Inverse function graphs and properties - Reciprocal Trig & Identities

Edexcel P3 2021 October Q8

7 marks Moderate -0.3

8. A curve $C$ has equation $y = \mathrm { f } ( x )$, where $$f ( x ) = \arcsin \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2 \quad - \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }$$

Sketch $C$.
Given $x = 2 \sin y$, show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { A - x ^ { 2 } } }$$ where $A$ is a constant to be found. The point $P$ lies on $C$ and has $y$ coordinate $\frac { \pi } { 4 }$
Find the equation of the tangent to $C$ at $P$. Write your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
(3)

OCR C3 Specimen Q9

11 marks Standard +0.3

9 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_424_707_1260_724} The diagram shows the curve $y = \tan ^ { - 1 } x$ and its asymptotes $y = \pm a$.

State the exact value of $a$.
Find the value of $x$ for which $\tan ^ { - 1 } x = \frac { 1 } { 2 } a$. The equation of another curve is $y = 2 \tan ^ { - 1 } ( x - 1 )$.
Sketch this curve on a copy of the diagram, and state the equations of its asymptotes in terms of $a$.
Verify by calculation that the value of $x$ at the point of intersection of the two curves is 1.54 , correct to 2 decimal places. Another curve (which you are not asked to sketch) has equation $y = \left( \tan ^ { - 1 } x \right) ^ { 2 }$.
Use Simpson's rule, with 4 strips, to find an approximate value for $\int _ { 0 } ^ { 1 } \left( \tan ^ { - 1 } x \right) ^ { 2 } \mathrm {~d} x$.

OCR MEI C3 2007 June Q6

8 marks Moderate -0.3

6 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{0ee3d87a-0d9e-4fa5-b8f5-8b28489e65b5-3_378_725_367_669} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Find the range of the function $\mathrm { f } ( x )$, giving your answer in terms of $\pi$.
Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$. Find the gradient of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ at the origin.
Hence write down the gradient of $y = \frac { 1 } { 2 } \arctan x$ at the origin.

OCR MEI C3 Q4

8 marks Standard +0.3

4 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1$.
Fig. 6 also shows the curve $y = \mathrm { g } ( x )$, where $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.
P is the point on the curve $y = \mathrm { f } ( x )$ with $x$-coordinate $\frac { 1 } { 2 }$.

[diagram]

Find the $y$-coordinate of P , giving your answer in terms of $\pi$. The point Q is the reflection of P in $y = x$.
Find $\mathrm { g } ( x )$ and its derivative $\mathrm { g } ^ { \prime } ( x )$. Hence determine the exact gradient of the curve $y = \mathrm { g } ( x )$ at the point Q . Write down the exact gradient of $y = \mathrm { f } ( x )$ at the point P .

OCR MEI C3 Q7

3 marks Moderate -0.8

7 Sketch the curve $y = 2 \arccos x$ for $- 1 \leqslant x \leqslant 1$.

OCR MEI C3 Q8

8 marks Standard +0.3

8 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{11877196-83d9-4283-9eef-e617bea50c63-4_379_722_467_715} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Find the range of the function $\mathrm { f } ( x )$, giving your answer in terms of $\pi$.
Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$. Find the gradient of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ at the origin.
Hence write down the gradient of $y = \frac { 1 } { 2 } \arctan x$ at the origin.

OCR FP2 2008 January Q2

5 marks Standard +0.3

2 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-2_577_700_577_721} The diagram shows parts of the curves with equations $y = \cos ^ { - 1 } x$ and $y = \frac { 1 } { 2 } \sin ^ { - 1 } x$, and their point of intersection $P$.

Verify that the coordinates of $P$ are $\left( \frac { 1 } { 2 } \sqrt { 3 } , \frac { 1 } { 6 } \pi \right)$.
Find the gradient of each curve at $P$.

OCR MEI C3 2009 June Q3

3 marks Moderate -0.8

3 Sketch the curve $y = 2 \arccos x$ for $- 1 \leqslant x \leqslant 1$.

OCR MEI C3 2012 June Q6

8 marks Standard +0.3

6 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 2 \arcsin x , - 1 \leqslant x \leqslant 1$.
Fig. 6 also shows the curve $y = \mathrm { g } ( x )$, where $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.
P is the point on the curve $y = \mathrm { f } ( x )$ with $x$-coordinate $\frac { 1 } { 2 }$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{7b77c646-2bc5-4166-b22e-3c1229abd722-3_711_693_466_685} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Find the $y$-coordinate of P , giving your answer in terms of $\pi$. The point Q is the reflection of P in $y = x$.
Find $\mathrm { g } ( x )$ and its derivative $\mathrm { g } ^ { \prime } ( x )$. Hence determine the exact gradient of the curve $y = \mathrm { g } ( x )$ at the point Q . Write down the exact gradient of $y = \mathrm { f } ( x )$ at the point P .

OCR FP2 2014 June Q4

6 marks Standard +0.8

4 The curves $y = \cos ^ { - 1 } x$ and $y = \tan ^ { - 1 } ( \sqrt { 2 } x )$ intersect at a point $A$.

Verify that the coordinates of $A$ are $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { 4 } \pi \right)$.
Determine whether the tangents to the curves at $A$ are perpendicular.