Moderate -0.3 Part (a) is a standard sketch of arcsin with given domain/range - pure recall. Part (b) requires rearranging to isolate g(x+1), then applying sin to both sides and solving a simple linear equation. This is a straightforward application of inverse function properties with minimal problem-solving, making it slightly easier than average but not trivial due to the algebraic manipulation required.
7. (a) For \(- \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }\), sketch the graph of \(y = \mathrm { g } ( x )\) where
$$g ( x ) = \arcsin x \quad - 1 \leqslant x \leqslant 1$$
(b) Find the exact value of \(x\) for which
$$3 g ( x + 1 ) + \pi = 0$$
Either one of the two sections with correct curvature passing through \((0,0)\), or both sections condoning dubious curvature passing through \((0,0)\) (but do not accept any negative gradients), or a curve with a different range or "extended range"
Correct position and curvature
A1
A curve only in quadrants one and three passing through \((0,0)\) with a gradient that is always positive. The gradient should appear to be approx \(\infty\) at each end. If range and domain are given then ignore.
Substitutes \(g(x+1) = \arcsin(x+1)\) in \(3g(x+1)+\pi=0\) and attempts to make \(\arcsin(x+1)\) the subject. Accept \(\arcsin(x+1) = \pm\dfrac{\pi}{3}\) or \(g(x+1) = \pm\dfrac{\pi}{3}\). Condone \(\dfrac{\pi}{3}\) in decimal form awrt 1.047
Proceeds by evaluating \(\sin\!\left(\pm\dfrac{\pi}{3}\right)\) and making \(x\) the subject. Accept \(x = \pm\dfrac{\sqrt{3}}{2}\pm 1\). Accept decimal such as \(-1.866\). Do not allow if candidate works in mixed modes (radians and degrees)
\(\Rightarrow x = -1 - \dfrac{\sqrt{3}}{2}\)
A1
\(-1-\dfrac{\sqrt{3}}{2}\) oe with no other solutions. Be careful: \(-\dfrac{2-\sqrt{3}}{2}\) and \(\dfrac{-2+\sqrt{3}}{2}\) are incorrect but \(-\dfrac{2+\sqrt{3}}{2}\) is correct
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct position **or** curvature | M1 | Either one of the two sections with correct curvature passing through $(0,0)$, or both sections condoning dubious curvature passing through $(0,0)$ (but do not accept any negative gradients), or a curve with a different range or "extended range" |
| Correct position **and** curvature | A1 | A curve only in quadrants one and three passing through $(0,0)$ with a gradient that is always positive. The gradient should appear to be approx $\infty$ at each end. If range and domain are given then ignore. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\arcsin(x+1) + \pi = 0 \Rightarrow \arcsin(x+1) = -\dfrac{\pi}{3}$ | M1 | Substitutes $g(x+1) = \arcsin(x+1)$ in $3g(x+1)+\pi=0$ and attempts to make $\arcsin(x+1)$ the subject. Accept $\arcsin(x+1) = \pm\dfrac{\pi}{3}$ or $g(x+1) = \pm\dfrac{\pi}{3}$. Condone $\dfrac{\pi}{3}$ in decimal form awrt 1.047 |
| $\Rightarrow (x+1) = \sin\!\left(-\dfrac{\pi}{3}\right)$ | dM1 | Proceeds by evaluating $\sin\!\left(\pm\dfrac{\pi}{3}\right)$ and making $x$ the subject. Accept $x = \pm\dfrac{\sqrt{3}}{2}\pm 1$. Accept decimal such as $-1.866$. Do not allow if candidate works in mixed modes (radians and degrees) |
| $\Rightarrow x = -1 - \dfrac{\sqrt{3}}{2}$ | A1 | $-1-\dfrac{\sqrt{3}}{2}$ oe with no other solutions. Be careful: $-\dfrac{2-\sqrt{3}}{2}$ and $\dfrac{-2+\sqrt{3}}{2}$ are incorrect but $-\dfrac{2+\sqrt{3}}{2}$ is correct |
---
7. (a) For $- \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }$, sketch the graph of $y = \mathrm { g } ( x )$ where
$$g ( x ) = \arcsin x \quad - 1 \leqslant x \leqslant 1$$
(b) Find the exact value of $x$ for which
$$3 g ( x + 1 ) + \pi = 0$$
\hfill \mbox{\textit{Edexcel C3 2016 Q7 [5]}}