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OCR C3 Q1

6 marks Moderate -0.3

Differentiate $x ^ { 3 } \ln x$ with respect to $x$.
Given that $$x = \frac { y + 1 } { 3 - 2 y }$$ find and simplify an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$.

OCR C3 Q2

7 marks Moderate -0.5

2. \includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-1_638_677_749_443} The diagram shows the curves $y = 3 + 2 \mathrm { e } ^ { x }$ and $y = \mathrm { e } ^ { x + 2 }$ which cross the $y$-axis at the points $A$ and $B$ respectively.

Write down the coordinates of $A$ and $B$. The two curves intersect at the point $C$.
Find an expression for the $x$-coordinate of $C$ and show that the $y$-coordinate of $C$ is $\frac { 3 \mathrm { e } ^ { 2 } } { \mathrm { e } ^ { 2 } - 2 }$.

OCR C3 Q3

8 marks Moderate -0.3

3. The functions f and g are defined by $$\begin{aligned} & \mathrm { f } ( x ) \equiv 6 x - 1 , \quad x \in \mathbb { R } , \\ & \mathrm {~g} ( x ) \equiv \log _ { 2 } ( 3 x + 1 ) , \quad x \in \mathbb { R } , \quad x > - \frac { 1 } { 3 } . \end{aligned}$$

Evaluate $\mathrm { gf } ( 1 )$.
Find an expression for $\mathrm { g } ^ { - 1 } ( x )$.
Find, in terms of natural logarithms, the solution of the equation $$\mathrm { fg } ^ { - 1 } ( x ) = 2$$

OCR C3 Q4

9 marks Moderate -0.3

Use the identity for $\cos ( A + B )$ to prove that $$\cos 2 x \equiv 2 \cos ^ { 2 } x - 1$$
Prove that, for $\cos x \neq 0$, $$2 \cos x - \sec x \equiv \sec x \cos 2 x$$
Hence, or otherwise, find the values of $x$ in the interval $0 \leq x \leq 180 ^ { \circ }$ for which $$2 \cos x - \sec x \equiv 2 \cos 2 x$$

OCR C3 Q5

9 marks Standard +0.8

Show that the equation $$2 \sin x + \sec \left( x + \frac { \pi } { 6 } \right) = 0$$ can be written as $$\sqrt { 3 } \sin x \cos x + \cos ^ { 2 } x = 0$$
Hence, or otherwise, find in terms of $\pi$ the solutions of the equation $$2 \sin x + \sec \left( x + \frac { \pi } { 6 } \right) = 0$$ for $x$ in the interval $0 \leq x \leq \pi$.

OCR C3 Q6

10 marks Standard +0.8

6. \includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-2_444_825_1571_516} The diagram shows the curve with equation $y = \sqrt { \frac { x } { x + 1 } }$.
The shaded region is bounded by the curve, the $x$-axis and the line $x = 3$.

Use Simpson's rule with six strips to estimate the area of the shaded region. The shaded region is rotated through four right angles about the $x$-axis.
Show that the volume of the solid formed is $\pi ( 3 - \ln 4 )$.

OCR C3 Q7

11 marks Standard +0.8

Sketch on the same diagram the graphs of $y = 4 a ^ { 2 } - x ^ { 2 }$ and $y = | 2 x - a |$, where $a$ is a positive constant. Show, in terms of $a$, the coordinates of any points where each graph meets the coordinate axes.
Find the exact solutions of the equation $$4 - x ^ { 2 } = | 2 x - 1 |$$

OCR C3 Q8

12 marks Standard +0.3

A curve has the equation $y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , x \neq 0$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
[0pt]
Show that the curve has a stationary point in the interval [1.3,1.4]. The point $A$ on the curve has $x$-coordinate 2 .
Show that the tangent to the curve at $A$ passes through the origin. The tangent to the curve at $A$ intersects the curve again at the point $B$.
The $x$-coordinate of $B$ is to be estimated using the iterative formula $$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$ with $x _ { 0 } = - 1$.
Find $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 7 significant figures and hence state the $x$-coordinate of $B$ to 5 significant figures.

OCR MEI C3 Q1

4 marks Moderate -0.5

1 Find $\int \sqrt [ 3 ] { 2 x - 1 } \mathrm {~d} x$.

OCR MEI C3 Q2

18 marks Standard +0.3

2 Fig. 8 shows the line $y = 1$ and the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }$. The curve touches the $x$-axis at $\mathrm { P } ( 2,0 )$ and has another turning point at the point Q . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{6ea594c5-52ba-4467-a098-cb66004b5a38-1_959_1469_748_317} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Show that $\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }$, and find $\mathrm { f } ^ { \prime \prime } ( x )$. Hence find the coordinates of Q and, using $\mathrm { f } ^ { \prime \prime } ( x )$, verify that it is a maximum point.
Verify that the line $y = 1$ meets the curve $y = \mathrm { f } ( x )$ at the points with $x$-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve $y = \mathrm { f } ( x )$ is now transformed by a translation with vector $\binom { - 1 } { - 1 }$. The resulting curve has equation $y = \mathrm { g } ( x )$.
Show that $\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }$.
Without further calculation, write down the value of $\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x$, justifying your answer.

OCR MEI C3 Q3

3 marks Moderate -0.8

3 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x$, giving your answer in exact form.

OCR MEI C3 Q4

18 marks Challenging +1.2

4 Fig. 9 shows the curve $y = x \mathrm { e } ^ { - 2 x }$ together with the straight line $y = m x$, where $m$ is a constant, with $0 < m < 1$. The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{6ea594c5-52ba-4467-a098-cb66004b5a38-2_431_977_728_602} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure}

Show that the $x$-coordinate of P is $- \frac { 1 } { 2 } \ln m$.
Find, in terms of $m$, the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the $x$-axis.
Show that $m = \mathrm { e } ^ { - 2 }$, and find the exact coordinates of P .
Find the exact area of the shaded region between the line OP and the curve.

OCR MEI C3 Q5

5 marks Standard +0.3

5 Using a suitable substitution or otherwise, show that $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2$.

OCR MEI C3 Q1

18 marks Standard +0.3

1 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }$, with its turning point P . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-1_722_817_450_642} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Write down the coordinates of the intercepts of $y = \mathrm { f } ( x )$ with the $x$ - and $y$-axes.
Find the exact coordinates of the turning point P .
Show that the exact area of the region enclosed by the curve and the $x$ - and $y$-axes is $\frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 } - 3 \right)$. The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)$.
Express $\mathrm { g } ( x )$ in terms of $x$. Sketch the curve $y = \mathrm { g } ( x )$ on the copy of Fig. 8, indicating the coordinates of its intercepts with the $x$ - and $y$-axes and of its turning point.
Write down the exact area of the region enclosed by the curve $y = \mathrm { g } ( x )$ and the $x$ - and $y$-axes.

OCR MEI C3 Q2

18 marks Standard +0.8

2 Fig. 9 shows the curve with equation $y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }$. It has an asymptote $x = a$ and turning point P . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-2_754_870_478_609} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure}

Write down the value of $a$.
Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }$. Hence find the coordinates of the turning point P , giving the $y$-coordinate to 3 significant figures.
Show that the substitution $u = 2 x - 1$ transforms $\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x$ to $\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u$. Hence find the exact area of the region enclosed by the curve $y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }$, the $x$-axis and the lines $x = 1$ and $x = 4.5$.

OCR MEI C3 Q3

18 marks Challenging +1.2

3 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]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-3_559_644_622_745} \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 $\ln \frac { 32 } { 27 }$. [Hint: consider its reflection in $y = x$.]

OCR MEI C3 Q1

5 marks Moderate -0.3

1 Show that $\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }$.

OCR MEI C3 Q2

23 marks Standard +0.3

2 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, which has a $y$-intercept at $\mathrm { P } ( 0,3 )$, a minimum point at $\mathrm { Q } ( 1,2 )$, and an asymptote $x = - 1$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f7049002-f97a-4c83-a7d6-eba28e3b589a-1_904_937_785_604} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure}

Find the coordinates of the images of the points P and Q when the curve $y = \mathrm { f } ( x )$ is transformed to
(A) $y = 2 \mathrm { f } ( x )$,
(B) $y = \mathrm { f } ( x + 1 ) + 2$. You are now given that $\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1$.
Find $\mathrm { f } ^ { \prime } ( x )$, and hence find the coordinates of the other turning point on the curve $y = \mathrm { f } ( x )$.
Show that $\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }$.
Find $\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x$ in terms of $a$ and $b$. Hence, by choosing suitable values for $a$ and $b$, find the exact area enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = 1$.

OCR MEI C3 Q3

3 marks Moderate -0.8

3 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x$.
[0pt] [3]

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]{f7049002-f97a-4c83-a7d6-eba28e3b589a-2_824_816_885_699} \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.

OCR MEI C3 Q1

6 marks Moderate -0.3

1 A curve has implicit equation $y ^ { 2 } + 2 x \ln y = x ^ { 2 }$.
Verify that the point $( 1,1 )$ lies on the curve, and find the gradient of the curve at this point.

OCR MEI C3 Q2

6 marks Standard +0.3

2 A curve has equation $x ^ { 2 } + 2 y ^ { 2 } = 4 x$.

By differentiating implicitly, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
[0pt]
Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]