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Find the composite functions $\mathrm { fg } ( x )$ and $\mathrm { gf } ( x )$.
Sketch the curves $y = \mathrm { f } ( x ) , y = \mathrm { fg } ( x )$ and $y = \mathrm { gf } ( x )$, indicating clearly which is which.

OCR MEI C3 Q3

3 marks Moderate -0.8

3 Given that $\mathrm { f } ( x ) = 1 - x$ and $\mathrm { g } ( x ) = | x |$, write down the composite function $\mathrm { gf } ( x )$. On separate diagrams, sketch the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { gf } ( x )$.

OCR MEI C3 Q4

7 marks Moderate -0.3

4 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 1 + 2 \sin x$ for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$.

Show that $\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)$ and state the domain of this function. Fig. 6 shows a sketch of the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
Write down the coordinates of the points $\mathrm { A } , \mathrm { B }$ and C .

OCR MEI C3 Q5

3 marks Easy -1.2

5 Given that $\arcsin x = \frac { 1 } { 6 } \pi$, find $x$. Find $\arccos x$ in terms of $\pi$.

OCR MEI C3 Q6

3 marks Moderate -0.8

6 The functions $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ are defined for the domain $x > 0$ as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function $\mathrm { fg } ( x )$ in terms of $\ln x$.
State the transformation which maps the curve $y = \mathrm { f } ( x )$ onto the curve $y = \mathrm { fg } ( x )$.

OCR MEI C3 Q7

19 marks Standard +0.3

7 Fig. 9 shows the line $y = x$ and the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)$. The line and the curve intersect at the origin and at the point $\mathrm { P } ( a , a )$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-3_694_886_430_604} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure}

Show that $\mathrm { e } ^ { a } = 1 + 2 a$.
Show that the area of the region enclosed by the curve, the $x$-axis and the line $x = a$ is $\frac { 1 } { 2 } a$. Hence find, in terms of $a$, the area enclosed by the curve and the line $y = x$.
Show that the inverse function of $\mathrm { f } ( x )$ is $\mathrm { g } ( x )$, where $\mathrm { g } ( x ) = \ln ( 1 + 2 x )$. Add a sketch of $y = \mathrm { g } ( x )$ to the copy of Fig. 9.
Find the derivatives of $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$. Hence verify that $\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }$. Give a geometrical interpretation of this result.

OCR MEI C3 Q8

8 marks Moderate -0.3

8 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 1 - 2 \sin x$ for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$. Fig. 3 shows the curve $y = \mathrm { f } ( x )$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-4_736_809_419_653} \captionsetup{labelformat=empty} \caption{Fig. 3}

\end{figure}

Write down the range of the function $\mathrm { f } ( x )$.
Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$.
Find $\mathrm { f } ^ { \prime } ( 0 )$. Hence write down the gradient of $y = \mathrm { f } ^ { - 1 } ( x )$ at the point $( 1,0 )$.

OCR MEI C3 Q1

18 marks Challenging +1.2

1 Fig. 9 shows the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$. The function $y = \mathrm { f } ( x )$ is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve $y = \mathrm { f } ( x )$ crosses the $x$-axis at P , and the line $x = 2$ at Q . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-1_555_641_573_748} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure}

Verify that the $x$-coordinate of P is 1 . Find the exact $y$-coordinate of Q .
Find the gradient of the curve at P . [Hint: use $\ln \frac { a } { b } = \ln a - \ln b$.] The function $\mathrm { g } ( x )$ is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve $y = \mathrm { g } ( x )$ crosses the $y$-axis at the point R .
Show that $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$. Write down the gradient of $y = \mathrm { g } ( x )$ at R.
Show, using the substitution $u = 2 - \mathrm { e } ^ { x }$ or otherwise, that $\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }$. Using this result, show that the exact area of the shaded region shown in Fig. 9 is $\ln \frac { 32 } { 27 }$.
[0pt] [Hint: consider its reflection in $y = x$.]

OCR MEI C3 Q2

18 marks Standard +0.3

2 Fig. 8 shows the line $y = x$ and parts of the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$, where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B, as shown. The curves and the line $y = x$ meet at the point C . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-2_811_893_609_655} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Find the exact coordinates of A and B . Verify that the coordinates of C are $( 1,1 )$.
Prove algebraically that $\mathrm { g } ( x )$ is the inverse of $\mathrm { f } ( x )$.
Evaluate $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$, giving your answer in terms of e .
Use integration by parts to find $\int \ln x \mathrm {~d} x$. Hence show that $\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }$.
Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .

OCR MEI C3 Q3

17 marks Moderate -0.3

3 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 1 + \sin 2 x$ for $- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-3_577_815_392_719} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

State a sequence of two transformations that would map part of the curve $y = \sin x$ onto the curve $y = \mathrm { f } ( x )$.
Find the area of the region enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis and the line $x = \frac { 1 } { 4 } \pi$.
Find the gradient of the curve $y = \mathrm { f } ( x )$ at the point ( 0,1 ). Hence write down the gradient of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ at the point $( 1,0 )$.
State the domain of $\mathrm { f } ^ { - 1 } ( x )$. Add a sketch of $y = \mathrm { f } ^ { - 1 } ( x )$ to a copy of Fig. 8.
Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.

OCR MEI C3 Q4

18 marks Standard +0.8

4 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }$, for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$.
P is the point on the curve with $x$-coordinate $\frac { 1 } { 3 } \pi$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-4_820_815_551_715} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Find the $y$-coordinate of P .
Find $\mathrm { f } ^ { \prime } ( x )$. Hence find the gradient of the curve at the point P .
Show that the derivative of $\frac { \sin x } { 1 + \cos x }$ is $\frac { 1 } { 1 + \cos x }$. Hence find the exact area of the region enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = \frac { 1 } { 3 } \pi$.
Show that $\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)$. State the domain of this inverse function, and add a sketch of $y = \mathrm { f } ^ { - 1 } ( x )$ to a copy of Fig. 8.