OCR MEI C3 (Core Mathematics 3)

Question 1

1 Fig. 8 shows part of the curve $y = \mathrm { f } ( x )$, where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-1_705_864_525_635} \captionsetup{labelformat=empty} \caption{Fig. 8}

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Find $\mathrm { f } ^ { \prime } ( x )$, and hence calculate the gradient of the curve $y = \mathrm { f } ( x )$ at the origin and at the point $( \ln 2,1 )$. The function $\mathrm { g } ( x )$ is defined by $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
Show that $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ are inverse functions. Hence sketch the graph of $y = \mathrm { g } ( x )$. Write down the gradient of the curve $y = \mathrm { g } ( x )$ at the point $( 1 , \ln 2 )$.
Show that $\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c$. Hence evaluate $\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x$, giving your answer in an exact form.
Using your answer to part (iii), calculate the area of the region enclosed by the curve $y = \mathrm { g } ( x )$, the $x$-axis and the line $x = 1$.

Question 2

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2 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]{6555136d-0444-41f6-9063-21960352089d-2_388_727_434_701} \captionsetup{labelformat=empty} \caption{Fig. 6}

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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.

Question 3

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3 The function $f ( x ) = \ln \left( 1 + x ^ { 2 } \right)$ has domain $- 3 \leqslant x \leqslant 3$.
Fig. 9 shows the graph of $y = f ( x )$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-3_495_867_519_607} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure} (1) Show algebraically that the function is even. State how this property relates to the shape of the curve.
(ii) Find the gradient of the curve at the point $\mathrm { P } ( 2 , \ln 5 )$.
(iii) Explain why the function does not have an inverse for the domain $- 3 \leqslant x \leqslant 3$. The domain of $f ( x )$ is now restricted to $0 \leqslant x \leqslant 3$. The inverse of $f ( x )$ is the function $g ( x )$,
(iv) Sketch the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$ on the same axes. State the domain of the function $g ( x )$.
Show that $\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }$.
(v) Differentiate $\mathrm { g } ( x )$. Hence verify that $\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }$. Explain the commection between this result and your answer to part (ii).