \frac{1}{2x^2}\) or \(\frac{1}{2}x^{-2}\)
\*M1
Differentiate \(-\frac{1}{2x}\). M0 for \(2x^{-2}\). No errors.
\([y=]\frac{1}{2x^2}x - \frac{1}{2x^2} = -\frac{1}{2x}\) or \(\frac{1}{x} = \frac{1}{2x^2}\ [\Rightarrow 2x^2 - x = 0]\)
DM1
Sub *their* \(\frac{dy}{dx}\) into equation of line or set gradient \(= k\) to form equation in \(x\).
\(x = \frac{1}{2}\) only
A1
If DM0 then \(x = \frac{1}{2}\), award A0XP then B0 B0.
\(y = \left[2\times\frac{1}{2} - 2\right] = -1\)
B1
\(k = 2\)
B1
5
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $kx - k = -\frac{1}{2x} \Rightarrow 2kx^2 - 2kx + 1 [= 0]$ OR quadratic in $y$: $x = \frac{y+k}{k} \Rightarrow y = -\frac{1}{2\!\left(\frac{y+k}{k}\right)} \Rightarrow 2y^2 + 2ky + k = 0$ | **\*M1** | OE e.g. $kx^2 - kx + \frac{1}{2}[=0]$, $x^2 - x + \frac{1}{2k}[=0]$. Equate line and curve to form 3-term quadratic (all terms on one side). |
| $b^2 - 4ac[=0] \Rightarrow ([-]2k)^2 - 4(2k)(1)[=0]$ or $4k^2 - 8k\ [=0] \Rightarrow 4k(k-2)=0$ OR using equation in $y$: $(2k)^2 - 4(2)(k) = 0$ | **DM1** | Use discriminant correctly with their $a,b,c$ not in quadratic formula. DM0 if $x$ still present. May see $k^2 - 4(k)\!\left(\frac{1}{2}\right)=0$ or $1 - 4\!\left(\frac{1}{2k}\right)=0$. |
| $k = 2$ only | **A1** | If DM0 then $k=2$, award A0 XP then B0 B0. Allow A1 even if divides by $k$ to solve. If $k=0$ also present but uses $k=2$, award A1. |
| $4x^2 - 4x + 1 = 0 \Rightarrow (2x-1)^2 = 0 \Rightarrow x = \frac{1}{2}$ | **B1** | |
| $y = 2 \times \frac{1}{2} - 2 = -1$ | **B1** | |
**Alternative method for Q5:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{2x^2}$ or $\frac{1}{2}x^{-2}$ | **\*M1** | Differentiate $-\frac{1}{2x}$. M0 for $2x^{-2}$. No errors. |
| $[y=]\frac{1}{2x^2}x - \frac{1}{2x^2} = -\frac{1}{2x}$ or $\frac{1}{x} = \frac{1}{2x^2}\ [\Rightarrow 2x^2 - x = 0]$ | **DM1** | Sub *their* $\frac{dy}{dx}$ into equation of line or set gradient $= k$ to form equation in $x$. |
| $x = \frac{1}{2}$ only | **A1** | If DM0 then $x = \frac{1}{2}$, award A0XP then B0 B0. |
| $y = \left[2\times\frac{1}{2} - 2\right] = -1$ | **B1** | |
| $k = 2$ | **B1** | |
| | **5** | |
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5 The line with equation $y = k x - k$, where $k$ is a positive constant, is a tangent to the curve with equation $y = - \frac { 1 } { 2 x }$.
Find, in either order, the value of $k$ and the coordinates of the point where the tangent meets the curve. [5]\\
\hfill \mbox{\textit{CAIE P1 2023 Q5 [5]}}