By sketching the graphs of \(y = |2x + 1|\) and \(y = -x\) on the same axes, show that the equation \(|2x + 1| = -x\) has two roots. Find these roots. [4]
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Answer Marks
Guidance
Answer: Sketch of \(y =2x + 1
\)
Answer: \(y = −x\) and two intersections indicatedMarks: A1
Guidance: (none)
Answer: \(x = -1\) (not from ww, condone \((-1, 1)\))Marks: B1
Guidance: squaring: \((2x+1)^2 = x^2 \Rightarrow 3x^2 + 4x + 1 = 0 \Rightarrow (3x+1)(x+1) = 0\), \(x = -1, -1/3\)
Answer: \(x = -1/3\) (not from ww, condone \((-1/3, 1/3)\))Marks: B1
Guidance: (none)
Total Marks: [4]
Question 5(i)
Answer Marks
Guidance
Answer: \(dV/dh = 4 \frac{1}{2}(h^3 + 1)^{-1/2} \cdot 3h^2\)Marks: M1
Guidance: their deriv of \(4u^{1/2}\) × their deriv of \(h^3+1\)
Answer: (correct)Marks: A1
Guidance: (none)
Answer: Substituting \(h = 2\) into their derivativeMarks: M1
Guidance: (none)
Answer: when \(h = 2\), \(dV/dh = 8\)Marks: A1cao
Guidance: (none)
Total Marks: [4]
Question 5(ii)
Answer Marks
Guidance
Answer: \(dV/dt = 0.4\)Marks: B1
Guidance: soi
Answer: \(dV/dt = dV/dh \times dh/dt\)Marks: M1
Guidance: o.e. Any correct chain rule in \(V\), \(h\), \(t\) (or \(r\))
Answer: \(0.4 = 8 \times dh/dt \Rightarrow dh/dt = 0.05\) (m per min)Marks: A1cao
Guidance: 0.05 or 1/20
Total Marks: [3]
Question 6(i)
Answer Marks
Guidance
Answer: \(2\cos 2y \, dy/dx = 1\)Marks: M1
Guidance: \(k\cos 2y \, dy/dx = 1\) or \(dy/dy = k\cos 2y\), \(k\cos 2y \, dy = dx\)
Answer: \(\Rightarrow dy/dx = 1/(2\cos 2y)\)Marks: A1
Guidance: (none)
Answer: when \(x = 1\frac{1}{2}\), \(y = \pi/12\), \(dy/dx = 1/(2\cos(\pi/6)) = 1/\sqrt{3}\)Marks: M1*
Guidance: substituting \(y = \pi/12\) *dep 1st M1 or \(\sqrt{3}/3\)
Total Marks: [4]
Question 6(ii)
Answer Marks
Guidance
Answer: \(2y = \arcsin(x - 1)\)Marks: M1
Guidance: (none)
Answer: \(\Rightarrow y = \frac{1}{2}\arcsin(x - 1)\)Marks: A1
Guidance: or \(\frac{1}{2}\sin^{-1}(x-1)\)
Answer: translation of 1 unit in positive \(x\)-directionMarks: B1
Guidance: allow 'shift', but not 'move', vector only is B0
Answer: [one-way] stretch s.f. \(\frac{1}{2}\) in \(y\)-directionMarks: B1
Guidance: not 'shrink', 'squash' etc; transformations can be in either order
Total Marks: [4]
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**Answer:** Sketch of $y = |2x + 1|$ | **Marks:** M1 | **Guidance:** Condone no intercept labels, but must be a 'V' shape with vertex on −ve x axis
**Answer:** $y = −x$ and two intersections indicated | **Marks:** A1 | **Guidance:** (none)
**Answer:** $x = -1$ (not from ww, condone $(-1, 1)$) | **Marks:** B1 | **Guidance:** squaring: $(2x+1)^2 = x^2 \Rightarrow 3x^2 + 4x + 1 = 0 \Rightarrow (3x+1)(x+1) = 0$, $x = -1, -1/3$
**Answer:** $x = -1/3$ (not from ww, condone $(-1/3, 1/3)$) | **Marks:** B1 | **Guidance:** (none)
**Total Marks:** [4]
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## Question 5(i)
**Answer:** $dV/dh = 4 \frac{1}{2}(h^3 + 1)^{-1/2} \cdot 3h^2$ | **Marks:** M1 | **Guidance:** their deriv of $4u^{1/2}$ × their deriv of $h^3+1$
**Answer:** (correct) | **Marks:** A1 | **Guidance:** (none)
**Answer:** Substituting $h = 2$ into their derivative | **Marks:** M1 | **Guidance:** (none)
**Answer:** when $h = 2$, $dV/dh = 8$ | **Marks:** A1cao | **Guidance:** (none)
**Total Marks:** [4]
---
## Question 5(ii)
**Answer:** $dV/dt = 0.4$ | **Marks:** B1 | **Guidance:** soi
**Answer:** $dV/dt = dV/dh \times dh/dt$ | **Marks:** M1 | **Guidance:** o.e. Any correct chain rule in $V$, $h$, $t$ (or $r$)
**Answer:** $0.4 = 8 \times dh/dt \Rightarrow dh/dt = 0.05$ (m per min) | **Marks:** A1cao | **Guidance:** 0.05 or 1/20
**Total Marks:** [3]
---
## Question 6(i)
**Answer:** $2\cos 2y \, dy/dx = 1$ | **Marks:** M1 | **Guidance:** $k\cos 2y \, dy/dx = 1$ or $dy/dy = k\cos 2y$, $k\cos 2y \, dy = dx$
**Answer:** $\Rightarrow dy/dx = 1/(2\cos 2y)$ | **Marks:** A1 | **Guidance:** (none)
**Answer:** when $x = 1\frac{1}{2}$, $y = \pi/12$, $dy/dx = 1/(2\cos(\pi/6)) = 1/\sqrt{3}$ | **Marks:** M1* | **Guidance:** substituting $y = \pi/12$ *dep 1st M1 or $\sqrt{3}/3$
**Total Marks:** [4]
---
## Question 6(ii)
**Answer:** $2y = \arcsin(x - 1)$ | **Marks:** M1 | **Guidance:** (none)
**Answer:** $\Rightarrow y = \frac{1}{2}\arcsin(x - 1)$ | **Marks:** A1 | **Guidance:** or $\frac{1}{2}\sin^{-1}(x-1)$
**Answer:** translation of 1 unit in positive $x$-direction | **Marks:** B1 | **Guidance:** allow 'shift', but not 'move', vector only is B0
**Answer:** [one-way] stretch s.f. $\frac{1}{2}$ in $y$-direction | **Marks:** B1 | **Guidance:** not 'shrink', 'squash' etc; transformations can be in either order
**Total Marks:** [4]
---
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By sketching the graphs of $y = |2x + 1|$ and $y = -x$ on the same axes, show that the equation $|2x + 1| = -x$ has two roots. Find these roots. [4]
\hfill \mbox{\textit{OCR MEI C3 2016 Q4 [4]}}