Questions - OCR MEI - A-Level Maths

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OCR MEI C3 Q7

6 marks Moderate -0.5

7 An oil slick is circular with radius $r \mathrm {~km}$ and area $A \mathrm {~km} ^ { 2 }$. The radius increases with time at a rate given by $\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.5$, in kilometres per hour.

Show that $\frac { \mathrm { dA } } { \mathrm { d } t } = \pi r$.
Find the rate of increase of the area of the slick at a time when the radius is 6 km .

OCR MEI C3 Q8

18 marks Standard +0.3

8 Fig. 8 shows the graph of $y = x \sqrt { 1 + x }$. The point P on the curve is on the $x$-axis. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c7998a08-229a-40d2-ba34-b5f264139295-3_433_800_895_587} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Write down the coordinates of P .
Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x + 2 } { 2 \sqrt { 1 + x } }$.
Hence find the coordinates of the turning point on the curve. What can you say about the gradient of the curve at P ?
By using a suitable substitution, show that $\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$. Evaluate this integral, giving your answer in an exact form.
What does this value represent?
Use your answer to part (ii) to differentiate $y = x \sqrt { 1 + x } \sin 2 x$ with respect to $x$.
(You need not simplify your result.)

OCR MEI C3 Q9

18 marks Moderate -0.8

9 The functions $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ are defined by $$\mathrm { f } ( x ) = x ^ { 2 } , \quad \mathrm {~g} ( x ) = 2 x - 1$$ for all real values of $x$.

State the ranges of $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$. Explain why $\mathrm { f } ( x )$ has no inverse.
Find an expression for the inverse function $\mathrm { g } ^ { - 1 } ( x )$ in terms of $x$. Sketch the graphs of $y = \mathrm { g } ( x )$ and $y = \mathrm { g } ^ { - 1 } ( x )$ on the same axes.
Find expressions for $\operatorname { gf } ( x )$ and $\operatorname { fg } ( x )$.
Solve the equation $\operatorname { gf } ( x ) = \mathrm { fg } ( x )$. Sketch the graphs of $y = \operatorname { gf } ( x )$ and $y = \operatorname { fg } ( x )$ on the same axes to illustrate your answer.
Show that the equation $\mathrm { f } ( x + a ) = \mathrm { g } ^ { 2 } ( x )$ has no solution if $a > \frac { 1 } { 4 }$.

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.]

OCR MEI C3 Q3

4 marks Moderate -0.3

3 Given that $y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }$.

OCR MEI C3 Q4

8 marks Standard +0.8

4 Fig. 7 shows the curve $x ^ { 3 } + y ^ { 3 } = 3 x y$. The point P is a turning point of the curve. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{09d318c7-27b9-43aa-b4a0-e32ea8bd53c5-1_593_531_1573_805} \captionsetup{labelformat=empty} \caption{Fig. 7}

\end{figure}

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }$.
Hence find the exact $x$-coordinate of P .

Verify that the point $\mathrm { P } \left( \frac { 1 } { 6 } \pi , \frac { 1 } { 6 } \pi \right)$ lies on the curve.
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. Hence find the gradient of the curve at the point P .

OCR MEI C3 Q8

7 marks Moderate -0.3

Given that $y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }$, use the chain rule to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.
Given that $y ^ { 3 } = 1 + 3 x ^ { 2 }$, use implicit differentiation to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. Show that this result is equivalent to the result in part (i).

OCR MEI C3 Q1

7 marks Moderate -0.3

Differentiate $\sqrt { 1 + 3 x ^ { 2 } }$.
Hence show that the derivative of $x \sqrt { 1 + 3 x ^ { 2 } }$ is $\frac { 1 + 6 x ^ { 2 } } { \sqrt { 1 + 3 x ^ { 2 } } }$.

OCR MEI C3 Q2

7 marks Standard +0.3

2 Given that $y ^ { 3 } = x y - x ^ { 2 }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }$.
Hence show that the curve $y ^ { 3 } = x y - x ^ { 2 }$ has a stationary point when $x = \frac { 1 } { 8 }$.

Questions — OCR MEI (4301 questions)

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OCR MEI C3 Q7

OCR MEI C3 Q8

OCR MEI C3 Q9

OCR MEI C3 Q1

OCR MEI C3 Q2

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OCR MEI C3 Q4

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