1.07n Stationary points: find maxima, minima using derivatives

5 \includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-06_526_947_276_591} The diagram shows the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a maximum at the point \(C\).

1. Find the exact gradient of the curve at \(B\).
2. Find the exact coordinates of \(C\).

5 \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-06_526_947_276_591} The diagram shows the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a maximum at the point \(C\).

1. Find the exact gradient of the curve at \(B\).
2. Find the exact coordinates of \(C\).

1 A curve has equation \(\mathrm { y } = 2 \tan \mathrm { x } - 5 \sin \mathrm { x }\) for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).
Find the \(x\)-coordinate of the stationary point of the curve. Give your answer correct to 3 significant figures.

5 A curve has equation \(\mathrm { y } = \frac { 1 + \mathrm { e } ^ { 2 \mathrm { x } } } { 1 + 3 \mathrm { x } }\). The curve has exactly one stationary point \(P\).

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 6 } + \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }\).
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.35 and 0.45 .
3. Use an iterative formula based on the equation in part (a) to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{971a1d8d-a82e-4a3a-b72d-3c147e4f30bb-10_451_647_258_699} The diagram shows the curve with equation \(\mathrm { y } = \sqrt { \sin 2 \mathrm { x } + \sin ^ { 2 } 2 \mathrm { x } }\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\). The shaded region is bounded by the curve and the straight lines \(x = \frac { 1 } { 6 } \pi\) and \(y = 0\).

6 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685} The diagram shows the curve with equation \(y = \frac { \ln ( 2 x + 1 ) } { x + 3 }\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5\).
3. Show by calculation that the \(x\)-coordinate of \(M\) lies between 2.5 and 3.0 .
4. Use an iterative formula based on the equation in part (b) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 \includegraphics[max width=\textwidth, alt={}, center]{9cf008d5-c15f-4491-9e4d-4bd070f896d5-06_446_832_260_653} The diagram shows part of the curve with equation \(y = \frac { 5 x } { 4 x ^ { 3 } + 1 }\). The shaded region is bounded by the curve and the lines \(x = 1 , x = 3\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of the maximum point.
2. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
3. State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.

7 A curve has equation \(\mathrm { e } ^ { 2 x } y - \mathrm { e } ^ { y } = 100\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \mathrm { e } ^ { 2 x } y } { \mathrm { e } ^ { y } - \mathrm { e } ^ { 2 x } }\).
2. Show that the curve has no stationary points.
   It is required to find the \(x\)-coordinate of \(P\), the point on the curve at which the tangent is parallel to the \(y\)-axis.
3. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \ln 10 - \frac { 1 } { 2 } \ln ( 2 x - 1 )$$
4. Use an iterative formula, based on the equation in part (c), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

8 A curve has equation \(y = f ( x )\) where \(f ( x ) = \frac { 4 x ^ { 3 } + 8 x - 4 } { 2 x - 1 }\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the coordinates of each of the stationary points of the curve \(y = \mathrm { f } ( x )\).
2. Divide \(4 x ^ { 3 } + 8 x - 4\) by ( \(2 x - 1\) ), and hence find \(\int \mathrm { f } ( x ) \mathrm { d } x\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

4 \includegraphics[max width=\textwidth, alt={}, center]{c473f577-1e96-4d11-a0d5-cdfa4873c295-06_460_1445_260_349} The diagram shows the curve with equation \(y = \frac { x - 2 } { x ^ { 2 } + 8 }\). The shaded region is bounded by the curve and the lines \(x = 14\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence determine the exact \(x\)-coordinates of the stationary points.
2. Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures.

5 The equation of a curve is \(2 \mathrm { e } ^ { 2 x } y - y ^ { 3 } + 4 = 0\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 \mathrm { e } ^ { 2 x } y } { 3 y ^ { 2 } - 2 \mathrm { e } ^ { 2 x } }\).
2. The curve passes through the point \(( 0,2 )\). Find the equation of the tangent to the curve at this point, giving your answer in the form \(a x + b y + c = 0\).
3. Show that the curve has no stationary points.

4 The curve with equation \(y = x \mathrm { e } ^ { 2 x } + 5 \mathrm { e } ^ { - x }\) has a minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { 1 } { 3 } \ln 5 - \frac { 1 } { 3 } \ln ( 1 + 2 x )\).
2. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(M\) correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures.

3 The curve with equation $$y = 5 x - 2 \tan 2 x$$ has exactly one stationary point in the interval \(0 \leqslant x < \frac { 1 } { 4 } \pi\).
Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{389df578-e7a7-4d19-9416-5e580d107717-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

5 The curve with equation \(y = x \ln ( 4 x + 1 ) - 3 x\) has one stationary point \(P\).

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \frac { 2 x + 0.75 } { \ln ( 4 x + 1 ) } - 0.25$$
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 1.8 and 1.9.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{1cd04df5-3fe3-4573-b880-d49262afd16a-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

5 \includegraphics[max width=\textwidth, alt={}, center]{34177829-f05d-449e-8881-5ab4d852c4ce-3_458_643_285_751} The diagram shows the part of the curve \(y = x \mathrm { e } ^ { - x }\) for \(0 \leqslant x \leqslant 2\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 2 } x \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).

6 The parametric equations of a curve are $$x = 2 t + \ln t , \quad y = t + \frac { 4 } { t }$$ where \(t\) takes all positive values.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } - 4 } { t ( 2 t + 1 ) }\).
2. Find the equation of the tangent to the curve at the point where \(t = 1\).
3. The curve has one stationary point. Find the \(y\)-coordinate of this point, and determine whether this point is a maximum or a minimum.

6 \includegraphics[max width=\textwidth, alt={}, center]{08210e25-0f0e-405b-b72d-1bf989689b0a-3_641_865_264_641} The diagram shows the part of the curve \(y = \frac { \ln x } { x }\) for \(0 < x \leqslant 4\). The curve cuts the \(x\)-axis at \(A\) and its maximum point is \(M\).

1. Write down the coordinates of \(A\).
2. Show that the \(x\)-coordinate of \(M\) is e, and write down the \(y\)-coordinate of \(M\) in terms of e.
3. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 1 } ^ { 4 } \frac { \ln x } { x } \mathrm {~d} x$$ correct to 2 decimal places.
4. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii).

3 The equation of a curve is \(y = x + 2 \cos x\). Find the \(x\)-coordinates of the stationary points of the curve for \(0 \leqslant x \leqslant 2 \pi\), and determine the nature of each of these stationary points.

6 It is given that the curve \(y = ( x - 2 ) \mathrm { e } ^ { x }\) has one stationary point.

1. Find the exact coordinates of this point.
2. Determine whether this point is a maximum or a minimum point.

7 The equation of a curve is $$x ^ { 2 } + y ^ { 2 } - 4 x y + 3 = 0$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y - x } { y - 2 x }\).
2. Find the coordinates of each of the points on the curve where the tangent is parallel to the \(x\)-axis.

7 \includegraphics[max width=\textwidth, alt={}, center]{b9556031-871d-4dd3-9523-e3438a41339f-3_655_685_262_730} The diagram shows the curve \(y = x \mathrm { e } ^ { 2 x }\) and its minimum point \(M\).

1. Find the exact coordinates of \(M\).
2. Show that the curve intersects the line \(y = 20\) at the point whose \(x\)-coordinate is the root of the equation $$x = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x _ { n } } \right)$$ with initial value \(x _ { 1 } = 1.3\), to calculate the root correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

5 The equation of a curve is \(y = x ^ { 3 } \mathrm { e } ^ { - x }\).

1. Show that the curve has a stationary point where \(x = 3\).
2. Find the equation of the tangent to the curve at the point where \(x = 1\).

6 The curve \(y = 4 x ^ { 2 } \ln x\) has one stationary point.

1. Find the coordinates of this stationary point, giving your answers correct to 3 decimal places.
2. Determine whether this point is a maximum or a minimum point.

5 \includegraphics[max width=\textwidth, alt={}, center]{beb8df77-e091-4248-812b-20e885c42e37-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).