1.07l Derivative of ln(x): and related functions

2 Find the exact gradient of the curve \(y = \ln ( 1 - \cos 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\).

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

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. 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 \(g ( x )\).
4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
   Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

1 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] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. 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 .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R.
4. 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 \(\ln \frac { 32 } { 27 }\).
   [0pt] [Hint: consider its reflection in \(y = x\).]

2

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

3

1. Differentiate \(\frac { \ln x } { x ^ { 2 } }\), simplifying your answer.
2. Using integration by parts, show that \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c\).

2 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

1 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE .

12
0
5 \end{array} \right) + s \left( \begin{array} { r } 1
- 4
- 2 \end{array} \right) .$$

1. Show that the lines intersect.
2. Find the angle between the lines.

\\

1. Show that, if \(y = \operatorname { cosec } x\), then \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed as \(- \operatorname { cosec } x \cot x\).
2. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \sin x \tan x \cot t$$ given that \(x = \frac { 1 } { 6 } \pi\) when \(t = \frac { 1 } { 2 } \pi\).

8\\

1. Given that \(\frac { 2 t } { ( t + 1 ) ^ { 2 } }\) can be expressed in the form \(\frac { A } { t + 1 } + \frac { B } { ( t + 1 ) ^ { 2 } }\), find the values of the constants \(A\) and \(B\).
2. Show that the substitution \(t = \sqrt { 2 x - 1 }\) transforms \(\int \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\) to \(\int \frac { 2 t } { ( t + 1 ) ^ { 2 } } \mathrm {~d} t\).
3. Hence find the exact value of \(\int _ { 1 } ^ { 5 } \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\). 9 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 4 \sin \theta$$ and part of its graph is shown below. \includegraphics[max width=\textwidth, alt={}, center]{b8ba126f-c5fa-4828-9439-e5162a03ca5b-3_646_1150_1050_500}

1. Find the value of \(\theta\) at \(A\) and the value of \(\theta\) at \(B\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec \theta\).
3. At the point \(C\) on the curve, the gradient is 2 . Find the coordinates of \(C\), giving your answer in an exact form.

6.

1. Use the derivative of \(\cos x\) to prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x$$ The curve \(C\) has the equation \(y = \mathrm { e } ^ { 2 x } \sec x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\).
2. Find an equation for the tangent to \(C\) at the point where it crosses the \(y\)-axis.
3. Find, to 2 decimal places, the \(x\)-coordinate of the stationary point of \(C\).

5.

1. Given that $$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } }$$
2. Find an equation for the tangent to the curve \(y = \sqrt { 3 + 2 \cos x }\) at the point where \(x = \frac { \pi } { 3 }\).

1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model.

5 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model.

1 It is given that \(\mathrm { f } ( x ) = \ln ( 3 + x )\).

1. Find the exact values of \(f ( 0 )\) and \(f ^ { \prime } ( 0 )\), and show that \(f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 9 }\).
2. Hence write down the first three terms of the Maclaurin series for \(\mathrm { f } ( x )\), given that \(- 3 < x \leqslant 3\).

2 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-2_577_700_577_721} The diagram shows parts of the curves with equations \(y = \cos ^ { - 1 } x\) and \(y = \frac { 1 } { 2 } \sin ^ { - 1 } x\), and their point of intersection \(P\).

1. Verify that the coordinates of \(P\) are \(\left( \frac { 1 } { 2 } \sqrt { 3 } , \frac { 1 } { 6 } \pi \right)\).
2. Find the gradient of each curve at \(P\).

2

1. Given that \(y = \tan ^ { - 1 } x\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
2. Verify that \(y = \tan ^ { - 1 } x\) satisfies the equation $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$

4

1. Given that $$y = x \sqrt { 1 - x ^ { 2 } } - \cos ^ { - 1 } x$$ find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in a simplified form.
2. Hence, or otherwise, find the exact value of \(\int _ { 0 } ^ { 1 } 2 \sqrt { 1 - x ^ { 2 } } \mathrm {~d} x\).

7.The variable \(y\) is defined by $$y = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \text { for } 0 < x < \frac { \pi } { 2 } .$$ A student was asked to prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - 4 \cot 2 x .$$ The attempted proof was as follows: $$\begin{aligned} y & = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \\ & = \ln \left( \sec ^ { 2 } x \right) + \ln \left( \operatorname { cosec } ^ { 2 } x \right) \\ & = 2 \ln \sec x + 2 \ln \operatorname { cosec } x \\ \frac { \mathrm {~d} y } { \mathrm {~d} x } & = 2 \tan x - 2 \cot x \\ & = \frac { 2 \left( \sin ^ { 2 } x - \cos ^ { 2 } x \right) } { \sin x \cos x } \\ & = \frac { - 2 \cos 2 x } { \frac { 1 } { 2 } \sin 2 x } \\ & = - 4 \cot 2 x \end{aligned}$$

1. Identify the error in this attempt at a proof.
2. Give a correct version of the proof.
3. Find and simplify a general relationship between \(p\) and \(q\) ,where \(p\) and \(q\) are variables that depend on \(x\) ,such that the student would obtain the correct result when differentiating \(\ln ( p + q )\) with respect to \(x\) by the above incorrect method.
4. Given that \(p ( x ) = k \sec r x\) and \(q ( x ) = \operatorname { cosec } ^ { 2 } x\) ,where \(k\) and \(r\) are positive integers,find the values of \(k\) and \(r\) such that \(p\) and \(q\) satisfy the relationship found in part(c). \section*{END} Marks for presentation: 7
   TOTAL MARKS: 100

7. \includegraphics[max width=\textwidth, alt={}, center]{0df09d8a-7478-4679-b117-128ee226db6a-5_648_1590_296_275} The circle \(C _ { 1 }\) has centre \(O\) and radius \(R\). The tangents \(A P\) and \(B P\) to \(C _ { 1 }\) meet at the point \(P\) and angle \(A P B = 2 \alpha , 0 < \alpha < \frac { \pi } { 2 }\). A sequence of circles \(C _ { 1 } , C _ { 2 } , \ldots , C _ { n } , \ldots\) is drawn so that each new circle \(C _ { n + 1 }\) touches each of \(C _ { n } , A P\) and \(B P\) for \(n = 1,2,3 , \ldots\) as shown in Figure 2. The centre of each circle lies on the line \(O P\).

1. Show that the radii of the circles form a geometric sequence with common ratio $$\frac { 1 - \sin \alpha } { 1 + \sin \alpha }$$
2. Find, in terms of \(R\) and \(\alpha\), the total area enclosed by all the circles, simplifying your answer. The area inside the quadrilateral \(P A O B\), not enclosed by part of \(C _ { 1 }\) or any of the other circles, is \(S\).
3. Show that $$S = R ^ { 2 } \left( \alpha + \cot \alpha - \frac { \pi } { 4 } \operatorname { cosec } \alpha - \frac { \pi } { 4 } \sin \alpha \right) .$$
4. Show that, as \(\alpha\) varies, $$\frac { \mathrm { d } S } { \mathrm {~d} \alpha } = R ^ { 2 } \cot ^ { 2 } \alpha \left( \frac { \pi } { 4 } \cos \alpha - 1 \right)$$
5. Find, in terms of \(R\), the least value of \(S\) for \(\frac { \pi } { 6 } \leq \alpha \leq \frac { \pi } { 4 }\).

2. The curve \(C\) has equation \(y = x ^ { \sin x } , \quad x > 0\).

1. Find the equation of the tangent to \(C\) at the point where \(x = \frac { \pi } { 2 }\).
2. Prove that this tangent touches \(C\) at infinitely many points.

5.In this question u and v are functions of \(x\) .Given that \(\int \mathrm { u } \mathrm { d } x , \int \mathrm { v } \mathrm { d } x\) and \(\int \mathrm { uv } \mathrm { d } x\) satisfy $$\int \text { uv } \mathrm { d } x = \left( \int \mathrm { u } \mathrm {~d} x \right) \times \left( \int \mathrm { v } \mathrm {~d} x \right) \quad \text { uv } \neq 0$$

1. show that \(1 = \frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } + \frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) Given also that \(\frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } = \mathrm { sin } ^ { 2 } x\),
2. use part(a)to write down an expression,in terms of \(x\) ,for \(\frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) ,
3. show that $$\frac { 1 } { \mathrm { u } } \frac { \mathrm { du } } { \mathrm {~d} x } = \frac { 1 - 2 \sin x \cos x } { \sin ^ { 2 } x }$$
4. hence use integration to show that \(\mathrm { u } = A \mathrm { e } ^ { - \cot x } \operatorname { cosec } ^ { 2 } x\) ,where \(A\) is an arbitrary constant.
5. By differentiating \(\mathrm { e } ^ { \tan x }\) find a similar expression for v .

1

1. Given that \(y = \arctan \sqrt { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in terms of \(x\). Hence show that $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { x } ( x + 1 ) } \mathrm { d } x = \frac { \pi } { 2 }$$
2. A curve has cartesian equation $$x ^ { 2 } + y ^ { 2 } = x y + 1$$
   1. Show that the polar equation of the curve is $$r ^ { 2 } = \frac { 2 } { 2 - \sin 2 \theta }$$
   2. Determine the greatest and least positive values of \(r\) and the values of \(\theta\) between 0 and \(2 \pi\) for which they occur.
   3. Sketch the curve.

1

1. 1. Differentiate with respect to \(x\) the equation \(a \tan y = x\) (where \(a\) is a constant), and hence show that the derivative of \(\arctan \frac { x } { a }\) is \(\frac { a } { a ^ { 2 } + x ^ { 2 } }\).
   2. By first expressing \(x ^ { 2 } - 4 x + 8\) in completed square form, evaluate the integral \(\int _ { 0 } ^ { 4 } \frac { 1 } { x ^ { 2 } - 4 x + 8 } \mathrm {~d} x\), giving your answer exactly.
   3. Use integration by parts to find \(\int \arctan x \mathrm {~d} x\).
2. 1. A curve has polar equation \(r = 2 \cos \theta\), for \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). Show, by considering its cartesian equation, that the curve is a circle. State the centre and radius of the circle.
   2. Another circle has radius 2 and its centre, in cartesian coordinates, is ( 0,2 ). Find the polar equation of this circle.

1

1. Given that \(\mathrm { f } ( x ) = \arccos x\),
   1. sketch the graph of \(y = \mathrm { f } ( x )\),
   2. show that \(\mathrm { f } ^ { \prime } ( x ) = - \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\),
   3. obtain the Maclaurin series for \(\mathrm { f } ( x )\) as far as the term in \(x ^ { 3 }\).
2. A curve has polar equation \(r = \theta + \sin \theta , \theta \geqslant 0\).
   1. By considering \(\frac { \mathrm { d } r } { \mathrm {~d} \theta }\) show that \(r\) increases as \(\theta\) increases. Sketch the curve for \(0 \leqslant \theta \leqslant 4 \pi\).
   2. You are given that \(\sin \theta \approx \theta\) for small \(\theta\). Find in terms of \(\alpha\) the approximate area bounded by the curve and the lines \(\theta = 0\) and \(\theta = \alpha\), where \(\alpha\) is small.

5

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x - \tan ^ { - 1 } x \right) = \frac { x ^ { 2 } } { 1 + x ^ { 2 } }\).
2. Show that \(\int _ { 0 } ^ { \sqrt { 3 } } x \tan ^ { - 1 } x \mathrm {~d} x = \frac { 2 } { 3 } \pi - \frac { 1 } { 2 } \sqrt { 3 }\).
   

9 Given that \(\mathrm { f } ( x ) = \frac { 1 } { x } - \sqrt { x } + 3\),

1. find \(\mathrm { f } ^ { \prime } ( x )\),
2. find \(\mathrm { f } ^ { \prime \prime } ( 4 )\).