1.07n Stationary points: find maxima, minima using derivatives

7 The curve \(C\) has equation \(y = f ( x )\), where \(f ( x ) = \frac { x ^ { 2 } + 2 } { x ^ { 2 } - x - 2 }\).

1. Find the equations of the asymptotes of \(C\).
2. Find the coordinates of any stationary points on \(C\), giving your answers correct to 1 decimal place.
3. Sketch \(C\), stating the coordinates of any intersections with the axes.
4. Sketch the curve with equation \(\mathrm { y } = \frac { 1 } { \mathrm { f } ( \mathrm { x } ) }\).
5. Find the set of values for which \(\frac { 1 } { \mathrm { f } ( x ) } < \mathrm { f } ( x )\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

7 The curve \(C\) has equation \(y = f ( x )\), where \(f ( x ) = \frac { x ^ { 2 } } { x + 1 }\).

1. Find the equations of the asymptotes of \(C\).
2. Find the coordinates of any stationary points on \(C\).
3. Sketch \(C\).
4. Find the coordinates of any stationary points on the curve with equation \(\mathrm { y } = \frac { 1 } { \mathrm { f } ( \mathrm { x } ) }\).
5. Sketch the curve with equation \(y = \frac { 1 } { f ( x ) }\) and find, in exact form, the set of values for which $$\frac { 1 } { \mathrm { f } ( x ) } > \mathrm { f } ( x ) .$$ If you use the following page to complete the answer to any question, the question number must be clearly shown.

6 The curve \(C\) has equation \(y = \frac { x ^ { 2 } + 3 } { x ^ { 2 } + 1 }\).

1. Show that \(C\) has no vertical asymptotes and state the equation of the horizontal asymptote.
2. Show that \(1 < y \leqslant 3\) for all real values of \(x\).
3. Find the coordinates of any stationary points on \(C\). \includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-12_2718_42_107_2007} \includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-13_2720_40_106_18}
4. Sketch \(C\), stating the coordinates of any intersections with the axes and labelling the asymptote.
5. Sketch the curve with equation \(y = \frac { x ^ { 2 } + 1 } { x ^ { 2 } + 3 }\) and find the set of values of \(x\) for which \(\frac { x ^ { 2 } + 1 } { x ^ { 2 } + 3 } < \frac { 1 } { 2 }\).

7 The curve \(C\) has equation \(y = \frac { 2 x ^ { 2 } - 3 x - 2 } { x ^ { 2 } - 2 x + 1 }\).

1. State the equations of the asymptotes of \(C\).
2. Show that \(y \leqslant \frac { 25 } { 12 }\) at all points on \(C\).
3. Find the coordinates of any stationary points of \(C\).
4. Sketch \(C\), stating the coordinates of any intersections of \(C\) with the coordinate axes and the asymptotes.
5. Sketch the curve with equation \(y = \left| \frac { 2 x ^ { 2 } - 3 x - 2 } { x ^ { 2 } - 2 x + 1 } \right|\) and find the set of values of \(x\) for which \(\left| \frac { 2 x ^ { 2 } - 3 x - 2 } { x ^ { 2 } - 2 x + 1 } \right| < 2\).
   

3 Find the exact coordinates of the stationary point of the curve with equation \(y = \frac { 3 x } { \ln x }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{d53a2d6b-4c5e-4bc6-8aa1-587e97c87920-2_371_839_1409_651} The diagram shows the part of the curve \(y = 3 \mathrm { e } ^ { - x } \sin 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and the stationary point \(M\).

1. Find the equation of the tangent to the curve at the origin.
2. Find the coordinates of \(M\), giving each coordinate correct to 3 decimal places.

7 The equation of a curve is \(2 x ^ { 3 } + y ^ { 3 } = 24\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and show that the gradient of the curve is never positive.
2. Find the coordinates of the two points on the curve at which the gradient is - 2 .

7 The equation of a curve is $$2 x ^ { 2 } + 3 y ^ { 2 } - 2 x y = 10 .$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y - x }\).
2. Find the coordinates of the points on the curve where the tangent is parallel to the \(x\)-axis.

5 The curve with equation \(y = x ^ { 2 } \ln x\), where \(x > 0\), has one stationary point.

1. Find the \(x\)-coordinate of this point, giving your answer in terms of e .
2. Determine whether this point is a maximum or a minimum point.

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { - x }\). The point \(( 0,1 )\) lies on the curve.

1. Find the equation of the curve.
2. The curve has one stationary point. Find the \(x\)-coordinate of this point and determine whether it is a maximum or a minimum point.

6 \includegraphics[max width=\textwidth, alt={}, center]{4029c46c-50a1-4d23-bc29-589417a6b7f5-3_501_497_269_826} The diagram shows the part of the curve \(y = \frac { \mathrm { e } ^ { 2 x } } { x }\) for \(x > 0\), and its minimum point \(M\).

1. Find the coordinates of \(M\).
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 1 } ^ { 2 } \frac { \mathrm { e } ^ { 2 x } } { x } \mathrm {~d} x$$ giving your answer correct to 1 decimal place.
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).

1. Given that \(y = \tan 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence, or otherwise, show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sec ^ { 2 } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \sqrt { } 3$$ and, by using an appropriate trigonometrical identity, find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \tan ^ { 2 } 2 x \mathrm {~d} x\).
3. Use the identity \(\cos 4 x \equiv 2 \cos ^ { 2 } 2 x - 1\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \frac { 1 } { 1 + \cos 4 x } \mathrm {~d} x$$

4 The equation of a curve is \(y = 2 x - \tan x\), where \(x\) is in radians. Find the coordinates of the stationary points of the curve for which \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).

8 \includegraphics[max width=\textwidth, alt={}, center]{8f815127-61b2-4a7f-8687-747950ea6597-3_693_1061_262_541} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { - x }\) and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Show that the tangent to the curve at the point where \(x = 1\) passes through the origin.
3. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { 1 } ^ { 3 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.

6 The curve with equation \(y = x \ln x\) has one stationary point.

1. Find the exact coordinates of this point, giving your answers in terms of e .
2. Determine whether this point is a maximum or a minimum point.

7 \includegraphics[max width=\textwidth, alt={}, center]{e814d76c-8757-4cc4-a69c-e3636b4cab16-3_611_1084_648_532} The diagram shows the curve \(y = \frac { \ln x } { x ^ { 2 } }\) and its maximum point \(M\).

1. Find the exact coordinates of \(M\).
2. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 1 } ^ { 4 } \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.

3 \includegraphics[max width=\textwidth, alt={}, center]{322eb555-d40a-460c-8c71-5780f5772bcd-2_535_1041_573_552} The diagram shows the curve \(y = x - 2 \ln x\) and its minimum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 2 } ^ { 5 } ( x - 2 \ln x ) \mathrm { d } x$$ 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).

2 The curve with equation \(y = \frac { \sin 2 x } { \mathrm { e } ^ { 2 x } }\) has one stationary point in the interval \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Find the exact \(x\)-coordinate of this point.

3 The equation of a curve is \(y = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 5 \mathrm { e } ^ { x } + 4 x\). Find the exact \(x\)-coordinate of each of the stationary points of the curve and determine the nature of each stationary point.

2 The curve \(y = \frac { \mathrm { e } ^ { 3 x - 1 } } { 2 x }\) has one stationary point. Find the coordinates of this stationary point.

5 \includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-2_554_689_1354_726} The diagram shows part of the curve $$y = 2 \cos x - \cos 2 x$$ and its maximum point \(M\). The shaded region is bounded by the curve, the axes and the line through \(M\) parallel to the \(y\)-axis.

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. Find the exact value of the area of the shaded region.

2 A curve has equation $$y = \frac { 3 x + 1 } { x - 5 }$$ Find the coordinates of the points on the curve at which the gradient is - 4 .

4 The curve with equation \(y = \mathrm { e } ^ { 2 x } ( \sin x + 3 \cos x )\) has a stationary point in the interval \(0 \leqslant x \leqslant \pi\).

1. Find the \(x\)-coordinate of this point, giving your answer correct to 2 decimal places.
2. Determine whether the stationary point is a maximum or a minimum.

6 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-08_318_750_260_699} The diagram shows the curve \(y = \frac { x } { 1 + 3 x ^ { 4 } }\), for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
2. Using the substitution \(u = \sqrt { 3 } x ^ { 2 }\), find by integration the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 1\).

7 \includegraphics[max width=\textwidth, alt={}, center]{a257f49d-5c8f-4c23-be78-46619b746fde-10_353_689_262_726} The diagram shows the curve \(y = \frac { \tan ^ { - 1 } x } { \sqrt { x } }\) and its maximum point \(M\) where \(x = a\).

1. Show that \(a\) satisfies the equation $$a = \tan \left( \frac { 2 a } { 1 + a ^ { 2 } } \right)$$
2. Verify by calculation that \(a\) lies between 1.3 and 1.5.
3. Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

9 The equation of a curve is \(y = x ^ { - \frac { 2 } { 3 } } \ln x\) for \(x > 0\). The curve has one stationary point.

1. Find the exact coordinates of the stationary point.
2. Show that \(\int _ { 1 } ^ { 8 } y \mathrm {~d} x = 18 \ln 2 - 9\).