1.02w Graph transformations: simple transformations of f(x)

2 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).

1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ce82bfc4-90dd-4127-a11c-281cdcca70cf-1_621_934_1046_664} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure}
2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
   (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
   (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

3 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] \end{figure}

1. 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 \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1\).
2. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence find the coordinates of the other turning point on the curve \(y = \mathrm { f } ( x )\).
3. Show that \(\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }\).
4. 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 \(x = 1\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h] \end{figure}

1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).

1 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] \end{figure}

1. 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.
2. 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 \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

1 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).

1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-1_628_935_728_657} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure}
2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
   (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
   (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \cos ^ { 2 } x } , - \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), together with its asymptotes \(x = \frac { 1 } { 2 } \pi\) and \(x = - \frac { 1 } { 2 } \pi\). \begin{figure}[h] \end{figure}

1. Use the quotient rule to show that the derivative of \(\frac { \sin x } { \cos x }\) is \(\frac { 1 } { \cos ^ { 2 } x }\).
2. Find the area bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 4 } \pi\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \mathrm { f } \left( x + \frac { 1 } { 4 } \pi \right)\).
3. Verify that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) cross at \(( 0,1 )\).
4. State a sequence of two transformations such that the curve \(y = \mathrm { f } ( x )\) is mapped to the curve \(y = \mathrm { g } ( x )\). On the copy of Fig. 9, sketch the curve \(y = \mathrm { g } ( x )\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve.
5. Use your result from part (ii) to write down the area bounded by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = - \frac { 1 } { 4 } \pi\).

6 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,1 ). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

2 The curves in parts (i) and (ii) have equations of the form \(y = a + b \sin c x\), where \(a , b\) and \(c\) are constants. For each curve, find the values of \(a , b\) and \(c\).

1. \includegraphics[max width=\textwidth, alt={}, center]{11877196-83d9-4283-9eef-e617bea50c63-1_449_681_834_408}
2. \includegraphics[max width=\textwidth, alt={}, center]{11877196-83d9-4283-9eef-e617bea50c63-1_376_681_1344_408}

4

1. State the period of the function \(\mathrm { f } ( x ) = 1 + \cos 2 x\), where \(x\) is in degrees.
2. State a sequence of two geometrical transformations which maps the curve \(y = \cos x\) onto the curve \(y = \mathrm { f } ( x )\).
3. Sketch the graph of \(y = \mathrm { f } ( x )\) for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\).

3 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P . \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

2 \includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-2_341_1043_466_552} The diagram shows the curve \(y = \mathrm { f } ( x )\). The curve has a maximum point at ( 0,5 ) and crosses the \(x\)-axis at \(( - 2,0 ) , ( 3,0 )\) and \(( 4,0 )\). Sketch the curve \(y ^ { 2 } = \mathrm { f } ( x )\), showing clearly the coordinates of any turning points and of any points where this curve crosses the axes.

5.The function f is defined on the domain \([ - 2,2 ]\) by: $$f ( x ) = \left\{ \begin{array} { r l r } - k x ( 2 + x ) & \text { if } & - 2 \leq x < 0 , \\ k x ( 2 - x ) & \text { if } & 0 \leq x \leq 2 , \end{array} \right.$$ where \(k\) is a positive constant.
The function g is defined on the domain \([ - 2,2 ]\) by \(\mathrm { g } ( x ) = ( 2.5 ) ^ { 2 } - x ^ { 2 }\) .

1. Prove that there is a value of \(k\) such that the graph of f touches the graph of g .
2. For this value of \(k\) sketch the graphs of the functions f and g on the same axes,stating clearly where the graphs touch.
3. Find the exact area of the region bounded by the two graphs.

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = 1 + \frac { 4 } { x ( x - 3 ) }$$ The curve has a turning point at the point \(P\), and the lines with equations \(y = 1 , x = 0\) and \(x = a\) are asymptotes to the curve.

1. Write down the value of \(a\).
2. Find the coordinates of \(P\), justifying your answer.
3. Sketch the curve with equation \(y = \left| \mathrm { f } \left( x + \frac { 3 } { 2 } \right) \right| - 1\) On your sketch, you should show the coordinates of any points of intersection with the coordinate axes, the coordinates of any turning points and the equations of any asymptotes. \includegraphics[max width=\textwidth, alt={}, center]{4d5b914c-28b2-4485-a42e-627c95fa16e2-02_2255_50_311_1980}

7. $$\mathrm { f } ( x ) = \left[ 1 + \cos \left( x + \frac { \pi } { 4 } \right) \right] \left[ 1 + \sin \left( x + \frac { \pi } { 4 } \right) \right] , \quad 0 \leqslant x \leqslant 2 \pi$$

1. Show that \(\mathrm { f } ( x )\) may be written in the form $$f ( x ) = \left( \frac { 1 } { \sqrt { 2 } } + \cos x \right) ^ { 2 } , \quad 0 \leqslant x \leqslant 2 \pi$$
2. Find the range of the function \(\mathrm { f } ( x )\). The graph of \(y = \mathrm { f } ( x )\) is shown in Figure 2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0396f61a-b844-40ed-98d1-82ee2d8a6807-5_426_938_849_591} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure}
3. Find the coordinates of all the maximum and minimum points on this curve. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0396f61a-b844-40ed-98d1-82ee2d8a6807-5_432_942_1535_589} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} The region \(R\), bounded by \(y = 2\) and \(y = \mathrm { f } ( x )\), is shown shaded in Figure 3.
4. Find the area of \(R\).

7. (a) Find the set of values of \(k\) for which the equation $$\frac { x ^ { 2 } + 3 x + 8 } { x ^ { 2 } + x - 2 } = k$$ has no real roots. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C _ { 1 }\) with equation \(y = \mathrm { f } ( x )\) where \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 x + 8 } { x ^ { 2 } + x - 2 }\) The curve has asymptotes \(x = a , x = b\) and \(y = c\), where \(a , b\) and \(c\) are integers.
(b) Find the value of \(a\), the value of \(b\) and the value of \(c\).
(c) Find the coordinates of the points of intersection of \(C _ { 1 }\) with the line \(y = 2\) (d) Find all the integer pairs \(( r , s )\) that satisfy \(s = \frac { r ^ { 2 } + 3 r + 8 } { r ^ { 2 } + r - 2 }\) The curve \(C _ { 2 }\) has equation \(y = \mathrm { g } ( x )\) where \(\mathrm { g } ( x ) = \frac { 2 x ^ { 2 } - 4 x + 6 } { x ^ { 2 } - 3 x }\) (e) Show that, for suitable integers \(m\) and \(n , \mathrm {~g} ( x )\) can be written in the form \(\mathrm { f } ( x + m ) + n\).
(f) Sketch \(C _ { 2 }\) showing any asymptotes and stating their equations.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 4 ( x - 1 ) } { x ( x - 3 ) }$$ The curve cuts the \(x\)-axis at \(( a , 0 )\). The lines \(y = 0 , x = 0\) and \(x = b\) are asymptotes to the curve.

1. Write down the value of \(a\) and the value of \(b\).
   (2)
2. On separate axes, sketch the curves with the following equations. On your sketches, you should mark the coordinates of any intersections with the coordinate axes and state the equations of any asymptotes.
   1. \(y = \mathrm { f } ( x + 2 ) - 4\)
   2. \(y = \mathrm { f } ( | x | ) - 3\)

5 In this question, you are required to investigate the curve with equation $$y = x ^ { m } ( 1 - x ) ^ { n } , \quad 0 \leqslant x \leqslant 1 ,$$ for various positive values of \(m\) and \(n\).

1. On separate diagrams, sketch the curve in each of the following cases.
   (A) \(m = 1 , n = 1\),
   (B) \(m = 2 , n = 2\),
   (C) \(m = 2 , n = 4\),
   (D) \(m = 4 , n = 2\).
2. What feature does the curve have when \(m = n\) ? What is the effect on the curve of interchanging \(m\) and \(n\) when \(m \neq n\) ?
3. Describe how the \(x\)-coordinate of the maximum on the curve varies as \(m\) and \(n\) vary. Use calculus to determine the \(x\)-coordinate of the maximum.
4. Find the condition on \(m\) for the gradient to be zero when \(x = 0\). State a corresponding result for the gradient to be zero when \(x = 1\).
5. Use your calculator to investigate the shape of the curve for large values of \(m\) and \(n\). Hence conjecture what happens to the value of the integral \(\int _ { 0 } ^ { 1 } x ^ { m } ( 1 - x ) ^ { n } \mathrm {~d} x\) as \(m\) and \(n\) tend to infinity.
6. Use your calculator to investigate the shape of the curve for small values of \(m\) and \(n\). Hence conjecture what happens to the shape of the curve as \(m\) and \(n\) tend to zero. }{www.ocr.org.uk}) after the live examination series.
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4

1. Sketch the curve \(y = \frac { 1 } { x ^ { 2 } }\).
2. The curve \(y = \frac { 1 } { x ^ { 2 } }\) is translated by 3 units in the negative \(x\)-direction. State the equation of the curve after it has been translated.
3. The curve \(y = \frac { 1 } { x ^ { 2 } }\) is stretched parallel to the \(y\)-axis with scale factor 4 and, as a result, the point \(P ( 1,1 )\) is transformed to the point \(Q\). State the coordinates of \(Q\).

2 \includegraphics[max width=\textwidth, alt={}, center]{918d83c3-1608-4482-9d3d-8af05e65f353-2_330_681_390_731} The graph of \(y = \mathrm { f } ( x )\) for \(- 2 \leqslant x \leqslant 4\) is shown above.

1. Sketch the graph of \(y = 2 \mathrm { f } ( x )\) for \(- 2 \leqslant x \leqslant 4\) on the axes provided.
2. Describe the transformation which transforms the graph of \(y = \mathrm { f } ( x )\) to the graph of \(y = \mathrm { f } ( x - 1 )\).

5

1. Sketch the curve \(y = - x ^ { 3 }\).
2. The curve \(y = - x ^ { 3 }\) is translated by 3 units in the positive \(x\)-direction. Find the equation of the curve after it has been translated.
3. Describe a transformation that transforms the curve \(y = - x ^ { 3 }\) to the curve \(y = - 5 x ^ { 3 }\).

2 \includegraphics[max width=\textwidth, alt={}, center]{559e9f1a-340e-4478-adaa-a7361dd70fe8-2_325_479_468_794} The graph of \(y = \mathrm { f } ( x )\) for \(- 2 \leqslant x \leqslant 2\) is shown above.

1. Sketch the graph of \(y = \mathrm { f } ( - x )\) for \(- 2 \leqslant x \leqslant 2\).
2. Sketch the graph of \(y = \mathrm { f } ( x ) + 2\) for \(- 2 \leqslant x \leqslant 2\).

5

1. Expand and simplify \(( 2 x + 1 ) ( x - 3 ) ( x + 4 )\).
2. Find the coefficient of \(x ^ { 4 }\) in the expansion of $$x \left( x ^ { 2 } + 2 x + 3 \right) \left( x ^ { 2 } + 7 x - 2 \right) .$$

2

1. Sketch the curve \(y = - \frac { 1 } { x ^ { 2 } }\).
2. Sketch the curve \(y = 3 - \frac { 1 } { x ^ { 2 } }\).
3. The curve \(y = - \frac { 1 } { x ^ { 2 } }\) is stretched parallel to the \(y\)-axis with scale factor 2 . State the equation of the transformed curve.